This site is supported by donations to The OEIS Foundation.

# (1,k)-Pascal triangle

The (1,k)-Pascal triangle has its rightmost nonzero entries initialized to k and its leftmost nonzero entries (except the first row for n = 0) initialized to 1. Thus the rows of the (1,k)-Pascal triangle are the left-right reversal of the rows of the (k,1)-Pascal triangle, with the exception of the first row (for n = 0) which is now k instead of 1.

The (1,k)-Pascal triangle is a geometric arrangement of numbers produced recursively which generates (in its falling interior diagonals starting from the rightmost one) the (k+2)-gonal gnomonic numbers, the (k+2)-gonal numbers, the (k+2)-gonal pyramidal numbers and then the (k+2)-gonal hyperpyramidal numbers for dimension greather than 3 (The (k,1)-Pascal triangle , for the rectangular version, will generate those in its columns.) The original Pascal triangle, which is thus the (1,1)-Pascal triangle, generates (both in its columns for the rectangular version and in its falling interior diagonals starting from the rightmost one, since the (1,1)-Pascal triangle is symmetrical) the triangular gnomonic numbers (the natural numbers,) the triangular numbers, the tetrahedral numbers (the triangular pyramidal numbers) and then the hypertetrahedral numbers (the triangular hyperpyramidal numbers) for dimension greather than 3.

 n = 0 k 1 1 k 2 1 $\scriptstyle 1+k\,$ k 3 1 $\scriptstyle 2+k\,$ $\scriptstyle 1+2k\,$ k 4 1 $\scriptstyle 3+k\,$ $\scriptstyle 3+3k\,$ $\scriptstyle 1+3k\,$ k 5 1 $\scriptstyle 4+k\,$ $\scriptstyle 6+4k\,$ $\scriptstyle 4+6k\,$ $\scriptstyle 1+4k\,$ k 6 1 $\scriptstyle 5+k\,$ $\scriptstyle 10+5k\,$ $\scriptstyle 10+10k\,$ $\scriptstyle 5+10k\,$ $\scriptstyle 1+5k\,$ k 7 1 $\scriptstyle 6+k\,$ $\scriptstyle 15+6k\,$ $\scriptstyle 20+15k\,$ $\scriptstyle 15+20k\,$ $\scriptstyle 6+15k\,$ $\scriptstyle 1+6k\,$ k 8 1 $\scriptstyle 7+k\,$ $\scriptstyle 21+7k\,$ $\scriptstyle 35+21k\,$ $\scriptstyle 35+35k\,$ $\scriptstyle 21+35k\,$ $\scriptstyle 7+21k\,$ $\scriptstyle 1+7k\,$ k 9 1 $\scriptstyle 8+k\,$ $\scriptstyle 28+8k\,$ $\scriptstyle 56+28k\,$ $\scriptstyle 70+56k\,$ $\scriptstyle 56+70k\,$ $\scriptstyle 28+56k\,$ $\scriptstyle 8+28k\,$ $\scriptstyle 1+8k\,$ k 10 1 $\scriptstyle 9+k\,$ $\scriptstyle 36+9k\,$ $\scriptstyle 84+36k\,$ $\scriptstyle 126+84k\,$ $\scriptstyle 126+126k\,$ $\scriptstyle 84+126k\,$ $\scriptstyle 36+84k\,$ $\scriptstyle 9+36k\,$ $\scriptstyle 1+9k\,$ k 11 1 $\scriptstyle 10+k\,$ $\scriptstyle 45+10k\,$ $\scriptstyle 120+45k\,$ $\scriptstyle 210+120k\,$ $\scriptstyle 252+210k\,$ $\scriptstyle 210+252k\,$ $\scriptstyle 120+210k\,$ $\scriptstyle 45+120k\,$ $\scriptstyle 10+45k\,$ $\scriptstyle 1+10k\,$ k 12 1 $\scriptstyle 11+k\,$ $\scriptstyle 55+11k\,$ $\scriptstyle 165+55k\,$ $\scriptstyle 330+165k\,$ $\scriptstyle 462+330k\,$ $\scriptstyle 462+462k\,$ $\scriptstyle 330+462k\,$ $\scriptstyle 165+330k\,$ $\scriptstyle 55+165k\,$ $\scriptstyle 11+55k\,$ $\scriptstyle 1+11k\,$ k j = 0 1 2 3 4 5 6 7 8 9 10 11 12

In the equilateral version of the (1,k)-Pascal triangle, we start with a cell (row 0) initialized to k, with the leftmost nonzero cell in the row below initialized to 1, in a staggered array of empty (0) cells. We then recursively evaluate the cells as the sum of the two cells staggered above. The triangle thus grows into an equilateral triangle.

