Curly numbers

Curly numbers are numbers that can be represented using geometric shapes. This historical concept dates back to the Pythagoreans , who developed algebra on a geometric basis ; the echo of this approach has remained the expression "square a square or a cube." In number theory and combinatorics, curly numbers are associated with many other classes of integers - binomial coefficients , perfect numbers , Mersenne , Fermat , Fibonacci , Luc and others ^[1] .

Types of Curly Numbers

From the time of the Pythagoreans (VI century BC. E.) traditionally distinguished the following types of curly numbers (they are defined, for example, in the VII book of the " Beginnings " of Euclid ) ^[2] :

Linear numbers are numbers that are not factorizable, larger units, that is, it is a series of primes supplemented by a unit (Euclid uses the term " first numbers ", πρώτοι αριθμοί):
1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271 ... (sequence A008578 in OEIS )
Flat numbers are numbers that can be represented as the product of two factors greater than one, that is, composite :
4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49, 50, 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68, 69, 70, 72, 74, 75, 76, 77, 78, 80, 81, 82, 84, 85, 86, 87, 88 ... (sequence A002808 in OEIS )
- A special case is rectangular numbers (they are also called “ oblong ” in the sources), which are the product of two consecutive integers, that is, having the form $n(n+1).$ ${\ displaystyle n (n + 1).}$ $n(n+1).$
Corporal numbers - numbers represented by the product of three factors:
8, 12, 16, 18, 20, 24, 27, 28, 30, 32, 36, 40, 42, 44, 45, 48, 50, 52, 54, 56, 60, 63, 64, 66, 68, 70, 72, 75, 76, 78, 80, 81, 84, 88, 90, 92, 96, 98, 99, 100, 102, 104, 105, 108, 110, 112, 114, 116, 117, 120, 124, 125, 126, 128, 130, 132, 135, 136, 138, 140, 144 ... (sequence A033942 in OEIS )
Polygonal numbers - numbers associated with a particular polygon, see definition below .
Spatial polyhedral numbers - numbers associated with a particular polyhedron, see definition below .

Classic Polygonal Numbers

Definition and general view

General definition $k$ ${\ displaystyle k}$ -gonal number for any $k\geqslant 3$ ${\ displaystyle k \ geqslant 3}$ can be formulated as follows ^[3] :

$n$ ${\ displaystyle n}$ in order $k$ ${\ displaystyle k}$ angular number $P_{n}^{(k)}$ ${\ displaystyle P_ {n} ^ {(k)}}$ is the sum of the first $n$ ${\ displaystyle n}$ terms of arithmetic progression , in which the first term is 1, and the difference is $k-2$ ${\ displaystyle k-2}$ .

For example, triangular numbers are obtained as partial sums of a series $1+2+3+4\dots$ ${\ displaystyle 1 + 2 + 3 + 4 \ dots}$ , and the quadrangular (square) numbers correspond to the series $1+3+5+7\dots$ ${\ displaystyle 1 + 3 + 5 + 7 \ dots}$

The sequence of k- angular numbers has the form ^[4] :

1,k,3k-3,6k-8,10k-15,15k-24,21k-35,28k-48,36k-63,45k-80\dots

{\ displaystyle 1, k, 3k-3,6k-8,10k-15,15k-24,21k-35,28k-48,36k-63,45k-80 \ dots}

General formula for explicit calculation $n$ ${\ displaystyle n}$ in order $k$ ${\ displaystyle k}$ -gonal number $P_{n}^{(k)}$ ${\ displaystyle P_ {n} ^ {(k)}}$ it is easy to obtain by finding the sum of arithmetic progression by standard rules. This formula can be represented in several versions, obtained from one another by simple transformations ^[5] :

P_{n}^{(k)}=n+(k-2){\frac {n(n-1)}{2}}={\frac {(k-2)n^{2}-(k-4)n}{2}}={\frac {1}{2}}k(n^{2}-n)-n^{2}+2n

{\ displaystyle P_ {n} ^ {(k)} = n + (k-2) {\ frac {n (n-1)} {2}} = {\ frac {(k-2) n ^ {2} - (k-4) n} {2}} = {\ frac {1} {2}} k (n ^ {2} -n) -n ^ {2} + 2n}

(OKF)

You can also use the recurrence formula ^[5] :

P_{n+1}^{(k)}=P_{n}^{(k)}+(k-2)n+1;\ P_{1}^{(k)}=1

{\ displaystyle P_ {n + 1} ^ {(k)} = P_ {n} ^ {(k)} + (k-2) n + 1; \ P_ {1} ^ {(k)} = 1}

With an increase in the number of parties $k$ ${\ displaystyle k}$ per unit, the corresponding curly numbers are changed according to the Nicomache formula ^[6] :

P_{n}^{(k+1)}=P_{n}^{(k)}+P_{n-1}^{(3)},

{\ displaystyle P_ {n} ^ {(k + 1)} = P_ {n} ^ {(k)} + P_ {n-1} ^ {(3)},}

Where

n>1.

{\ displaystyle n> 1.}

(Nicomache)

Insofar as $P_{n}^{(k)}$ ${\ displaystyle P_ {n} ^ {(k)}}$ linearly dependent $k,$ ${\ displaystyle k,}$ the formula is valid:

P_{n}^{(k+s)}+P_{n}^{(k-s)}=2P_{n}^{(k)}

{\ displaystyle P_ {n} ^ {(k + s)} + P_ {n} ^ {(ks)} = 2P_ {n} ^ {(k)}}

where

s=0,1,2\dots k-3.

{\ displaystyle s = 0,1,2 \ dots k-3.}

In other words, each polygonal number is the arithmetic mean for equidistant from it by $k$ ${\ displaystyle k}$ polygonal numbers with the same number.

If a $k$ ${\ displaystyle k}$ Is a prime number , then the second $k$ ${\ displaystyle k}$ -gonal number equal to $k,$ ${\ displaystyle k,}$ also simple; this is the only situation where the polygonal number is prime. In fact, we write the general formula as follows:

P_{n}^{(k)}={\frac {2+(n-1)(k-2)}{2}}n

{\ displaystyle P_ {n} ^ {(k)} = {\ frac {2+ (n-1) (k-2)} {2}} n}

Let be $n>2.$ ${\ displaystyle n> 2.}$ If a $n$ ${\ displaystyle n}$ evenly, the figure number is divided by $n/2,$ ${\ displaystyle n / 2,}$ and if it’s odd, then it is divided by $(2+(n-1)(k-2))/2.$ ${\ displaystyle (2+ (n-1) (k-2)) / 2.}$ In both cases, the figure number is composite ^[7] .

Historical Review

Figured numbers, according to the Pythagoreans , play an important role in the structure of the universe. Therefore, many mathematicians of antiquity were engaged in their study: Eratosthenes , Gipsicle , Diophantus of Alexandria and others. Gypsicle (II century BC) gave a general definition $k$ ${\ displaystyle k}$ -gonal number $P_{n}^{(k)}$ ${\ displaystyle P_ {n} ^ {(k)}}$ as amounts $n$ ${\ displaystyle n}$ terms of arithmetic progression , in which the first term is 1, and the difference is $k-2;$ ${\ displaystyle k-2;}$ Gypsicle's definition is given in the book of Diophantus in the following form ^[8] ^[9] :

If you take any numbers, starting from a unit that have the same differences, then their sum, if the difference is one, will be a triangle, if it is a two, then it will be a quadrangle, and if it is a triple, it will be a pentagon. The number of angles is determined by the difference increased by two, and the side is determined by the number of numbers taken, including one.

Diophantus wrote a large study on the properties of polygonal numbers, fragments of which have survived to this day. Curly numbers are much discussed in the Pythagorean textbooks of arithmetic, created by Nikomakh Gerazsky and Theon Smirn (II century), which established a number of dependencies between curly numbers of different dimensions. Indian mathematicians and the first mathematicians of medieval Europe ( Fibonacci , Pacioli , Cardano and others) showed great interest in figure numbers ^[10] ^[1] .

In modern times, Fermat , Wallis , Euler , Lagrange , Gauss and others dealt with polygonal numbers. In September 1636 ^[11] Fermat formulated in a letter to Mersenne a remarkable theorem, which today is called Fermat's polygonal number theorem ^[10] :

I was the first to discover a very beautiful and completely general theorem that each number is either triangular or the sum of two or three triangular numbers; each number is either square, or is the sum of two, three, or four squares; or pentagonal, or is the sum of two, three, four or five pentagonal numbers, etc. to infinity, whether for hexagonal, heptagonal or any polygonal numbers. I cannot give a proof here, which depends on the numerous and confusing secrets of numbers, for I intend to devote a whole book to this topic and get amazing achievements in this part of arithmetic compared to previously known limits.

Contrary to the promise, Fermat never published a proof of this theorem, which in a letter to Pascal (1654) called his main achievement in mathematics ^[11] . The problem was addressed by many outstanding mathematicians - in 1770 Lagrange proved a theorem for square numbers ( Lagrange's sum of four squares theorem ), in 1796 Gauss gave a proof for triangular. Cauchy was able to give a complete proof of the theorem in 1813 ^[12] ^[13] .

Triangular numbers

The sequence of triangular numbers :

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, 231, 253, 276, 300, 325, 351, 378, 406, 435, 465, 496, 528, 561, 595, 630, 666, 703, 741, 780, 820, 861, 903, 946, 990, 1035, 1081, 1128, 1176, 1225, 1275, 1326, 1378, 1431 ...,

{\frac {n(n+1)}{2}}

{\ displaystyle {\ frac {n (n + 1)} {2}}}

... (sequence A000217 in OEIS )

Properties:

The parity of an element of a sequence changes with period 4: odd, odd, even, even.
Denote for brevity $n$ ${\ displaystyle n}$ triangular number: $T_{n}=P_{n}^{(3)}={\frac {n(n+1)}{2}}.$ ${\ displaystyle T_ {n} = P_ {n} ^ {(3)} = {\ frac {n (n + 1)} {2}}.}$ Then the recurrence formulas ^[14] are valid:

T_{2n}=3T_{n}+T_{n-1}

{\ displaystyle T_ {2n} = 3T_ {n} + T_ {n-1}}

T_{2n+1}=3T_{n}+T_{n+1}

{\ displaystyle T_ {2n + 1} = 3T_ {n} + T_ {n + 1}}

Bachet de Meziriac formula : the general formula of a polygonal number can be transformed so that it shows the expression of any polygonal number in terms of triangular ^[15] :

P_{n}^{(k)}=n+(k-2){\frac {n(n-1)}{2}}=(k-2)T_{n-1}+n=(k-3)T_{n-1}+T_{n}

{\ displaystyle P_ {n} ^ {(k)} = n + (k-2) {\ frac {n (n-1)} {2}} = (k-2) T_ {n-1} + n = (k-3) T_ {n-1} + T_ {n}}

(Bashe)

The sum of consecutive triangular numbers forms a square number

The sum of two consecutive triangular numbers gives the full square ( square number ):

T_{n}+T_{n+1}=(n+1)^{2}=P_{n+1}^{(4)}.

