# Polygonal number

In mathematics, a polygonal number is a number represented as dots or pebbles arranged in the shape of a regular polygon. The dots are thought of as alphas (units). These are one type of 2-dimensional figurate numbers.

## Definition and examples

The number 10 for example, can be arranged as a triangle (see triangular number):

But 10 cannot be arranged as a square. The number 9, on the other hand, can be (see square number):

Some numbers, like 36, can be arranged both as a square and as a triangle (see square triangular number):

By convention, 1 is the first polygonal number for any number of sides. The rule for enlarging the polygon to the next size is to extend two adjacent arms by one point and to then add the required extra sides between those points. In the following diagrams, each extra layer is shown as in red.

### Square numbers

Polygons with higher numbers of sides, such as pentagons and hexagons, can also be constructed according to this rule, although the dots will no longer form a perfectly regular lattice like above.

## Formula

An s-gonal number can be decomposed into s−2 triangular numbers and a natural number

If s is the number of sides in a polygon, the formula for the nth s-gonal number P(s,n) is

${\displaystyle P(s,n)={\frac {(s-2)n^{2}-(s-4)n}{2}}}$

or

${\displaystyle P(s,n)=(s-2){\frac {n(n-1)}{2}}+n}$

The nth s-gonal number is also related to the triangular numbers Tn as follows:

${\displaystyle P(s,n)=(s-2)T_{n-1}+n=(s-3)T_{n-1}+T_{n}\,.}$

Thus:

{\displaystyle {\begin{aligned}P(s,n+1)-P(s,n)&=(s-2)n+1\,,\\P(s+1,n)-P(s,n)&=T_{n-1}={\frac {n(n-1)}{2}}\,,\\P(s+k,n)-P(s,n)&=kT_{n-1}=k{\frac {n(n-1)}{2}}\,.\end{aligned}}}

For a given s-gonal number P(s,n) = x, one can find n by

${\displaystyle n={\frac {{\sqrt {8(s-2)x+{(s-4)}^{2}}}+(s-4)}{2(s-2)}}}$

and one can find s by

${\displaystyle s=2+{\frac {2}{n}}\cdot {\frac {x-n}{n-1}}}$ .

### Every hexagonal number is also a triangular number

Applying the formula above:

${\displaystyle P(s,n)=(s-2)T_{n-1}+n}$

to the case of 6 sides gives:

${\displaystyle P(6,n)=4T_{n-1}+n}$

but since:

${\displaystyle T_{n-1}={\frac {n(n-1)}{2}}}$

it follows that:

${\displaystyle P(6,n)={\frac {4n(n-1)}{2}}+n={\frac {2n(2n-1)}{2}}=T_{2n-1}}$

This shows that the nth hexagonal number P(6,n) is also the (2n − 1)th triangular number T2n−1. We can find every hexagonal number by simply taking the odd-numbered triangular numbers:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, ...

## Table of values

The first 6 values in the column "sum of reciprocals", for triangular to octagonal numbers, come from a published solution to the general problem, which also gives a general formula for any number of sides, in terms of the digamma function.[1]

s Name Formula n Sum of reciprocals[1][2] OEIS number
1 2 3 4 5 6 7 8 9 10
2 Natural (line segment) 1/2(0n2 + 2n) = n 1 2 3 4 5 6 7 8 9 10 ∞ (diverges) A000027
3 Triangular 1/2(n2 + n) 1 3 6 10 15 21 28 36 45 55 2[1] A000217
4 Square 1/2(2n2 − 0n)
= n2
1 4 9 16 25 36 49 64 81 100 π2/6[1] A000290
5 Pentagonal 1/2(3n2n) 1 5 12 22 35 51 70 92 117 145 3 ln 3 − π3/3[1] A000326
6 Hexagonal 1/2(4n2 − 2n)
= 2n2 - n
1 6 15 28 45 66 91 120 153 190 2 ln 2[1] A000384
7 Heptagonal 1/2(5n2 − 3n) 1 7 18 34 55 81 112 148 189 235 ${\displaystyle {\begin{matrix}{\tfrac {2}{3}}\ln 5\\+{\tfrac {{1}+{\sqrt {5}}}{3}}\ln {\tfrac {\sqrt {10-2{\sqrt {5}}}}{2}}\\+{\tfrac {{1}-{\sqrt {5}}}{3}}\ln {\tfrac {\sqrt {10+2{\sqrt {5}}}}{2}}\\+{\tfrac {\pi {\sqrt {25-10{\sqrt {5}}}}}{15}}\end{matrix}}}$ [1] A000566
8 Octagonal 1/2(6n2 − 4n)
= 3n2 - 2n
1 8 21 40 65 96 133 176 225 280 3/4 ln 3 + π3/12[1] A000567
9 Nonagonal 1/2(7n2 − 5n) 1 9 24 46 75 111 154 204 261 325 A001106
10 Decagonal 1/2(8n2 − 6n)
= 4n2 - 3n
1 10 27 52 85 126 175 232 297 370 ln 2 + π/6 A001107
11 Hendecagonal 1/2(9n2 − 7n) 1 11 30 58 95 141 196 260 333 415 A051682
12 Dodecagonal 1/2(10n2 − 8n) 1 12 33 64 105 156 217 288 369 460 A051624
13 Tridecagonal 1/2(11n2 − 9n) 1 13 36 70 115 171 238 316 405 505 A051865
14 Tetradecagonal 1/2(12n2 − 10n) 1 14 39 76 125 186 259 344 441 550 2/5 ln 2 + 3/10 ln 3 + π3/10 A051866
15 Pentadecagonal 1/2(13n2 − 11n) 1 15 42 82 135 201 280 372 477 595 A051867
16 Hexadecagonal 1/2(14n2 − 12n) 1 16 45 88 145 216 301 400 513 640 A051868
17 Heptadecagonal 1/2(15n2 − 13n) 1 17 48 94 155 231 322 428 549 685 A051869
18 Octadecagonal 1/2(16n2 − 14n) 1 18 51 100 165 246 343 456 585 730 4/7 ln 2 − 2/14 ln (3 − 22) + π(1 + 2)/14 A051870
19 Enneadecagonal 1/2(17n2 − 15n) 1 19 54 106 175 261 364 484 621 775 A051871
20 Icosagonal 1/2(18n2 − 16n) 1 20 57 112 185 276 385 512 657 820 A051872
21 Icosihenagonal 1/2(19n2 − 17n) 1 21 60 118 195 291 406 540 693 865 A051873
22 Icosidigonal 1/2(20n2 − 18n) 1 22 63 124 205 306 427 568 729 910 A051874
23 Icositrigonal 1/2(21n2 − 19n) 1 23 66 130 215 321 448 596 765 955 A051875
24 Icositetragonal 1/2(22n2 − 20n) 1 24 69 136 225 336 469 624 801 1000 A051876
... ... ... ... ... ... ... ... ... ... ... ... ... ... ...
10000 Myriagonal 1/2(9998n2 − 9996n) 1 10000 29997 59992 99985 149976 209965 279952 359937 449920 A167149