In the rectangular version of the (1,k)-Pascal triangle, we start with a cell (row 0) initialized to k, with the cell below it initialized to 1, in a regular array of empty (0) cells. We then recursively evaluate the cells as the sum of the one above left and the one directly above. The triangle thus grows into a rectangular triangle.

The rightmost nonzero cells on each rows are therefore set to k and the leftmost nonzero cells on each rows except the first one (for n = 0) are set to 1. All the interior cells are necessarily greater than or equal to k+1. The number of cells from rows 0 to n which are equal to 1 is n. (Cf. A001477(n),) the number of cells from rows 0 to n which are equal to k is n+1 (Cf. A001477(n+1),) and the number of cells from rows 0 to n which are greater than or equal to k+1 is $\scriptstyle P^{(2)}_{3}(n-1)\,$, the (n-1)th triangular number.

## Recursion rule

The (1,k)-Pascal triangle recursion rule is:

$T_{(1,k)}(n, n) = k,\ n \ge 0,\,$
$T_{(1,k)}(n, 0) = 1,\ n \ge 1,\,$
$T_{(1,k)}(n, j) = T_{(1,k)}(n - 1, j - 1) + T_{(1,k)}(n - 1, j),\ 0 < j < n.\,$

## Formulae

### Formulae in terms of binomial coefficients

$T_{(1,k)}(0, 0) = k,\,$
$T_{(1,k)}(n, j) = k \binom{n-1}{j-1} + \binom{n-1}{j} = k T_{(1,1)}(n-1, j-1) + T_{(1,1)}(n-1, j),\ n \ge 1,\,$

where $\scriptstyle \binom{n}{r} \equiv 0\,$ when n < 0, r < 0 or n - r < 0,[1] and $\scriptstyle T_{(1,1)}(n, j)\,$ is cell (n, j) of Pascal's triangle.

### Formulae in terms of (hyper)pyramidal numbers

$T_{(1,k)}(0, 0) = k,\,$
$T_{(1,k)}(n, j) = Y^{(n-j)}_{k+(n-j)}(j+1),\ n+j \ge 1\,$ [2]

where

$Y^{(d)}_{k+d}(1) = 1,\ d \ge 1,\,$
$Y^{(0)}_{k}(n) = k,\ n \ge 1,\,$
$Y^{(d)}_{k+d}(m) = \sum_{n=1}^{m} Y^{(d-1)}_{k+d-1}(n),\,$ [2]

and where

$Y^{(1)}_{k+1}(n) = P^{(1)}_{k+1}(n),\ n \ge 0,\,$

is the the nth k-step or (k+2)-gonal gnomonic number, and

$Y^{(2)}_{k+2}(n) = P^{(2)}_{k+2}(n),\ n \ge 0,\,$

is the the nth (k+2)-gonal number.

## (1,k)-Pascal triangle rows

The (1,k)-Pascal triangle rows give an infinite sequence of finite sequences:

{{$\scriptstyle k\,$}, {$\scriptstyle k \binom{0}{-1} + \binom{0}{0},\ k \binom{0}{0} + \binom{0}{1}\,$}, {$\scriptstyle k \binom{1}{-1} + \binom{1}{0},\ k \binom{1}{0} + \binom{1}{1},\ k \binom{1}{1} + \binom{1}{2}\,$}, {$\scriptstyle k \binom{2}{-1} + \binom{2}{0},\ k \binom{2}{0} + \binom{2}{1},\ k \binom{2}{1} + \binom{2}{2},\ k \binom{2}{2} + \binom{2}{3}\,$}, ...}

The concatenation of the infinite sequence of finite sequences gives the infinite sequence:

{$\scriptstyle k,\ k \binom{0}{-1} + \binom{0}{0},\ k \binom{0}{0} + \binom{0}{1},\ k \binom{1}{-1} + \binom{1}{0},\ k \binom{1}{0} + \binom{1}{1},\ k \binom{1}{1} + \binom{1}{2},\ k \binom{2}{-1} + \binom{2}{0},\ k \binom{2}{0} + \binom{2}{1},\ k \binom{2}{1} + \binom{2}{2},\ k \binom{2}{2} + \binom{2}{3}, ...\,$}