{\ displaystyle T_ {n} + T_ {n + 1} = (n + 1) ^ {2} = P_ {n + 1} ^ {(4)}.}

.

The sum of a finite series of triangular numbers is calculated by the formula:

S_{m-1}=1+3+6+\dots +{\frac {(m-1)m}{2}}={\frac {m^{3}-m}{6}}

{\ displaystyle S_ {m-1} = 1 + 3 + 6 + \ dots + {\ frac {(m-1) m} {2}} = {\ frac {m ^ {3} -m} {6} }}

.

A series of inverse triangular numbers converges:

1+{1 \over 3}+{1 \over 6}+{1 \over 10}+{1 \over 15}+\dots =2\sum _{n=1}^{\infty }\left({1 \over n}-{1 \over n+1}\right)=2

{\ displaystyle 1+ {1 \ over 3} + {1 \ over 6} + {1 \ over 10} + {1 \ over 15} + \ dots = 2 \ sum _ {n = 1} ^ {\ infty} \ left ({1 \ over n} - {1 \ over n + 1} \ right) = 2}

Doubled triangular numbers give a sequence (defined above) of rectangular numbers .
Natural number $N$ ${\ displaystyle N}$ is triangular if and only if the number $8N+1$ ${\ displaystyle 8N + 1}$ is a full square . This is easy to prove directly, but it is easier to deduce from the general methodology (problem 2) ..
There are infinitely many triangular numbers that are simultaneously square (“ square triangular numbers ”) ^[16] : $1,36,1225,41616,1413721\dots$ ${\ displaystyle 1.36.1225.41616.1413721 \ dots}$ (OEIS | A001110}}).
The mysterious “ number of the beast ” (666) is the 36th triangular. It is the smallest triangular number, which is representable as the sum of the squares of triangular numbers ^[17] : $666=15^{2}+21^{2}.$ ${\ displaystyle 666 = 15 ^ {2} + 21 ^ {2}.}$
The triangular numbers form the third diagonal line of the Pascal triangle ; see details below .

Square Numbers


$0+\color {blue}1\color {black}=1$ ${\ displaystyle 0+ \ color {blue} 1 \ color {black} = 1}$	$1+\color {blue}3\color {black}=4$ ${\ displaystyle 1+ \ color {blue} 3 \ color {black} = 4}$	$4+\color {blue}5\color {black}=9$ ${\ displaystyle 4+ \ color {blue} 5 \ color {black} = 9}$	$9+\color {blue}7\color {black}=16$ ${\ displaystyle 9+ \ color {blue} 7 \ color {black} = 16}$

Square numbers are the product of two identical natural numbers, that is, they are full squares:

1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, 961, 1024, 1089, 1156, 1225, 1296, 1369, 1444, 1521, 1600, 1681, 1764, 1849, 1936, 2025, 2116, 2209, 2304, 2401, 2500 ... ,

n^{2}

{\ displaystyle n ^ {2}}

... (sequence A000290 in OEIS )

Every square number, except one, is the sum of two consecutive triangular numbers:

n^{2}=T_{n-1}+T_{n}

{\ displaystyle n ^ {2} = T_ {n-1} + T_ {n}}

. Examples:

4=1+3;\quad 9=3+6;\quad 16=6+10

{\ displaystyle 4 = 1 + 3; \ quad 9 = 3 + 6; \ quad 16 = 6 + 10}

etc.

The sum of the squares of the first $n$ ${\ displaystyle n}$ of natural numbers is calculated by the formula ^[18] :

1^{2}+2^{2}+3^{2}+...+n^{2}={\frac {n(n+1)(2n+1)}{6}}

{\ displaystyle 1 ^ {2} + 2 ^ {2} + 3 ^ {2} + ... + n ^ {2} = {\ frac {n (n + 1) (2n + 1)} {6} }}

A number of inverse squares converge ^[19] :

\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}={\frac {1}{1^{2}}}+{\frac {1}{2^{2}}}+\dots +{\frac {1}{n^{2}}}+\dots ={\frac {\pi ^{2}}{6}}

{\ displaystyle \ sum _ {n = 1} ^ {\ infty} {\ frac {1} {n ^ {2}}} = {\ frac {1} {1 ^ {2}}} + {\ frac { 1} {2 ^ {2}}} + \ dots + {\ frac {1} {n ^ {2}}} + \ dots = {\ frac {\ pi ^ {2}} {6}}}

Each natural number can be represented as a sum of not more than four squares ( Lagrange's theorem on the sum of four squares ).

Pentagonal Numbers

1, 5, 12, 22, 35, 51, 70, 92, 117, 145, 176, 210, 247, 287, 330, 376, 425, 477, 532, 590, 651, 715, 782, 852, 925, 1001, 1080, 1162, 1247, 1335, 1426, 1520, 1617, 1717, 1820, 1926, 2035, 2147, 2262, 2380, 2501, 2625, 2752, 2882, 3015, 3151 ...,

{\frac {n(3n-1)}{2}}

{\ displaystyle {\ frac {n (3n-1)} {2}}}

... (sequence A000326 in OEIS )

If in the formula ${\frac {n(3n-1)}{2}}$ ${\ displaystyle {\ frac {n (3n-1)} {2}}}$ indicate for $n$ ${\ displaystyle n}$ more general sequence:

n=0,1,-1,2,-2,3,-3\dots

{\ displaystyle n = 0,1, -1,2, -2,3, -3 \ dots}

then we get the so-called generalized pentagonal numbers :

0, 1, 2, 5, 7, 12, 15, 22, 26, 35, 40, 51, 57, 70, 77, 92, 100, 117, 126, 145, 155, 176, 187, 210, 222, 247, 260, 287, 301, 330, 345, 376, 392, 425, 442, 477, 495, 532, 551, 590, 610, 651, 672, 715, 737, 782, 805, 852, 876, 925, 950, 1001, 1027, 1080, 1107, 1162, 1190, 1247, 1276, 1335 ... (sequence A001318 in OEIS )

Leonhard Euler discovered generalized pentagonal numbers in the following identity:

(1-x)(1-x^{2})(1-x^{3})\ldots =1-x-x^{2}+x^{5}+x^{7}-x^{12}-x^{15}+x^{22}+x^{26}-x^{35}-x^{40}+\ldots

{\ displaystyle (1-x) (1-x ^ {2}) (1-x ^ {3}) \ ldots = 1-xx ^ {2} + x ^ {5} + x ^ {7} -x ^ {12} -x ^ {15} + x ^ {22} + x ^ {26} -x ^ {35} -x ^ {40} + \ ldots}

Degrees $x$ ${\ displaystyle x}$ on the right side of the identity form a sequence of generalized pentagonal numbers ^[20] . See more details: The Euler Pentagonal Theorem .

Hexagonal numbers

1, 6, 15, 28, 45, 66, 91, 120, 153, 190, 231, 276, 325, 378, 435, 496, 561, 630, 703, 780, 861, 946, 1035, 1128, 1225, 1326, 1431, 1540, 1653, 1770, 1891, 2016, 2145, 2278, 2415, 2556, 2701, 2850, 3003, 3160, 3321, 3486, 3655, 3828, 4005, 4186, 4371, 4560 ...,

2n^{2}-n

{\ displaystyle 2n ^ {2} -n}

... (sequence A000384 in OEIS )

Obviously, a sequence of hexagonal numbers is obtained from a sequence of triangular numbers by deleting elements with even numbers: $P_{n}^{(6)}=P_{2n-1}^{(3)}.$ ${\ displaystyle P_ {n} ^ {(6)} = P_ {2n-1} ^ {(3)}.}$

Natural number $N$ ${\ displaystyle N}$ is hexagonal if and only if the number ${\frac {{\sqrt {8N+1}}+1}{4}}$ ${\ displaystyle {\ frac {{\ sqrt {8N + 1}} + 1} {4}}}$ is natural. This is easy to prove directly, but it is easier to deduce from the general methodology (problem 2) ..

Twelve- Numbers

Twelve-cornered numbers are calculated by the formula $5n^{2}-4n$ ${\ displaystyle 5n ^ {2} -4n}$ :

1, 12, 33, 64, 105, 156, 217, 288, 369, 460, 561, 672, 793, 924, 1065, 1216, 1377, 1548, 1729, 1920, 2121, 2332, 2553, 2784, 3025, 3276, 3537, 3808, 4089, 4380, 4681, 4992, 5313, 5644, 5985, 6336, 6697, 7068, 7449, 7840, 8241, 8652, 9073, 9504, 9945 ... (sequence A051624 in OEIS )

In decimal system $n$ ${\ displaystyle n}$ the twelfth hexagonal number ends with the same digit as the number itself $n$ ${\ displaystyle n}$ .

Determining if a given number is polygonal

Problem 1 (often called the Diophantine problem): a natural number is given $N>2,$ ${\ displaystyle N> 2,}$ need to determine if it is a polygonal number $P_{n}^{(k)}$ ${\ displaystyle P_ {n} ^ {(k)}}$ and if so, for what values $k, n .$ ${\ displaystyle k, n.}$ Diophantus formulated this problem as follows: " find out how many times a given number occurs among all kinds of polygonal numbers ." The algorithm for solving this problem is as follows ^[21] .

We write out all the natural divisors of the number $2N$ ${\ displaystyle 2N}$ (including 1 and itself $2N$ ${\ displaystyle 2N}$ )
We write out all the natural divisors of the number $2N-2.$ ${\ displaystyle 2N-2.}$
We select from the first set those numbers that are 1 more than any number from the second set. These numbers correspond $n .$ ${\ displaystyle n.}$
For each selected $n$ ${\ displaystyle n}$ count $k={\frac {2N-2}{n-1}}-{\frac {2N}{n}}+2.$ ${\ displaystyle k = {\ frac {2N-2} {n-1}} - {\ frac {2N} {n}} + 2.}$
Cross off the pairs $(n,k),$ ${\ displaystyle (n, k),}$ in which $k<3$ ${\ displaystyle k <3}$ .