The On-Line Encyclopedia of Integer Sequences eschews terms using Greek prefixes (e.g., "octagonal") in favor of terms using numerals (i.e., "8-gonal").

A property of this table can be expressed by the following identity (see A086270):

${\displaystyle 2\,P(s,n)=P(s+k,n)+P(s-k,n),}$

with

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

## Combinations

Some numbers, such as 36 which is both square and triangular, fall into two polygonal sets. The problem of determining, given two such sets, all numbers that belong to both can be solved by reducing the problem to Pell's equation. The simplest example of this is the sequence of square triangular numbers.

The following table summarizes the set of s-gonal t-gonal numbers for small values of s and t.

s t Sequence OEIS number
4 3 1, 36, 1225, 41616, 1413721, 48024900, 1631432881, 55420693056, 1882672131025, 63955431761796, 2172602007770041, 73804512832419600, 2507180834294496361, 85170343853180456676, 2893284510173841030625, 98286503002057414584576, 3338847817559778254844961, ... A001110
5 3 1, 210, 40755, 7906276, 1533776805, 297544793910, 57722156241751, 11197800766105800, 2172315626468283465, … A014979
5 4 1, 9801, 94109401, 903638458801, 8676736387298001, 83314021887196947001, 799981229484128697805801, ... A036353
6 3 All hexagonal numbers are also triangular. A000384
6 4 1, 1225, 1413721, 1631432881, 1882672131025, 2172602007770041, 2507180834294496361, 2893284510173841030625, 3338847817559778254844961, 3853027488179473932250054441, ... A046177
6 5 1, 40755, 1533776805, … A046180
7 3 1, 55, 121771, 5720653, 12625478965, 593128762435, 1309034909945503, 61496776341083161, 135723357520344181225, 6376108764003055554511, 14072069153115290487843091, … A046194
7 4 1, 81, 5929, 2307361, 168662169, 12328771225, 4797839017609, 350709705290025, 25635978392186449, 9976444135331412025, … A036354
7 5 1, 4347, 16701685, 64167869935, … A048900
7 6 1, 121771, 12625478965, … A048903
8 3 1, 21, 11781, 203841, … A046183
8 4 1, 225, 43681, 8473921, 1643897025, 318907548961, 61866420601441, 12001766689130625, 2328280871270739841, 451674487259834398561, 87622522247536602581025, 16998317641534841066320321, … A036428
8 5 1, 176, 1575425, 234631320, … A046189
8 6 1, 11781, 113123361, … A046192
8 7 1, 297045, 69010153345, … A048906
9 3 1, 325, 82621, 20985481, … A048909
9 4 1, 9, 1089, 8281, 978121, 7436529, 878351769, 6677994961, 788758910641, 5996832038649, 708304623404049, 5385148492712041, 636056763057925561, ... A036411
9 5 1, 651, 180868051, … A048915
9 6 1, 325, 5330229625, … A048918
9 7 1, 26884, 542041975, … A048921
9 8 1, 631125, 286703855361, … A048924

In some cases, such as s = 10 and t = 4, there are no numbers in both sets other than 1.

The problem of finding numbers that belong to three polygonal sets is more difficult. A computer search for pentagonal square triangular numbers has yielded only the trivial value of 1, though a proof that there are no other such numbers has yet to be found.[3]

The number 1225 is hecatonicositetragonal (s = 124), hexacontagonal (s = 60), icosienneagonal (s = 29), hexagonal, square, and triangular.

The only polygonal set that is contained entirely in another polygonal set is the set of hexagonal numbers, which is contained in the set of triangular numbers.[citation needed]

## Notes

1. "Archived copy" (PDF). Archived from the original (PDF) on 2011-06-15. Retrieved 2010-06-13.{{cite web}}: CS1 maint: archived copy as title (link)
2. ^ "Beyond the Basel Problem: Sums of Reciprocals of Figurate Numbers" (PDF). Archived from the original (PDF) on 2013-05-29. Retrieved 2010-05-13.
3. ^