### (1,k)-Pascal triangle rows sums

The sums of the respective finite sequences give the infinite sequence:

{k, k+1, 2(k+1), 4(k+1), 8(k+1), 16(k+1), 32(k+1), 64(k+1), ...}

with members given by the formula:

$\sum_{j=0}^{n} T_{(1,k)}(n, j) = \frac{(k+1) 2^n + (k-1) 0^n}{2},\,$

where:

$0^0 = 1,\,$
$0^n = 0,\ n \ge 1.\,$

The generating function is:

$G_{\{\sum_{j=0}^{n} T_{(1,k)}(n, j)\}} = \frac{(k+1) G_{2^n}(x) + (k-1) G_{0^n}(x) }{2} = \frac{(k+1) (\frac{1}{1-2x}) + (k-1) (1) }{2} = \frac{k + (1-k)x}{1-2x}\,$

### (1,k)-Pascal triangle rows alternating sign sums

The alternating sign sums of the respective finite sequences give the infinite sequence:

{k, 1-k, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...}

with members given by the formula:

$\sum_{j=0}^{n} (-1)^j T_{(1,k)}(n, j) = k\ 0^{|n|} + (1-k)\ 0^{|n-1|},\,$

where:

$0^0 = 1,\,$
$0^n = 0,\ n \ge 1.\,$

The generating function is:

$G_{\{\sum_{j=0}^{n} (-1)^j T_{(1,k)}(n, j)\}} = k\ G_{0^{|n|}}(x) + (1-k)\ G_{0^{|n-1|}} = k\ (1) + (1-k)\ (x) = k+(1-k)x\,$

## (1,k)-Pascal (rectangular) triangle columns (and Chebyshev polynomials?)

### Table of columns sequences

The i th, i ≥ 0, member of column j appears in row j+i.

(1,k)-Pascal triangle columns sequences
j $T_{(1,k)}(j+i, j),\ i \ge 0\,$ sequences OEIS

number

0 {k, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...}
1 {$\scriptstyle Y^{(i)}_{k+i}(2)|_{i=0}^{\infty}\,$}
2 {$\scriptstyle Y^{(i)}_{k+i}(3)|_{i=0}^{\infty}\,$}
3 {$\scriptstyle Y^{(i)}_{k+i}(4)|_{i=0}^{\infty}\,$}
4 {$\scriptstyle Y^{(i)}_{k+i}(5)|_{i=0}^{\infty}\,$}
5 {$\scriptstyle Y^{(i)}_{k+i}(6)|_{i=0}^{\infty}\,$}
6 {$\scriptstyle Y^{(i)}_{k+i}(7)|_{i=0}^{\infty}\,$}
7 {$\scriptstyle Y^{(i)}_{k+i}(8)|_{i=0}^{\infty}\,$}
8 {$\scriptstyle Y^{(i)}_{k+i}(9)|_{i=0}^{\infty}\,$}
9 {$\scriptstyle Y^{(i)}_{k+i}(10)|_{i=0}^{\infty}\,$}
10 {$\scriptstyle Y^{(i)}_{k+i}(11)|_{i=0}^{\infty}\,$}
11 {$\scriptstyle Y^{(i)}_{k+i}(12)|_{i=0}^{\infty}\,$}
12 {$\scriptstyle Y^{(i)}_{k+i}(13)|_{i=0}^{\infty}\,$}

### Table of columns sequences related formulae

The i th, i ≥ 0, member of column j appears in row j+i.

(1,k)-Pascal triangle columns sequences related formulae
j Formulae

$T_{(1,k)}(j+i, j) =\,$

$\binom{i+j-1}{j-1} \frac{(i+kj)}{j},\,$

$j \ge 1\,$

Generating

function

for i th (i ≥ 0)

member of column

$G_{T_{(1,k)}}(x, j) =\,$

$\frac{k+(1-k)x}{(1-x)^{j+1}}\,$

Order

of basis

$g_{T_{(1,k)}}(j) =\,$

Differences

$T_{(1,k)}(j+i, j) - \,$

$T_{(1,k)}(j+i-1, j) =\,$

Partial sums

$\sum_{i=0}^m T_{(1,k)}(j+i, j) =\,$

Partial sums of reciprocals

$\sum_{i=0}^m {1\over{T_{(1,k)}(j+i, j)}} =$

Sum of Reciprocals[3][4]