Then all matching the remaining pairs $P_{n}^{(k)}$ ${\ displaystyle P_ {n} ^ {(k)}}$ are equal $N .$ ${\ displaystyle N.}$

Example ^[21] . Let be $N=105.$ ${\ displaystyle N = 105.}$ .

Dividers $2N=210\colon \quad 1,2,3,5,6,7,10,14,15,21,30,35,42,70,105,210.$ ${\ displaystyle 2N = 210 \ colon \ quad 1,2,3,5,6,7,10,14,15,21,30,35,42,70,105,210.}$
Dividers $2N-2=208\colon \quad 1,2,4,8,13,16,26,52,104,208.$ ${\ displaystyle 2N-2 = 208 \ colon \ quad 1,2,4,8,13,16,26,52,104,208.}$
We select $n=2,3,5,14,105.$ ${\ displaystyle n = 2,3,5,14,105.}$
Respectively $k=105,36,12,14,2.$ ${\ displaystyle k = 105,36,12,14,2.}$ We discard the last value.

Answer: $105$ ${\ displaystyle 105}$ meets as $P_{2}^{(105)},P_{3}^{(36)},P_{5}^{(12)},P_{14}^{(14)},$ ${\ displaystyle P_ {2} ^ {(105)}, P_ {3} ^ {(36)}, P_ {5} ^ {(12)}, P_ {14} ^ {(14)},}$ that is, as the 2nd 105th coal, 3rd 36th coal, 5th 12th coal and 14th 14th coal number.

Task 2 : given a natural number $N>2,$ ${\ displaystyle N> 2,}$ required to determine whether it is $k-$ ${\ displaystyle k-}$ coal number $P_{n}^{(k)}.$ ${\ displaystyle P_ {n} ^ {(k)}.}$ Unlike task 1, here $k$ ${\ displaystyle k}$ set.

For the solution, we can use the Diophantine identity ^[22] :

8(k-2)P_{n}^{(k)}+(k-4)^{2}=(2n(k-2)-(k-4))^{2}

{\ displaystyle 8 (k-2) P_ {n} ^ {(k)} + (k-4) ^ {2} = (2n (k-2) - (k-4)) ^ {2}}

This identity is easily obtained from the above general formula for $P_{n}^{(k)}$ ${\ displaystyle P_ {n} ^ {(k)}}$ and is tantamount to her. The identity implies the solution of problem 2: if $N$ ${\ displaystyle N}$ there is $k-$ ${\ displaystyle k-}$ coal number, i.e. $N=P_{n}^{(k)}$ ${\ displaystyle N = P_ {n} ^ {(k)}}$ for some $n,$ ${\ displaystyle n,}$ then $8(k-2)N+(k-4)^{2}$ ${\ displaystyle 8 (k-2) N + (k-4) ^ {2}}$ there is some square number $R^{2}$ ${\ displaystyle R ^ {2}}$ , and back. In this case, the number $n$ ${\ displaystyle n}$ is found by the formula ^[22] :

n={\frac {R+k-4}{2k-4}}

{\ displaystyle n = {\ frac {R + k-4} {2k-4}}}

Example ^[22] . Determine if the number is $1540$ ${\ displaystyle 1540}$ 10-sided. Value $8(k-2)N+(k-4)^{2}$ ${\ displaystyle 8 (k-2) N + (k-4) ^ {2}}$ here is equal $98596=314^{2},$ ${\ displaystyle 98596 = 314 ^ {2},}$ therefore the answer is yes. $n=20,$ ${\ displaystyle n = 20,}$ Consequently, $1540$ ${\ displaystyle 1540}$ is the 20th 10-digit number.

Generating Function

Power series whose coefficients are $k$ ${\ displaystyle k}$ -gonal numbers converges at $|x|<1$ ${\ displaystyle | x | <1}$ :

P_{1}^{(k)}x+P_{2}^{(k)}x^{2}+P_{3}^{(k)}x^{3}+\dots ={\frac {x(1+(k-3)x)}{(1-x)^{3}}}

{\ displaystyle P_ {1} ^ {(k)} x + P_ {2} ^ {(k)} x ^ {2} + P_ {3} ^ {(k)} x ^ {3} + \ dots = {\ frac {x (1+ (k-3) x)} {(1-x) ^ {3}}}}

The expression on the right is a generating function for the sequence $k$ ${\ displaystyle k}$ -gonal numbers ^[23] .

The apparatus of generating functions allows the use of mathematical analysis methods in number theory and combinatorics. The above formula also explains the appearance of $k$ ${\ displaystyle k}$ -gonal numbers among the coefficients of the Taylor series for various rational fractions. Examples:

At

k=3

{\ displaystyle k = 3}

:

\qquad {\frac {x}{(1-x)^{3}}}=P_{1}^{(3)}x+P_{2}^{(3)}3x^{2}+P_{3}^{(3)}x^{3}+\dots +P_{n}^{(3)}x^{n}+\dots

{\ displaystyle \ qquad {\ frac {x} {(1-x) ^ {3}}} = P_ {1} ^ {(3)} x + P_ {2} ^ {(3)} 3x ^ {2 } + P_ {3} ^ {(3)} x ^ {3} + \ dots + P_ {n} ^ {(3)} x ^ {n} + \ dots}

At

k=4

{\ displaystyle k = 4}

:

\qquad {\frac {x(x+1)}{(1-x)^{3}}}=P_{1}^{(4)}x+P_{2}^{(4)}3x^{2}+P_{3}^{(4)}x^{3}+\dots +P_{n}^{(4)}x^{n}+\dots

{\ displaystyle \ qquad {\ frac {x (x + 1)} {(1-x) ^ {3}}} = P_ {1} ^ {(4)} x + P_ {2} ^ {(4) } 3x ^ {2} + P_ {3} ^ {(4)} x ^ {3} + \ dots + P_ {n} ^ {(4)} x ^ {n} + \ dots}

At

k=5

{\ displaystyle k = 5}

:

\qquad {\frac {x(2x+1)}{(1-x)^{3}}}=P_{1}^{(5)}x+P_{2}^{(5)}3x^{2}+P_{3}^{(5)}x^{3}+\dots +P_{n}^{(5)}x^{n}+\dots

{\ displaystyle \ qquad {\ frac {x (2x + 1)} {(1-x) ^ {3}}} = P_ {1} ^ {(5)} x + P_ {2} ^ {(5) } 3x ^ {2} + P_ {3} ^ {(5)} x ^ {3} + \ dots + P_ {n} ^ {(5)} x ^ {n} + \ dots}

etc.

For some classes of polygonal numbers, there are their own specific generating functions. For example, for square triangular numbers $1,36,1225,41616,1413721\dots$ ${\ displaystyle 1.36.1225.41616.1413721 \ dots}$ the generating function has the following form ^[24] :

{\frac {x(1+x)}{(1-x)(1-34x+x^{2})}}=x+36x^{2}+1225x^{3}+\dots

{\ displaystyle {\ frac {x (1 + x)} {(1-x) (1-34x + x ^ {2})}} = x + 36x ^ {2} + 1225x ^ {3} + \ dots }

; the series converges at

|x|<17-12{\sqrt {2}}

{\ displaystyle | x | <17-12 {\ sqrt {2}}}

PivotTable

k	Number type	General formula	n										Sum of Inverse Values ^[25]	OEIS Number
k	Number type	General formula	one	2	3	four	five	6	7	eight	9	ten	Sum of Inverse Values ^[25]	OEIS Number
3	Triangular	1 2 ( n ² + n )	one	3	6	ten	15	21	28	36	45	55	2	A000217
four	Square	1 2 (2 n ² - 0 n ) = n ²	one	four	9	sixteen	25	36	49	64	81	100	$\pi$ ${\ displaystyle \ pi}$ ² 6	A000290
five	Pentagonal	1 2 (3 n ² - n )	one	five	12	22	35	51	70	92	117	145	$3\ln 3-{\frac {\pi {\sqrt {3}}}{3}}$ ${\ displaystyle 3 \ ln 3 - {\ frac {\ pi {\ sqrt {3}}} {3}}}$	A000326
6	Hexagonal	1 2 (4 n ² - 2 n )	one	6	15	28	45	66	91	120	153	190	2 ln 2	A000384
7	Heptagonal	1 2 (5 n ² - 3 n )	one	7	18	34	55	81	112	148	189	235	${\begin{matrix}{\tfrac {2}{3}}\ln 5\\+{\tfrac {{1}+{\sqrt {5}}}{3}}\ln {\tfrac {\sqrt {10-2{\sqrt {5}}}}{2}}\\+{\tfrac {{1}-{\sqrt {5}}}{3}}\ln {\tfrac {\sqrt {10+2{\sqrt {5}}}}{2}}\\+{\tfrac {\pi {\sqrt {25-10{\sqrt {5}}}}}{15}}\end{matrix}}$ ${\ displaystyle {\ begin {matrix} {\ tfrac {2} {3}} \ ln 5 \\ + {\ tfrac {{1} + {\ sqrt {5}}} {3}} \ ln {\ tfrac {\ sqrt {10-2 {\ sqrt {5}}}} {2}} \\ + {\ tfrac {{1} - {\ sqrt {5}}} {3}} \ ln {\ tfrac {\ sqrt {10 + 2 {\ sqrt {5}}}} {2}} \\ + {\ tfrac {\ pi {\ sqrt {25-10 {\ sqrt {5}}}}} {15}} \ end {matrix}}}$	A000566
eight	Octagonal	1 2 (6 n ² - 4 n )	one	eight	21	40	65	96	133	176	225	280	3 4 ln 3 + $\pi$ ${\ displaystyle \ pi}$ √ 3 12	A000567
9	Nine	1 2 (7 n ² - 5 n )	one	9	24	46	75	111	154	204	261	325		A001106
ten	Decagonal	1 2 (8 n ² - 6 n )	one	ten	27	52	85	126	175	232	297	370	ln 2 + $\pi$ ${\ displaystyle \ pi}$ 6	A001107
eleven	11-sided	1 2 (9 n ² - 7 n )	one	eleven	thirty	58	95	141	196	260	333	415		A051682
12	12 angle	1 2 (10 n ² - 8 n )	one	12	33	64	105	156	217	288	369	460		A051624
13	13-angle	1 2 (11 n ² - 9 n )	one	13	36	70	115	171	238	316	405	505		A051865
14	14-sided	1 2 (12 n ² - 10 n )	one	14	39	76	125	186	259	344	441	550	2 5 ln 2 + 3 10 ln 3 + $\pi$ ${\ displaystyle \ pi}$ √ 3 10	A051866
15	15-sided	1 2 (13 n ² - 11 n )	one	15	42	82	135	201	280	372	477	595		A051867
sixteen	16 angle	1 2 (14 n ² - 12 n )	one	sixteen	45	88	145	216	301	400	513	640		A051868
17	17-coal	1 2 (15 n ² - 13 n )	one	17	48	94	155	231	322	428	549	685		A051869
18	18-sided	1 2 (16 n ² - 14 n )	one	18	51	100	165	246	343	456	585	730	4 7 ln 2 - √ 2 14 ln (3 - 2 √ 2 ) + $\pi$ ${\ displaystyle \ pi}$ (1 + √ 2 ) 14	A051870
nineteen	19-coal	1 2 (17 n ² - 15 n )	one	nineteen	54	106	175	261	364	484	621	775		A051871
20	20 coal	1 2 (18 n ² - 16 n )	one	20	57	112	185	276	385	512	657	820		A051872
...	...	...	...	...	...	...	...	...	...	...	...	...	...	...
10,000		1 2 (9998 n ² - 9996 n )	one	10,000	29997	59992	99985	149976	209965	279952	359937	449920		A167149