$\sum_{i=0}^\infty{1\over{T_{(1,k)}(j+i, j)}} =$

0 $\,$ $\displaystyle \frac{k+(1-k)x}{(1-x)}\,$ $\infty\,$ $\,$ $\,$ $\,$ $\,$
1 $\displaystyle \binom{i+0}{0} \frac{(i+k)}{1}\,$

$\textstyle \frac{(i+k)}{1}\,$

$\displaystyle \frac{k+(1-k)x}{(1-x)^2}\,$ $1\,$

(for $\mathbb{N}_{\ge k}\,$)

$\,$ $\,$ $\,$ $\,$
2 $\displaystyle \binom{i+1}{1} \frac{(i+2k)}{2}\,$

$\textstyle \frac{(i+1)(i+2k)}{2}\,$

$\displaystyle \frac{k+(1-k)x}{(1-x)^3}\,$ $\,$ $\,$ $\,$ $\,$ $\,$
3 $\displaystyle \binom{i+2}{2} \frac{(i+3k)}{3}\,$

$\textstyle \frac{(i+1)(i+2)(i+3k)}{6}\,$

$\displaystyle \frac{k+(1-k)x}{(1-x)^4}\,$ $\,$ $\,$ $\,$ $\,$ $\,$
4 $\displaystyle \binom{i+3}{3} \frac{(i+4k)}{4}\,$ $\displaystyle \frac{k+(1-k)x}{(1-x)^5}\,$ $\,$ $\,$ $\,$ $\,$ $\,$
5 $\displaystyle \binom{i+4}{4} \frac{(i+5k)}{5}\,$ $\displaystyle \frac{k+(1-k)x}{(1-x)^6}\,$ $\,$ $\,$ $\,$ $\,$ $\,$
6 $\displaystyle \binom{i+5}{5} \frac{(i+6k)}{6}\,$ $\displaystyle \frac{k+(1-k)x}{(1-x)^7}\,$ $\,$ $\,$ $\,$ $\,$ $\,$
7 $\displaystyle \binom{i+6}{6} \frac{(i+7k)}{7}\,$ $\displaystyle \frac{k+(1-k)x}{(1-x)^8}\,$ $\,$ $\,$ $\,$ $\,$ $\,$
8 $\displaystyle \binom{i+7}{7} \frac{(i+8k)}{8}\,$ $\displaystyle \frac{k+(1-k)x}{(1-x)^9}\,$ $\,$ $\,$ $\,$ $\,$ $\,$
9 $\displaystyle \binom{i+8}{8} \frac{(i+9k)}{9}\,$ $\displaystyle \frac{k+(1-k)x}{(1-x)^{10}}\,$ $\,$ $\,$ $\,$ $\,$ $\,$
10 $\displaystyle \binom{i+9}{9} \frac{(i+10k)}{10}\,$ $\displaystyle \frac{k+(1-k)x}{(1-x)^{11}}\,$ $\,$ $\,$ $\,$ $\,$ $\,$
11 $\displaystyle \binom{i+10}{10} \frac{(i+11k)}{11}\,$ $\displaystyle \frac{k+(1-k)x}{(1-x)^{12}}\,$ $\,$ $\,$ $\,$ $\,$ $\,$
12 $\displaystyle \binom{i+11}{11} \frac{(i+12k)}{12}\,$ $\displaystyle \frac{k+(1-k)x}{(1-x)^{13}}\,$ $\,$ $\,$ $\,$ $\,$ $\,$

## (1,k)-Pascal (rectangular) triangle falling diagonals and (k+2)-gonal (hyper)pyramidal numbers

If you look at the (1,k)-Pascal triangle, it seems that the convention giving the most symmetry in the triangle would be to have 1 (for the initial dot) for n = 0, so n would coincide with j. With this convention, n would indicate the number of nondegenerate subfigures of the figurate number.

 n = 0 k 1 1 [6] k 2 1 [7] $\scriptstyle Y^{(1)}_{k+1}(2)\,$ k 3 1 [2] $\scriptstyle Y^{(2)}_{k+2}(2)\,$ $\scriptstyle Y^{(1)}_{k+1}(3)\,$ k 4 1 $\scriptstyle Y^{(3)}_{k+3}(2)\,$ $\scriptstyle Y^{(2)}_{k+2}(3)\,$ $\scriptstyle Y^{(1)}_{k+1}(4)\,$ k 5 1 $\scriptstyle Y^{(4)}_{k+4}(2)\,$ $\scriptstyle Y^{(3)}_{k+3}(3)\,$ $\scriptstyle Y^{(2)}_{k+2}(4)\,$ $\scriptstyle Y^{(1)}_{k+1}(5)\,$ k 6 1 $\scriptstyle Y^{(5)}_{k+5}(2)\,$ $\scriptstyle Y^{(4)}_{k+4}(3)\,$ $\scriptstyle Y^{(3)}_{k+3}(4)\,$ $\scriptstyle Y^{(2)}_{k+2}(5)\,$ $\scriptstyle Y^{(1)}_{k+1}(6)\,$ k 7 1 $\scriptstyle Y^{(6)}_{k+6}(2)\,$ $\scriptstyle Y^{(5)}_{k+5}(3)\,$ $\scriptstyle Y^{(4)}_{k+4}(4)\,$ $\scriptstyle Y^{(3)}_{k+3}(5)\,$ $\scriptstyle Y^{(2)}_{k+2}(6)\,$ $\scriptstyle Y^{(1)}_{k+1}(7)\,$ k 8 1 $\scriptstyle Y^{(7)}_{k+7}(2)\,$ $\scriptstyle Y^{(6)}_{k+6}(3)\,$ $\scriptstyle Y^{(5)}_{k+5}(4)\,$ $\scriptstyle Y^{(4)}_{k+4}(5)\,$ $\scriptstyle Y^{(3)}_{k+3}(6)\,$ $\scriptstyle Y^{(2)}_{k+2}(7)\,$ $\scriptstyle Y^{(1)}_{k+1}(8)\,$ k 9 1 $\scriptstyle Y^{(8)}_{k+8}(2)\,$ $\scriptstyle Y^{(7)}_{k+7}(3)\,$ $\scriptstyle Y^{(6)}_{k+6}(4)\,$ $\scriptstyle Y^{(5)}_{k+5}(5)\,$ $\scriptstyle Y^{(4)}_{k+4}(6)\,$ $\scriptstyle Y^{(3)}_{k+3}(7)\,$ $\scriptstyle Y^{(2)}_{k+2}(8)\,$ $\scriptstyle Y^{(1)}_{k+1}(9)\,$ k 10 1 $\scriptstyle Y^{(9)}_{k+9}(2)\,$ $\scriptstyle Y^{(8)}_{k+8}(3)\,$ $\scriptstyle Y^{(7)}_{k+7}(4)\,$ $\scriptstyle Y^{(6)}_{k+6}(5)\,$ $\scriptstyle Y^{(5)}_{k+5}(6)\,$ $\scriptstyle Y^{(4)}_{k+4}(7)\,$ $\scriptstyle Y^{(3)}_{k+3}(8)\,$ $\scriptstyle Y^{(2)}_{k+2}(9)\,$ $\scriptstyle Y^{(1)}_{k+1}(10)\,$ k 11 1 $\scriptstyle Y^{(10)}_{k+10}(2)\,$ $\scriptstyle Y^{(9)}_{k+9}(3)\,$ $\scriptstyle Y^{(8)}_{k+8}(4)\,$ $\scriptstyle Y^{(7)}_{k+7}(5)\,$ $\scriptstyle Y^{(6)}_{k+6}(6)\,$ $\scriptstyle Y^{(5)}_{k+5}(7)\,$ $\scriptstyle Y^{(4)}_{k+4}(8)\,$ $\scriptstyle Y^{(3)}_{k+3}(9)\,$ $\scriptstyle Y^{(2)}_{k+2}(10)\,$ $\scriptstyle Y^{(1)}_{k+1}(11)\,$ k 12 1 $\scriptstyle Y^{(11)}_{k+11}(2)\,$ $\scriptstyle Y^{(10)}_{k+10}(3)\,$ $\scriptstyle Y^{(9)}_{k+9}(4)\,$ $\scriptstyle Y^{(8)}_{k+8}(5)\,$ $\scriptstyle Y^{(7)}_{k+7}(6)\,$ $\scriptstyle Y^{(6)}_{k+6}(7)\,$ $\scriptstyle Y^{(5)}_{k+5}(8)\,$ $\scriptstyle Y^{(4)}_{k+4}(9)\,$ $\scriptstyle Y^{(3)}_{k+3}(10)\,$ $\scriptstyle Y^{(2)}_{k+2}(11)\,$ $\scriptstyle Y^{(1)}_{k+1}(12)\,$ k j = 0 1 2 3 4 5 6 7 8 9 10 11 12