Centered Polygonal Numbers

Definition

Centered $k$ ${\ displaystyle k}$ -gonal numbers ( $k\geqslant 3$ ${\ displaystyle k \ geqslant 3}$ ) Is a class of curly numbers obtained by the following geometric construction. First, a certain central point is fixed on the plane. Then the correct one is built around it. $k$ ${\ displaystyle k}$ -gon with $k$ ${\ displaystyle k}$ vertex points, each side contains two points (see figure). Next, new layers are built outside $k$ ${\ displaystyle k}$ -gon, and each side on a new layer contains one point more than in the previous layer, that is, starting from the second layer, each next layer contains $k$ ${\ displaystyle k}$ more points than the previous one. The total number of points inside each layer and is taken as a centered polygonal number (the point in the center is considered the initial layer) ^[26] .

Examples of building centered polygonal numbers:

Triangular	Square	Pentagonal	Hexagonal

From the construction it is seen that centered polygonal numbers are obtained as partial sums of the following series: $1+k+2k+3k+4k+\dots$ ${\ displaystyle 1 + k + 2k + 3k + 4k + \ dots}$ (e.g. centered square numbers for which $k=4,$ ${\ displaystyle k = 4,}$ form a sequence: $1,5,13,25,41\dots$ ${\ displaystyle 1,5,13,25,41 \ dots}$ ) This series can be written as $1+k(1+2+3+4+\dots ).$ ${\ displaystyle 1 + k (1 + 2 + 3 + 4 + \ dots).}$ , whence it is seen that in parentheses is the generating series for classical triangular numbers (see above ). Consequently, each sequence of centered $k$ ${\ displaystyle k}$ -gonal numbers, starting from the 2nd element, can be represented as $kT_{n}+1,$ ${\ displaystyle kT_ {n} +1,}$ Where $T_{n}(n=1,2,3\dots )$ ${\ displaystyle T_ {n} (n = 1,2,3 \ dots)}$ Is a sequence of triangular numbers. For example, centered square numbers are quadruple triangular numbers plus 1, the generating series for them has the form: $1+4+8+12\dots$ ${\ displaystyle 1 + 4 + 8 + 12 \ dots}$ ^[27]

From the above formula for triangular numbers, we obtain the general formula for $n$ ${\ displaystyle n}$ centered $k$ ${\ displaystyle k}$ -gonal number $C_{n}^{(k)}$ ${\ displaystyle C_ {n} ^ {(k)}}$ ^[27] :

C_{n}^{(k)}=1+k{\frac {n(n-1)}{2}}={\frac {kn^{2}-kn+2}{2}};\ n=1,2,3\dots

{\ displaystyle C_ {n} ^ {(k)} = 1 + k {\ frac {n (n-1)} {2}} = {\ frac {kn ^ {2} -kn + 2} {2} }; \ n = 1,2,3 \ dots}

(OTsF)

The generating function for centered polygonal numbers has the form ^[28] :

f(x)={\frac {x(1+(k-2)x+x^{2})}{(1-x)^{3}}};\quad |x|<1

{\ displaystyle f (x) = {\ frac {x (1+ (k-2) x + x ^ {2})} {(1-x) ^ {3}}}; \ quad | x | <1 }

Types of Centered Polygonal Numbers

Centered Triangular Numbers

$n$ ${\ displaystyle n}$ the centered triangular number in order is given by the formula:

C_{n}^{(3)}={\frac {3n^{2}-3n+2}{2}}

{\ displaystyle C_ {n} ^ {(3)} = {\ frac {3n ^ {2} -3n + 2} {2}}}

.

The first elements of a sequence of centered triangular numbers:

1, 4, 10, 19, 31, 46, 64, 85, 109, 136, 166, 199, 235, 274, 316, 361, 409, 460, 514, 571, 631, 694, 760, 829, 901, 976, 1054, 1135, 1219, 1306, 1396, 1489, 1585, 1684, 1786, 1891, 1999, 2110, 2224, 2341, 2461, 2584, 2710, 2839, 2971 ...,

{\frac {3n^{2}-3n+2}{2}}

{\ displaystyle {\ frac {3n ^ {2} -3n + 2} {2}}}

(sequence A005448 in OEIS )

Some properties

Each centered triangular number, starting at 10, is the sum of three consecutive classic triangular numbers: $C_{n}^{(3)}=P_{n}^{(3)}+P_{n-1}^{(3)}+P_{n-2}^{(3)}.$ ${\ displaystyle C_ {n} ^ {(3)} = P_ {n} ^ {(3)} + P_ {n-1} ^ {(3)} + P_ {n-2} ^ {(3)} .}$
Each centered triangular number $C_{n}^{(3)}$ ${\ displaystyle C_ {n} ^ {(3)}}$ when divided by 3, gives the remainder 1, and the quotient (if positive) is the classical triangular number $T_{n-1}$ ${\ displaystyle T_ {n-1}}$ .
Some centered triangular numbers are prime ^[7] : 19, 31, 109, 199, 409 ... (sequence A125602 in OEIS ).

Centered Square Numbers

one		five		13		25

$n$ ${\ displaystyle n}$ In order, the centered 4-angle (square) number is given by the formula:

C_{n}^{(4)}={(2n-1)^{2}+1 \over 2}=2n^{2}-2n+1=(n-1)^{2}+n^{2}

{\ displaystyle C_ {n} ^ {(4)} = {(2n-1) ^ {2} +1 \ over 2} = 2n ^ {2} -2n + 1 = (n-1) ^ {2} + n ^ {2}}

The first elements of a sequence of centered square numbers:

1, 5, 13, 25, 41, 61, 85, 113, 145, 181, 221, 265, 313, 365, 421, 481, 545, 613, 685, 761, 841, 925, 1013, 1105, 1201, 1301, 1405, 1513, 1625, 1741, 1861, 1985, 2113, 2245, 2381, 2521, 2665, 2813, 2965, 3121, 3281, 3445, 3613, 3785, 3961, 4141, 4325 ...,

n^{2}+(n-1)^{2}\dots

{\ displaystyle n ^ {2} + (n-1) ^ {2} \ dots}

(sequence A001844 in OEIS )

Some properties

As can be seen from the general formula , a centered square number is the sum of two consecutive squares.
All centered square numbers are odd, and the last digit in their decimal representation changes in a cycle: 1-5-3-5-1.
All centered square numbers and their divisors give a remainder of 1 when divided by 4, and when divided by 6, 8 or 12, give a remainder of 1 or 5.
All centered square numbers, with the exception of 1, represent the length of the hypotenuse in one of the Pythagorean triples (e.g. 3-4-5, 5-12-13). Thus, each centered square number is equal to the number of points within a given distance in blocks from the center point on the square lattice.
The difference between two consecutive classic octagonal numbers is a centered square number.
Some centered square numbers are prime (obviously, classic square numbers are always composite). Examples of simple centered square numbers:

5, 13, 41, 61, 113, 181, 313, 421, 613, 761, 1013, 1201, 1301, 1741, 1861, 2113, 2381, 2521, 3121, 3613 ... (sequence A027862 in OEIS ).

Centered Pentagonal Numbers

$n$ ${\ displaystyle n}$ -th in order, the centered pentagonal number is given by the formula:

C_{n}^{(5)}={\frac {5(n-1)^{2}+5(n-1)+2}{2}}

{\ displaystyle C_ {n} ^ {(5)} = {\ frac {5 (n-1) ^ {2} +5 (n-1) +2} {2}}}

.

The first few centered pentagonal numbers are:

1, 6, 16, 31, 51, 76, 106, 141, 181, 226, 276, 331, 391, 456, 526, 601, 681, 766, 856, 951, 1051, 1156, 1266, 1381, 1501, 1626, 1756, 1891, 2031, 2176, 2326, 2481, 2641, 2806, 2976 ...,

{\frac {5(n-1)^{2}+5(n-1)+2}{2}}

{\ displaystyle {\ frac {5 (n-1) ^ {2} +5 (n-1) +2} {2}}}

... (sequence A005891 in OEIS )

The parity of centered pentagonal numbers changes according to the rule: even-even-odd-odd, and the last decimal digit changes in a cycle: 6-6-1-1.

Some centered pentagonal numbers are prime ^[7] : 31, 181, 331, 391, 601. . . (sequence A145838 in OEIS )

Centered Hexagonal Numbers

Presentation of the formula as

1+6(n(n-1)/2)

{\ displaystyle 1 + 6 (n (n-1) / 2)}

shows that

n

{\ displaystyle n}

the centered hexagonal number is 1 greater than the six-fold value

(n-1)

{\ displaystyle (n-1)}

th triangular number.

$n$ ${\ displaystyle n}$ In order, the centered hexagonal number is given by the formula:

C_{n}^{(6)}=n^{3}-(n-1)^{3}=3n(n-1)+1

{\ displaystyle C_ {n} ^ {(6)} = n ^ {3} - (n-1) ^ {3} = 3n (n-1) +1}

.

The first few centered hexagonal numbers:

1, 7, 19, 37, 61, 91, 127, 169, 217, 271, 331, 397, 469, 547, 631, 721, 817, 919 ...

1+3n(n-1)

{\ displaystyle 1 + 3n (n-1)}

... (sequence A003215 in OEIS )

Some properties

The last decimal place of the centered hexagonal numbers changes in the cycle 1-7-9-7-1.
The sum of the first n centered hexagonal numbers is equal to the " cubic number " $n^{3}.$ ${\ displaystyle n ^ {3}.}$
Fair recurrence equality: $C_{n}^{(6)}=2C_{n-1}^{(6)}-C_{n-2}^{(6)}+6$ ${\ displaystyle C_ {n} ^ {(6)} = 2C_ {n-1} ^ {(6)} - C_ {n-2} ^ {(6)} + 6}$
Some centered hexagonal numbers are prime ^[7] : 7, 19, 37, 61, 127 ... (sequence A002407 in OEIS ).

Centered Heptagonal Numbers

$n$ ${\ displaystyle n}$ in order, the centered heptagonal number is given by ${\frac {7n^{2}-7n+2}{2}}$ ${\ displaystyle {\ frac {7n ^ {2} -7n + 2} {2}}}$ . It can also be calculated by multiplying the triangular number $(n-1)$ ${\ displaystyle (n-1)}$ by 7 with the addition of 1.

The first few centered heptagonal numbers:

1, 8, 22, 43, 71, 106, 148, 197, 253, 316, 386, 463, 547, 638, 736, 841, 953 ...,

{\frac {7n^{2}-7n+2}{2}}

{\ displaystyle {\ frac {7n ^ {2} -7n + 2} {2}}}

... (sequence A069099 in OEIS )

The parity of the centered heptagonal numbers changes in the odd-even-even-odd cycle.

Some centered heptagonal numbers are prime ^[7] :

43, 71, 197, 463, 547, 953, 1471, 1933, 2647, 2843, 3697 ... (sequence A144974 in OEIS )

There are also centered heptagonal numbers in pairs of twin primes :

43, 71, 197, 463, 1933, 5741, 8233, 9283, 11173, 14561, 34651 ... (sequence A144975 in OEIS )

Centered Octagonal Numbers

$n$ ${\ displaystyle n}$ the centered octagonal number is given by the formula $(2n-1)^{2}=4n^{2}-4n+1.$ ${\ displaystyle (2n-1) ^ {2} = 4n ^ {2} -4n + 1.}$

The first few centered octagonal numbers:

1, 9, 25, 49, 81, 121, 169, 225, 289, 361, 441, 529, 625, 729, 841, 961, 1089.

Some properties

All centered octagonal numbers are odd, and their last decimal digit changes in the cycle 1-9-5-9-1.
The centered octagonal number matches the classic square number with an odd number: $C_{n}^{(6)}=P_{2n-1}^{(4)}.$ ${\ displaystyle C_ {n} ^ {(6)} = P_ {2n-1} ^ {(4)}.}$ In other words, an odd number is a centered octagonal number if and only if it is the square of an integer.
From the previous property it follows that all centered octagonal numbers, except 1, are compound.

Centered Nentagonal Numbers

$n$ ${\ displaystyle n}$ the centered ninth number in order is determined by the general formula ${\frac {(3n-2)(3n-1)}{2}}.$ ${\ displaystyle {\ frac {(3n-2) (3n-1)} {2}}.}$

Multiplying $(n-1)$ ${\ displaystyle (n-1)}$ th triangular number by 9 and adding 1, we get $n$ ${\ displaystyle n}$ -th centered octagonal number, but there is also a simpler connection with triangular numbers - every third triangular number (1st, 4th, 7th, etc.) is also a centered octagonal number, and so you can get everything centered octagonal numbers. Formal Record: $C_{n}^{(9)}=P_{3n-2}^{(3)}.$ ${\ displaystyle C_ {n} ^ {(9)} = P_ {3n-2} ^ {(3)}.}$

The first centered octagonal numbers:

1, 10, 28, 55, 91, 136, 190, 253, 325, 406, 496, 595, 703, 820, 946 ... (sequence A060544 in OEIS )

With the exception of 6, all even perfect numbers are also centered octagonal numbers. In 1850, the amateur mathematician Frederick Pollock ( Sir Frederick Pollock ) made an assumption that has not yet been proved and not refuted that any natural number is the sum of a maximum of eleven centered octagonal numbers ^[29] .

From the general formula it follows that all centered octagonal numbers, except 1, are compound.

Centered Decagonal Numbers

$n$ ${\ displaystyle n}$ the centered decagonal number in order is given by the formula $5(n^{2}-n)+1$ ${\ displaystyle 5 (n ^ {2} -n) +1}$ .

The first representatives of centered decagonal numbers:

1, 11, 31, 61, 101, 151, 211, 281, 361, 451, 551, 661, 781, 911, 1051 ...

5(n^{2}-n)+1

{\ displaystyle 5 (n ^ {2} -n) +1}

(sequence A062786 in OEIS )

Like others $k$ ${\ displaystyle k}$ -gonal numbers $n$ ${\ displaystyle n}$ the centered decagonal number can be calculated by multiplying $(n-1)$ ${\ displaystyle (n-1)}$ th triangular number on $k$ ${\ displaystyle k}$ , in our case 10, then adding 1. As a result, centered decagonal numbers can be obtained simply by adding 1 to the decimal representation of the number. Thus, all centered decagonal numbers are odd and always end with 1 in decimal.

Some centered decagonal chiseos are simple, for example:

11, 31, 61, 101, 151, 211, 281, 661, 911, 1051, 1201, 1361, 1531, 1901, 2311, 2531, 3001, 3251, 3511, 4651, 5281 ... (sequence A090562 in OEIS )

Polygonal numbers, both classic and centered

Some centered polygonal numbers coincide with the classic ones, for example: $1,10,25,51;$ ${\ displaystyle 1,10,25,51;}$ for brevity, we will call such polygonal numbers double .

1. Double numbers with a common parameter

k

{\ displaystyle k}

(number of angles): the identity ^[30] holds:

C_{k}^{(k)}=P_{k+1}^{(k)}\quad

{\ displaystyle C_ {k} ^ {(k)} = P_ {k + 1} ^ {(k)} \ quad}

2. Double triangular numbers with different

k .

{\ displaystyle k.}

Example:

1,10,136,1891,26335\dots

{\ displaystyle 1,10,136,1891,26335 \ dots}

(sequence A128862 in OEIS ). To find them, it is necessary to solve the Diophantine equation :

m^{2}+m=3n^{2}-3n+2,

{\ displaystyle m ^ {2} + m = 3n ^ {2} -3n + 2,}

then

P_{m}^{(3)}=C_{n}^{(3)}.

{\ displaystyle P_ {m} ^ {(3)} = C_ {n} ^ {(3)}.}

Some solutions:

m=1,4,16,61,229\dots

{\ displaystyle m = 1,4,16,61,229 \ dots}

(sequence A133161 in OEIS ), respectively:

n=1,3,10,36,133\dots

{\ displaystyle n = 1,3,10,36,133 \ dots}

(sequence A102871 in OEIS )

3. Classic square, which are centered triangular numbers. They are determined by the Diophantine equation:

m^{2}={\frac {3n^{2}-3n+2}{2}}.\quad

{\ displaystyle m ^ {2} = {\ frac {3n ^ {2} -3n + 2} {2}}. \ quad}

Then

C_{m}^{(3)}=P_{n}^{(4)}.

{\ displaystyle C_ {m} ^ {(3)} = P_ {n} ^ {(4)}.}

Solutions:

m=1,2,8,19,79\dots

{\ displaystyle m = 1,2,8,19,79 \ dots}

(sequence A129445 in OEIS ), respectively

n=1,2,7,16,65\dots

{\ displaystyle n = 1,2,7,16,65 \ dots}

The first numbers are:

1,4,64,361,6241\dots

{\ displaystyle 1,4,64,361,6241 \ dots}

4. Classic triangular, which are centered hexagonal numbers. The first numbers are:

1,91,8911,873181,85562821\dots

{\ displaystyle 1.91.8911.873181.85562821 \ dots} (sequence A006244 in OEIS ). They are determined by the Diophantine equation:

{\frac {m(m+1)}{2}}=3n^{2}+3n+1.\quad

{\ displaystyle {\ frac {m (m + 1)} {2}} = 3n ^ {2} + 3n + 1. \ quad}

Then

P_{m}^{(3)}=C_{n+1}^{(6)}

{\ displaystyle P_ {m} ^ {(3)} = C_ {n + 1} ^ {(6)}}

Solutions:

m=1,13,133,1321,13081\dots

{\ displaystyle m = 1,13,133,1321,13081 \ dots}

(sequence A031138 in OEIS )

n=0,5,54,539,5340\dots

{\ displaystyle n = 0,5,54,539,5340 \ dots}

(sequence A087125, in OEIS )

5. Classic square, which are centered hexagonal numbers. The first numbers are:

1,169,32761,6355441,1232922769\dots

{\ displaystyle 1,169,32761,6355441,11232922769 \ dots}

(sequence A006051 in OEIS ). They are determined by the Diophantine equation:

m^{2}=3n^{2}+3n+1.\quad

{\ displaystyle m ^ {2} = 3n ^ {2} + 3n + 1. \ quad}

Then

P_{m}^{(4)}=C_{n+1}^{(6)}

{\ displaystyle P_ {m} ^ {(4)} = C_ {n + 1} ^ {(6)}}

Solutions:

m=1,13,181,2521,35113\dots

{\ displaystyle m = 1,13,181,2521,35113 \ dots}

(sequence A001570 in OEIS )

n=0,7,104,1455,20272\dots

{\ displaystyle n = 0,7,104,1455,20272 \ dots}

(sequence A001921, in OEIS )

Spatial Curly Numbers

Along with the figured numbers considered above for planar figures, it is possible to determine their spatial or even multidimensional analogues. Already ancient mathematicians studied tetrahedral and square pyramidal numbers. It is easy to determine the numbers associated with the pyramids , which are based on any other polygon, for example:

.
.
.

Other classes of spatial curly numbers are associated with classical polyhedra .