## (1,k)-Pascal triangle central elements

The central elements (for row 2m, m ≥ 0) of the (1,k)-Pascal triangle give the sequence:

$\{k, 1+k, 3(1+k), 10(1+k), 35(1+k), 126(1+k), 462(1+k), 1716(1+k), 6435(1+k), \ldots\} = \{k, (1+k)\ \tfrac{A000984(m)}{2}|_{m=1}^{\infty}\},\,$

where A000984(n), n ≥ 0, are the central binomial coefficients, $\scriptstyle C(2n,n) = \frac{(2n)!}{(n!)^2}\,$:

{1, 2, 6, 20, 70, 252, 924, 3432, 12870, 48620, 184756, 705432, 2704156, 10400600, 40116600, 155117520, 601080390, 2333606220, 9075135300, 35345263800, ...}

They central elements are given by the formulae:

$T_{(1,k)}(0, 0) = k\,$
$T_{(1,k)}(2m, m) = (1+k) \frac{T_{(1,1)}(2m, m)}{2} = (1+k) (m+1) \frac{C_m}{2},\ m \ge 1,\,$

where:

$C_m = \frac{T_{(1,1)}(2m, m)}{m+1} = \frac{\binom{2m}{m}}{m+1} = \frac{(2m)!}{m!(m+1)!}\,$

is the mth, m ≥ 0, Catalan number (also called Segner numbers) (Cf. A000108(m)):

{1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700, 1767263190, 6564120420, ...}

The generating function, for m ≥ 1, is:

$G_{\{T_{(1,k)}(2m, m)\}}(x) = \frac{k+1}{2}\{G_{\{m\ C_m\}}(x) + G_{\{C_m\}}(x)\} = \frac{k+1}{2}\{xC'(x) + C(x)\} = \frac{k+1}{2} \bigg\{\frac{1}{1-2xC(x)}\bigg\} = (k+1) \frac{1}{2\sqrt{1-4x}},\ m \ge 1,\,$

and the generating function, for m ≥ 0, is:

$G_{\{T_{(1,k)}(2m, m)\}}(x) = k + (k+1) \frac{1}{2} \bigg(\frac{1}{\sqrt{1-4x}}-1\bigg),\ m \ge 0\,$

where $\scriptstyle C(x)\,$ is the generating function of the Catalan numbers:

$C(x) = G_{\{C_m\}}(x) = \frac{1 - \sqrt{1 - 4x}}{2x},\,$

and

$C'(x) = \frac{dC(x)}{dx} = \frac{1}{x \sqrt{1-4x}} - \frac{1-\sqrt{1-4x}}{2 x^2}\,$

## Notes

1. Weisstein, Eric W., Binomial Coefficient, From MathWorld--A Wolfram Web Resource.
2. 2.0 2.1 2.2 Where $\scriptstyle Y^{(d)}_{[(k+2)+(d-2)]}(n) = Y^{(d)}_{k+d}(n)\,$, k ≥ 1, n ≥ 0, is the d-dimensional, d ≥ 0, (k+2)-gonal base (hyper)pyramidal number where, for d ≥ 2, $\scriptstyle N_0 = [(k+2)+(d-2)]\,$ is the number of vertices (including the $\scriptstyle d-2\,$ apex vertices) of the polygonal base (hyper)pyramid.
3. Downey, Lawrence M., Ong, Boon W., and Sellers, James A., Beyond the Basel Problem: Sums of Reciprocals of Figurate Numbers, 2008.
4. PSYCHEDELIC GEOMETRY, INVERSE POLYGONAL NUMBERS SERIES.
5. Weisstein, Eric W., Figurate Number Triangle, From MathWorld--A Wolfram Web Resource.
6. Where $\scriptstyle Y^{(1)}_{k+1}(n) = P^{(1)}_{k+1}(n)\,$, k ≥ 1, n ≥ 0, is the nth k-step or (k+2)-gonal gnomonic number.
7. Where $\scriptstyle Y^{(2)}_{k+2}(n) = P^{(2)}_{k+2}(n)\,$, k ≥ 1, n ≥ 0, is the nth (k+2)-gonal number.