Pyramidal numbers

Geometric representation of a square pyramidal number:

\Pi _{4}^{(4)}=1+4+9+16=30

{\ displaystyle \ Pi _ {4} ^ {(4)} = 1 + 4 + 9 + 16 = 30}

.

Pyramidal numbers are defined as follows.

$n$ ${\ displaystyle n}$ in order $k$ ${\ displaystyle k}$ -gonal pyramidal number $\Pi _{n}^{(k)}$ ${\ displaystyle \ Pi _ {n} ^ {(k)}}$ is the sum of the first $n$ ${\ displaystyle n}$ flat curly numbers with the same number of angles $k$ ${\ displaystyle k}$ :

\Pi _{n}^{(k)}=P_{1}^{(k)}+P_{2}^{(k)}+P_{3}^{(k)}+\dots +P_{n}^{(k)}

{\ displaystyle \ Pi _ {n} ^ {(k)} = P_ {1} ^ {(k)} + P_ {2} ^ {(k)} + P_ {3} ^ {(k)} + \ dots + P_ {n} ^ {(k)}}

Geometrically Pyramidal Number $\Pi _{n}^{(k)}$ ${\ displaystyle \ Pi _ {n} ^ {(k)}}$ can be imagined as a pyramid of $n$ ${\ displaystyle n}$ layers (see figure), each of which contains from 1 (top layer) to $P_{n}^{(k)}$ ${\ displaystyle P_ {n} ^ {(k)}}$ (bottom) balls.

By induction, it is not difficult to prove the general formula for the pyramidal number, known even to Archimedes ^[31] :

\Pi _{n}^{(k)}={\frac {n(n+1)((k-2)n-k+5)}{6}}

{\ displaystyle \ Pi _ {n} ^ {(k)} = {\ frac {n (n + 1) ((k-2) n-k + 5)} {6}}}

(OPF)

The right side of this formula can also be expressed in terms of flat polygonal numbers:

\Pi _{n}^{(k)}={\frac {(k-2)n-k+5}{3}}P_{n}^{(3)}={\frac {n+1}{6}}(2P_{n}^{(k)}+n)

{\ displaystyle \ Pi _ {n} ^ {(k)} = {\ frac {(k-2) n-k + 5} {3}} P_ {n} ^ {(3)} = {\ frac { n + 1} {6}} (2P_ {n} ^ {(k)} + n)}

There is a three-dimensional analogue of the Nicomache formula for pyramidal numbers ^[32] :

\Pi _{n}^{(k+1)}=\Pi _{n}^{(k)}+\Pi _{n-1}^{(3)}

{\ displaystyle \ Pi _ {n} ^ {(k + 1)} = \ Pi _ {n} ^ {(k)} + \ Pi _ {n-1} ^ {(3)}}

The generating function of the pyramidal numbers has the form ^[33] :

f(x)={\frac {x(1+(k-3)x)}{(1-x)^{4}}};\quad |x|<1

{\ displaystyle f (x) = {\ frac {x (1+ (k-3) x)} {(1-x) ^ {4}}}; \ quad | x | <1}

Triangular pyramidal (tetrahedral) numbers

A tetrahedron with a side length of 5 contains 35 spheres. Each layer represents one of the first five triangular numbers .

Triangular pyramidal numbers, also called tetrahedral numbers, are curly numbers that represent a tetrahedron , that is, a pyramid with a triangle at its base. The first few tetrahedral numbers are:

1, 4, 10, 20, 35, 56, 84, 120, 165, 220, 286, 364, 455, 560, 680, 816, 969 ... (sequence A000292 in OEIS )

Interestingly, the fifth number is equal to the sum of all the previous ones.

The general formula for a tetrahedral number is: $\Pi _{n}^{(3)}={\frac {n(n+1)(n+2)}{6}}.$ ${\ displaystyle \ Pi _ {n} ^ {(3)} = {\ frac {n (n + 1) (n + 2)} {6}}.}$

There is a three-dimensional analogue of the Bachet de Meziriac formula , namely the decomposition of an arbitrary pyramidal number into tetrahedral ones ^[32] :

\Pi _{n}^{(k)}=\Pi _{n}^{(3)}+(k-3)\Pi _{n-1}^{(3)}

{\ displaystyle \ Pi _ {n} ^ {(k)} = \ Pi _ {n} ^ {(3)} + (k-3) \ Pi _ {n-1} ^ {(3)}}

Five tetrahedral numbers are simultaneously triangular (sequence A027568 in OEIS ):

1, 10, 120, 1540, 7140

Only three tetrahedral numbers are square numbers:

1^{2}=1

{\ displaystyle 1 ^ {2} = 1}

,

2^{2}=4

{\ displaystyle 2 ^ {2} = 4}

,

140^{2}=19\,600

{\ displaystyle 140 ^ {2} = 19 \, 600}

.

One of Pollock ’s hypotheses (1850): each natural number is representable as the sum of no more than five tetrahedral numbers. It has not yet been proven, although verified for all numbers less than 10 billion ^[34] ^[35] .

Square Pyramidal Numbers

Square pyramidal numbers are often briefly called simply pyramidal. For them, the pyramid has a square base. Initial Sequence:

1, 5, 14, 30, 55, 91, 140, 204, 285, 385, 506, 650, 819 ... (sequence A000330 in OEIS ).

The general formula for a square pyramidal number is: $\Pi _{n}^{(4)}={\frac {n(n+1)(2n+1)}{6}}.$ ${\ displaystyle \ Pi _ {n} ^ {(4)} = {\ frac {n (n + 1) (2n + 1)} {6}}.}$

Square pyramidal number $\Pi _{n}^{(4)}$ ${\ displaystyle \ Pi _ {n} ^ {(4)}}$ also expresses the total number of squares in a square grid $n\times n$ ${\ displaystyle n \ times n}$ .

Between square and triangular pyramidal numbers there is the following relationship ^[36] :

4\Pi _{n}^{(4)}=\Pi _{2n}^{(3)}

{\ displaystyle 4 \ Pi _ {n} ^ {(4)} = \ Pi _ {2n} ^ {(3)}}

It was noted above that the sum of consecutive triangular numbers is a square number; similarly, the sum of consecutive tetrahedral numbers is the square pyramidal number ^[36] : $\Pi _{n}^{(4)}=\Pi _{n}^{(3)}+\Pi _{n-1}^{(3)}$ ${\ displaystyle \ Pi _ {n} ^ {(4)} = \ Pi _ {n} ^ {(3)} + \ Pi _ {n-1} ^ {(3)}}$

Polyhedral numbers

By analogy with square, you can enter "cubic numbers" $Q_{n}=n^{3},$ ${\ displaystyle Q_ {n} = n ^ {3},}$ as well as numbers corresponding to other regular and irregular polyhedra - for example, Platonic solids :

Octahedral number
Dodecahedral number
Icosahedral number

Their centered options are also provided.

Cubic Numbers

Cubic numbers $Q_{n}$ ${\ displaystyle Q_ {n}}$ are a product of three identical natural numbers and have a general form $Q_{n}=n^{3}.$ ${\ displaystyle Q_ {n} = n ^ {3}.}$ Initial Values:

1, 8, 27, 64, 125, 216, 343, 512, 729, 1000. . . (sequence A000578 in OEIS )

The cubic number can be expressed as the difference of the squares of successive triangular numbers ^[37] :

Q_{n}=(T_{n})^{2}-(T_{n-1})^{2},n\geqslant 2

{\ displaystyle Q_ {n} = (T_ {n}) ^ {2} - (T_ {n-1}) ^ {2}, n \ geqslant 2}

Consequence: the sum of the first $n$ ${\ displaystyle n}$ cubic numbers is squared $n$ ${\ displaystyle n}$ triangular number:

Q_{1}+Q_{2}+Q_{3}+\dots +Q_{n}=(T_{n})^{2}

{\ displaystyle Q_ {1} + Q_ {2} + Q_ {3} + \ dots + Q_ {n} = (T_ {n}) ^ {2}}

The difference between two adjacent cubic numbers is a centered hexagonal number. Consequence: the sum of the first $n$ ${\ displaystyle n}$ centered hexagonal numbers there is a cubic number $Q_{n}$ ${\ displaystyle Q_ {n}}$ ^[37] .

The expression of the cubic number in terms of tetrahedral ^[37] :

Q_{n}=\Pi _{n}^{(3)}+4\Pi _{n-1}^{(3)}+\Pi _{n-2}^{(3)}

{\ displaystyle Q_ {n} = \ Pi _ {n} ^ {(3)} + 4 \ Pi _ {n-1} ^ {(3)} + \ Pi _ {n-2} ^ {(3) }}

where

n>2.

{\ displaystyle n> 2.}

One of Pollock ’s hypotheses (1850): each natural number is representable as a sum of not more than nine cubic numbers. Proven at the beginning of the XX century. Seven cubes are usually sufficient, but 15 numbers require eight (15, 22, 50, 114, 167, 175, 186, 212, 231, 238, 303, 364, 420, 428, 454, sequence A018889 in OEIS ), and two numbers all nine are needed: 23 and 239. If, in addition to addition, subtraction is allowed, then five cubes are enough (it is possible that even four cubes, but this has not yet been proved) ^[38] .

The generating function of cubic numbers has the form ^[37] :

f(x)={\frac {x(x^{2}+4x+1)}{(x-1)^{4}}};\quad |x|<1

{\ displaystyle f (x) = {\ frac {x (x ^ {2} + 4x + 1)} {(x-1) ^ {4}}}; \ quad | x | <1}

Octahedral numbers

Dodecahedral numbers

Icosahedral numbers

Multidimensional Generalizations

The three-dimensional constructions described above can be generalized to four or more dimensions. An analog of tetrahedral numbers in $d$ ${\ displaystyle d}$ -dimensional space are " simplex numbers", also called hypertetrahedral :

S_{n}^{[d]}={\frac {(n-1+d)!}{(n-1)!\ d!}}

{\ displaystyle S_ {n} ^ {[d]} = {\ frac {(n-1 + d)!} {(n-1)! \ d!}}}

Their special case are:

$S_{n}^{[2]}$ ${\ displaystyle S_ {n} ^ {[2]}}$ Are triangular numbers .
$S_{n}^{[3]}$ ${\ displaystyle S_ {n} ^ {[3]}}$ - tetrahedral numbers .
$S_{n}^{[4]}$ ${\ displaystyle S_ {n} ^ {[4]}}$ - pentatope numbers

Another class of multidimensional numbers is hypercubic : $Q_{n}^{[d]}=n^{d}.$ ${\ displaystyle Q_ {n} ^ {[d]} = n ^ {d}.}$ Four-dimensional hypercubic numbers ( $d=4$ ${\ displaystyle d = 4}$ ) are called biquadratic .

Numbers from more than one class

Some curly numbers can be included in more than one class of plane and / or multidimensional numbers; examples for plane numbers have already been given above. For multidimensional numbers, this is a rather rare situation ^[39] .

Five numbers $1,10,120,1540,7140$ ${\ displaystyle 1,10,120,1540,7140}$ (and only they) are simultaneously triangular and tetrahedral (sequence A027568 in OEIS ).
Four numbers $1,55,91,208335$ ${\ displaystyle 1.55.91.208335}$ both triangular and square pyramidal (sequence A039596 in OEIS ).
Three numbers $1,4,19600$ ${\ displaystyle 1,4,19600}$ both planar square and tetrahedral (sequence A003556 in OEIS ).
Two numbers $1,4900$ ${\ displaystyle 1,4900}$ both square flat and square pyramidal. This statement became known as the “ Lucas hypothesis” or “ the cannonball problem ” (1875). The full decision was given in 1918 by George Neville Watson ^[40] .
No natural number, except 1, can be simultaneously:
- triangular and cubic;
- centered hexagonal and cubic.

In 1988, F. Bakers and J. Top proved that no number but 1 can be simultaneously tetrahedral and square pyramidal ^[41] . It is also proved that there are no numbers that are simultaneously:

tetrahedral and cubic;
square pyramidal and cubic;
tetrahedral and biquadratic;
square pyramidal and biquadratic.

Role in Number Theory

Pascal's Triangle

Numbers from the Pascal triangle show a connection with many types of curly numbers.

{\begin{array}{c}1\\1\quad 1\\1\quad 2\quad 1\\{\color {red}1}\quad 3\quad 3\quad 1\\1\quad {\color {red}4}\quad 6\quad 4\quad 1\\1\quad 5\quad {\color {red}10}\quad 10\quad 5\quad 1\\1\quad 6\quad 15\quad {\color {red}20}\quad 15\quad 6\quad 1\\1\quad 7\quad 21\quad 35\quad {\color {red}35}\quad 21\quad 7\quad 1\\\end{array}}

{\ displaystyle {\ begin {array} {c} 1 \\ 1 \ quad 1 \\ 1 \ quad 2 \ quad 1 \\ {\ color {red} 1} \ quad 3 \ quad 3 \ quad 1 \\ 1 \ quad {\ color {red} 4} \ quad 6 \ quad 4 \ quad 1 \\ 1 \ quad 5 \ quad {\ color {red} 10} \ quad 10 \ quad 5 \ quad 1 \\ 1 \ quad 6 \ quad 15 \ quad {\ color {red} 20} \ quad 15 \ quad 6 \ quad 1 \\ 1 \ quad 7 \ quad 21 \ quad 35 \ quad {\ color {red} 35} \ quad 21 \ quad 7 \ quad 1 \\\ end {array}}}

Tetrahedral numbers in Pascal's triangle (highlighted in red)

On the third line in the Pascal triangle are triangular numbers, and on the fourth - tetrahedral numbers (see figure). This is explained by $n$ ${\ displaystyle n}$ the tetrahedral number is the sum of the first $n$ ${\ displaystyle n}$ triangular numbers that are located on the third line. Similarly, on the fifth line are four-dimensional pentatope numbers , etc. All of them, like other numbers inside the Pascal triangle, are binomial coefficients .

Thus, all the internal elements of the Pascal triangle are curly numbers, and their various types are presented. Along each line, from left to right, are hypertetrahedral numbers of increasing dimension. It is known that the sum of all numbers $n$ ${\ displaystyle n}$ th row equals $2^{n},$ ${\ displaystyle 2 ^ {n},}$ it follows that the sum of all numbers of the first $n$ ${\ displaystyle n}$ rows equal to the Mersenne number $M_{n}.$ ${\ displaystyle M_ {n}.}$ Therefore, the Mersenne number can be represented as the sum of hypertetrahedral numbers ^[42] .

Other Applications

Many theorems of number theory can be stated in terms of curly numbers. For example, the Catalan hypothesis claims that among hypercubic numbers of arbitrary dimensions, only one pair differs by 1: $3^{2}=2^{3}+1$ ${\ displaystyle 3 ^ {2} = 2 ^ {3} +1}$ (proved in 2002) ^[43] .

Every even perfect number is triangular ^[44] (and at the same time hexagonal, moreover, the number of the hexagonal number is a power of two). Such a number cannot simultaneously be a square, cubic, or other hypercubic number ^[45] .

Legendre hypothesis (1808, it is the third problem of Edmund Landau ): between consecutive square numbers there is always a prime number . Still not proven.

Amount First $n$ ${\ displaystyle n}$ centered triangular numbers $(n>2)$ ${\ displaystyle (n> 2)}$ there is a “magic constant” for a magic square of dimension $n\times n$ ${\ displaystyle n \ times n}$ . Other ways to get the same constant is through a triangular number ${\frac {T_{n^{2}}}{n}},$ ${\ displaystyle {\ frac {T_ {n ^ {2}}} {n}},}$ or add all the natural numbers from $T_{n-1}$ ${\ displaystyle T_ {n-1}}$ before $T_{n}$ ${\ displaystyle T_ {n}}$ inclusive ^[46] .

A Mersenne number greater than 1 cannot be square, cubic, or other hypercubic, but it can be triangular. Треугольных чисел Мерсенна всего четыре: $1,3,5,4095,$ ${\displaystyle 1,3,5,4095,}$ их поиск эквивалентен решению в натуральных числах : $2^{n}-7=x^{2}.$ $2^{n}-7=x^{2}.$ As it turned out, a solution to this equation exists only for $n=3,4,5,7,15$ ${\ displaystyle n = 3,4,5,7,15}$ (sequence A060728 in OEIS ), and when $n>3$ ${\ displaystyle n> 3}$ corresponding Mersenne number $M_{n-3}$ ${\ displaystyle M_ {n-3}}$ will then be triangular ^[42] .

Getting Fibonacci numbers from Pascal's triangle

Fermat's number also cannot be square, cubic or other hypercubic, but in the only case it can be triangular: $F_{0}=3.$ ${\ displaystyle F_ {0} = 3.}$ The Fermat number also cannot be tetrahedral and hypertetrahedral of any dimension higher than the 2nd ^[42] .

Among the Fibonacci numbers, there are only three square numbers (0, 1 and 144) and four triangular numbers (1, 3, 21, 55, sequence A039595 in OEIS ). If you rotate the Pascal triangle, as shown in the figure, then the Fibonacci numbers can be obtained as sums along the ascending diagonals; this fact gives an expansion of the Fibonacci number in hypertetrahedral numbers ^[47] .

Among the Lucas numbers, there are two square numbers (1 and 4) and three triangular numbers (1, 3, 5778) ^[47] .

Catalan numbers $Cat_{n}$ ${\ displaystyle Cat_ {n}}$ expressed in terms of hypertetrahedral numbers as follows ^[48] :

Cat_{n}=S_{n+1}^{[n]}-S_{n+2}^{[n-1]}

{\ displaystyle Cat_ {n} = S_ {n + 1} ^ {[n]} - S_ {n + 2} ^ {[n-1]}}

Another class of numbers closely related to curly numbers is Stirling numbers of the second kind. $S(n,m).$ ${\ displaystyle S (n, m).}$ This class includes all triangular numbers: $T_{n}=S(n+1,n),$ ${\ displaystyle T_ {n} = S (n + 1, n),}$ and the expression $S(n,2)+1$ ${\ displaystyle S (n, 2) +1}$ equals 2nd in order $n$ ${\ displaystyle n}$ -dimensional hypercubic number $Q_{2}^{[n]}.$ ${\ displaystyle Q_ {2} ^ {[n]}.}$ Finally, all $n$ ${\ displaystyle n}$ -dimensional hypercubic number decomposes in $S(n,m)$ ${\ displaystyle S (n, m)}$ as follows ^[48] :

Q_{n}^{[d]}=\sum _{m=0}^{d}{S(d,m)x(x-1)\dots (x-m+1)}

{\ displaystyle Q_ {n} ^ {[d]} = \ sum _ {m = 0} ^ {d} {S (d, m) x (x-1) \ dots (x-m + 1)}}

Notes

↑ ¹ ² Desa E., Desa M., 2016 , p. ten.
↑ Gaidenko P.P. The evolution of the concept of science (the formation and development of the first scientific programs) , chapter 1. M .: Nauka, 1980.
↑ Ozhigova E.P. What is the theory of numbers. - M .: Knowledge, 1970. - S. 56-57.
↑ Arithmetic series // Mathematical Encyclopedia (in 5 volumes) . - M .: Soviet Encyclopedia , 1982.- T. 1.
↑ ¹ ² Desa E., Desa M., 2016 , p. 15.
↑ Behind the pages of a mathematics textbook, 1996 , p. 50.
↑ ¹ ² ³ ⁴ ⁵ Desa E., Desa M., 2016 , p. 217.
↑ Desa E., Desa M., 2016 , p. 14.
↑ Diophantus of Alexandria . Arithmetic and a book about polygonal numbers. / Per. I.N. Veselovsky; Ed. and comment. I. G. Bashmakova. - M.: Science, GRFML, 1974 . - 328 p. S. 48.
↑ ¹ ² Matvievskaya G.P. The doctrine of number in the medieval Near and Middle East. - Tashkent: FAN, 1967. - S. 22-23. - 344 p. Contrary to the title, the book traces the history of the concept of number from the most ancient times.
↑ ¹ ² Desa E., Desa M., 2016 , p. 237.
↑ Vilenkin N. Ya. Popular combinatorics . - M .: Nauka, 1975 .-- S. 10-11. - 208 p.
↑ Desa E., Desa M., 2016 , p. ten.
↑ Desa E., Desa M., 2016 , p. 20-21.
↑ Desa E., Desa M., 2016 , p. 24.
↑ Desa E., Desa M., 2016 , p. 25–33.
↑ Desa E., Desa M., 2016 , p. 225.
↑ Some finite number series (neopr.) . Math24.ru . Date of treatment June 14, 2019.
↑ Kokhas K.P. Sum of inverse squares // Mathematical education. - 2004. - Vol. 8 . - S. 142-163 .
↑ Weinstein F.V. Partition of numbers. // Journal "Quantum". - 1988. - No. 11.
↑ ¹ ² Desa E., Desa M., 2016 , p. 37-39.
↑ ¹ ² ³ Desa E., Desa M., 2016 , p. 39-39.
↑ Desa E., Desa M., 2016 , p. 17-19.
↑ Desa E., Desa M., 2016 , p. 33.
↑ Lawrence Downey, Boon W. Ong . Beyond the Basel Problem: Sums of Reciprocals of Figurate Numbers
↑ Desa E., Desa M., 2016 , p. 39-40.
↑ ¹ ² Desa E., Desa M., 2016 , p. 40–41.
↑ Desa E., Desa M., 2016 , p. 42.
↑ Dickson, LE (2005), Diophantine Analysis , vol. 2, History of the Theory of Numbers , New York: Dover, p. 22-23 , < https://books.google.com/books?id=eNjKEBLt_tQC&pg=PA22 > .
↑ Desa E., Desa M., 2016 , p. 48.
↑ Desa E., Desa M., 2016 , p. 70-71.
↑ ¹ ² Desa E., Desa M., 2016 , p. 76.
↑ Desa E., Desa M., 2016 , p. 74-75.
↑ Desa E., Desa M., 2016 , p. 239.
↑ Frederick Pollock. On the extension of the principle of Fermat's theorem on the polygonal numbers to the higher order of series whose ultimate differences are constant. With a new theorem proposed, applicable to all the orders ( Abstract ) // Abstracts of the Papers Communicated to the Royal Society of London: journal. - 1850. - Vol. 5 . - P. 922-924 .
↑ ¹ ² Desa E., Desa M., 2016 , p. 75.
↑ ¹ ² ³ ⁴ Desa E., Desa M., 2016 , p. 78-81.
↑ Desa E., Desa M., 2016 , p. 231-232.
↑ Desa E., Desa M., 2016 , p. 77-78.
↑ Watson GN The Problem of the Square Pyramid // Messenger. Math. 1918. Vol. 48. P. 1–16.
↑ Beukers F., Top J. On oranges and integral points on certain plane cubic curves // Nieuw Arch. Wisk. (four). 1988. Vol. 6, No. 3. P. 203–210.
↑ ¹ ² ³ Desa E., Desa M., 2016 , p. 203-205.
↑ Desa E., Desa M., 2016 , p. 196-197.
↑ Behind the pages of a mathematics textbook, 1996 , p. 51.
↑ Desa E., Desa M., 2016 , p. 200-201.
↑ Desa E., Desa M., 2016 , p. 222-223.
↑ ¹ ² Desa E., Desa M., 2016 , p. 208.
↑ ¹ ² Desa E., Desa M., 2016 , p. 214-215.

Literature

Vilenkin N. Ya., Shibasov L.P. Shibasova 3. F. Behind the pages of a mathematics textbook: Arithmetic. Algebra. Geometry. - M .: Education, 1996.- S. 30. - 320 p. - ISBN 5-09-006575-6 .
Gleizer G.I. History of mathematics at school . - M .: Enlightenment, 1964 .-- 376 p.
Desa E., Desa M. Curly numbers. - M .: ICMMO, 2016 .-- 349 p. - ISBN 978-5-4439-2400-7 .
Depman I. Ya. History of arithmetic. A manual for teachers . - Second Edition. - M .: Enlightenment, 1965. - S. 150-155.
Matvievskaya G.P. Notes on polygonal numbers in Euler notebooks // Historical and mathematical research . - M .: Nauka , 1983 .-- No. 27 . - S. 27-49 .
Sierpinski V. Pythagorian triangles. - M .: Uchpedgiz, 1959. - 111 p.
Stillwell D. Chapter 3. Greek number theory // Mathematics and its history. - Moscow-Izhevsk: Institute for Computer Research, 2004.

Links

[DD10-1] ¹ ² Desa E., Desa M., 2016 , p. ten.

[2] Gaidenko P.P. The evolution of the concept of science (the formation and development of the first scientific programs) , chapter 1. M .: Nauka, 1980.

[3] Ozhigova E.P. What is the theory of numbers. - M .: Knowledge, 1970. - S. 56-57.

[4] Arithmetic series // Mathematical Encyclopedia (in 5 volumes) . - M .: Soviet Encyclopedia , 1982.- T. 1.

[DD15-5] ¹ ² Desa E., Desa M., 2016 , p. 15.

[_f26532716c1a07da-6] Behind the pages of a mathematics textbook, 1996 , p. 50.

[DD217-7] ¹ ² ³ ⁴ ⁵ Desa E., Desa M., 2016 , p. 217.

[_cef8a15096c5ef58-8] Desa E., Desa M., 2016 , p. 14.

[9] Diophantus of Alexandria . Arithmetic and a book about polygonal numbers. / Per. I.N. Veselovsky; Ed. and comment. I. G. Bashmakova. - M.: Science, GRFML, 1974 . - 328 p. S. 48.

[MATV-10] ¹ ² Matvievskaya G.P. The doctrine of number in the medieval Near and Middle East. - Tashkent: FAN, 1967. - S. 22-23. - 344 p. Contrary to the title, the book traces the history of the concept of number from the most ancient times.

[DD237-11] ¹ ² Desa E., Desa M., 2016 , p. 237.

[12] Vilenkin N. Ya. Popular combinatorics . - M .: Nauka, 1975 .-- S. 10-11. - 208 p.

[_cef8a15096c5ef5c-13] Desa E., Desa M., 2016 , p. ten.

[_33a67ac429eca262-14] Desa E., Desa M., 2016 , p. 20-21.

[_cef8a45096c5f4b1-15] Desa E., Desa M., 2016 , p. 24.

[_72c0946749498450-16] Desa E., Desa M., 2016 , p. 25–33.

[_7673ebf0325ed2c0-17] Desa E., Desa M., 2016 , p. 225.

[18] Some finite number series (neopr.) . Math24.ru . Date of treatment June 14, 2019.

[19] Kokhas K.P. Sum of inverse squares // Mathematical education. - 2004. - Vol. 8 . - S. 142-163 .

[20] Weinstein F.V. Partition of numbers. // Journal "Quantum". - 1988. - No. 11.

[DD37-21] ¹ ² Desa E., Desa M., 2016 , p. 37-39.

[DD39-22] ¹ ² ³ Desa E., Desa M., 2016 , p. 39-39.

[_40c17f687550b3bd-23] Desa E., Desa M., 2016 , p. 17-19.

[_cef8a35096c5f2e1-24] Desa E., Desa M., 2016 , p. 33.

[siam_07-003s-25] Lawrence Downey, Boon W. Ong . Beyond the Basel Problem: Sums of Reciprocals of Figurate Numbers

[_b488700c70e4626b-26] Desa E., Desa M., 2016 , p. 39-40.

[DD40-27] ¹ ² Desa E., Desa M., 2016 , p. 40–41.

[_cef89e5096c5ea61-28] Desa E., Desa M., 2016 , p. 42.

[29] Dickson, LE (2005), Diophantine Analysis , vol. 2, History of the Theory of Numbers , New York: Dover, p. 22-23 , < https://books.google.com/books?id=eNjKEBLt_tQC&pg=PA22 > .

[_cef89e5096c5ea6b-30] Desa E., Desa M., 2016 , p. 48.

[_e7e8729a69c8fae0-31] Desa E., Desa M., 2016 , p. 70-71.

[DD76-32] ¹ ² Desa E., Desa M., 2016 , p. 76.

[_0c7edbd5fea9dae0-33] Desa E., Desa M., 2016 , p. 74-75.

[_7673eaf0325ed17b-34] Desa E., Desa M., 2016 , p. 239.

[35] Frederick Pollock. On the extension of the principle of Fermat's theorem on the polygonal numbers to the higher order of series whose ultimate differences are constant. With a new theorem proposed, applicable to all the orders ( Abstract ) // Abstracts of the Papers Communicated to the Royal Society of London: journal. - 1850. - Vol. 5 . - P. 922-924 .

[DD75-36] ¹ ² Desa E., Desa M., 2016 , p. 75.

[DD78-37] ¹ ² ³ ⁴ Desa E., Desa M., 2016 , p. 78-81.

[_054bc6dedf1dda70-38] Desa E., Desa M., 2016 , p. 231-232.

[_fd15251747dd2228-39] Desa E., Desa M., 2016 , p. 77-78.

[40] Watson GN The Problem of the Square Pyramid // Messenger. Math. 1918. Vol. 48. P. 1–16.

[41] Beukers F., Top J. On oranges and integral points on certain plane cubic curves // Nieuw Arch. Wisk. (four). 1988. Vol. 6, No. 3. P. 203–210.

[DD203-42] ¹ ² ³ Desa E., Desa M., 2016 , p. 203-205.

[DD196-43] Desa E., Desa M., 2016 , p. 196-197.

[_f26532716c1a07db-44] Behind the pages of a mathematics textbook, 1996 , p. 51.

[_a199fb30d6dc30b2-45] Desa E., Desa M., 2016 , p. 200-201.

[_9466d1c5aef430e2-46] Desa E., Desa M., 2016 , p. 222-223.

[DD208-47] ¹ ² Desa E., Desa M., 2016 , p. 208.

[DD214-48] ¹ ² Desa E., Desa M., 2016 , p. 214-215.

Curly numbers

Types of Curly Numbers

Classic Polygonal Numbers

Definition and general view

Historical Review

Triangular numbers

Square Numbers

Pentagonal Numbers

Hexagonal numbers

Twelve- Numbers

Determining if a given number is polygonal

Generating Function

PivotTable

Centered Polygonal Numbers

Definition

Types of Centered Polygonal Numbers

Centered Triangular Numbers

Centered Square Numbers

Centered Pentagonal Numbers

Centered Hexagonal Numbers

Centered Heptagonal Numbers

Centered Octagonal Numbers

Centered Nentagonal Numbers

Centered Decagonal Numbers

Polygonal numbers, both classic and centered

Spatial Curly Numbers

Pyramidal numbers

Triangular pyramidal (tetrahedral) numbers

Square Pyramidal Numbers

Polyhedral numbers

Cubic Numbers

Octahedral numbers

Dodecahedral numbers

Icosahedral numbers

Multidimensional Generalizations

Numbers from more than one class

Role in Number Theory

Pascal's Triangle

Other Applications

Notes

Literature

Links

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