(k,1)-Pascal triangle

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

The (k,1)-Pascal triangle is a geometric arrangement of numbers produced recursively which generates (in its columns, for the rectangular version) 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 (1,k)-Pascal triangle will generate those in its falling interior diagonals starting from the rightmost one.) 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.

The rectangular version of the (k,1)-Pascal triangle

n = 0

1

1

k

1

2

k

$\scriptstyle k+1\,$

1

3

k

$\scriptstyle 2k+1\,$

$\scriptstyle k+2\,$

1

4

k

$\scriptstyle 3k+1\,$

$\scriptstyle 3k+3\,$

$\scriptstyle k+3\,$

1

5

k

$\scriptstyle 4k+1\,$

$\scriptstyle 6k+4\,$

$\scriptstyle 4k+6\,$

$\scriptstyle k+4\,$

1

6

k

$\scriptstyle 5k+1\,$

$\scriptstyle 10k+5\,$

$\scriptstyle 10k+10\,$

$\scriptstyle 5k+10\,$

$\scriptstyle k+5\,$

1

7

k

$\scriptstyle 6k+1\,$

$\scriptstyle 15k+6\,$

$\scriptstyle 20k+15\,$

$\scriptstyle 15k+20\,$

$\scriptstyle 6k+15\,$

$\scriptstyle k+6\,$

1

8

k

$\scriptstyle 7k+1\,$

$\scriptstyle 21k+7\,$

$\scriptstyle 35k+21\,$

$\scriptstyle 35k+35\,$

$\scriptstyle 21k+35\,$

$\scriptstyle 7k+21\,$

$\scriptstyle k+7\,$

1

9

k

$\scriptstyle 8k+1\,$

$\scriptstyle 28k+8\,$

$\scriptstyle 56k+28\,$

$\scriptstyle 70k+56\,$

$\scriptstyle 56k+70\,$

$\scriptstyle 28k+56\,$

$\scriptstyle 8k+28\,$

$\scriptstyle k+8\,$

1

10

k

$\scriptstyle 9k+1\,$

$\scriptstyle 36k+9\,$

$\scriptstyle 84k+36\,$

$\scriptstyle 126k+84\,$

$\scriptstyle 126k+126\,$

$\scriptstyle 84k+126\,$

$\scriptstyle 36k+84\,$

$\scriptstyle 9k+36\,$

$\scriptstyle k+9\,$

1

11

k

$\scriptstyle 10k+1\,$

$\scriptstyle 45k+10\,$

$\scriptstyle 120k+45\,$

$\scriptstyle 210k+120\,$

$\scriptstyle 252k+210\,$

$\scriptstyle 210k+252\,$

$\scriptstyle 120k+210\,$

$\scriptstyle 45k+120\,$

$\scriptstyle 10k+45\,$

$\scriptstyle k+10\,$

1

12

k

$\scriptstyle 11k+1\,$

$\scriptstyle 55k+11\,$

$\scriptstyle 165k+55\,$

$\scriptstyle 330k+165\,$

$\scriptstyle 462k+330\,$

$\scriptstyle 462k+462\,$

$\scriptstyle 330k+462\,$

$\scriptstyle 165k+330\,$

$\scriptstyle 55k+165\,$

$\scriptstyle 11k+55\,$

$\scriptstyle k+11\,$

1

j = 0

1

2

3

4

5

6

7

8

9

10

11

12

In the equilateral version of the (k,1)-Pascal triangle, we start with a cell (row 0) initialized to 1, with all the leftmost nonzero cells in the rows below initialized to k, 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 (k,1)-Pascal triangle, we start with a cell (row 0) initialized to 1, with all the cells below it initialized to k, 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 1+0 = 1. The leftmost nonzero cells on each rows except the first one (for n = 0) are initialized to k. All the interior cells are necessarily greater than k. The number of cells from rows 0 to n which are equal to 1 is n+1. (Cf. A000027(n+1),) the number of cells from rows 0 to n which are equal to k is n (Cf. A000027(n),) and the number of cells from rows 0 to n which are greater than k is $\scriptstyle P_{3}^{(2)}(n-1)\,$, the (n-1)^th triangular number.

Recursion rule

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

$T_{(k,1)}(n,n)=1,\ n\geq 0,\,$

$T_{(k,1)}(n,0)=k,\ n\geq 1,\,$

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

Formulae

Formulae in terms of binomial coefficients

$T_{(k,1)}(0,0)=1,\,$

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

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

Formulae in terms of (hyper)pyramidal numbers

$T_{(k,1)}(0,0)=1,\,$

$T_{(k,1)}(n,d)=Y_{k+d}^{(d)}(n-d+1),\ n+d\geq 1,\,$

where

$Y_{k+d}^{(d)}(1)=1,\ d\geq 1,\,$

$Y_{k}^{(0)}(n)=k,\ n\geq 1,\,$

$Y_{k+d}^{(d)}(m)=\sum _{n=1}^{m}Y_{k+d-1}^{(d-1)}(n),\,$

and where

$Y_{k+1}^{(1)}(n)=P_{k+1}^{(1)}(n),\ n\geq 0,\,$

is the the n^th k-step or (k+2)-gonal gnomonic number, and

$Y_{k+2}^{(2)}(n)=P_{k+2}^{(2)}(n),\ n\geq 0,\,$

is the the n^th (k+2)-gonal number.

(k,1)-Pascal triangle rows

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

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

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

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

(k,1)-Pascal triangle rows sums

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

{1, 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:

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

where:

$0^{0}=1,\,$

$0^{n}=0,\ n\geq 1.\,$

The generating function is:

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

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

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

{1, k-1, 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:

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

where:

$0^{0}=1,\,$

$0^{n}=0,\ n\geq 1.\,$

The generating function is:

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

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

If you look at the (k,1)-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, the index of the j^th falling diagonal (from the right), j ≥ 0, which corresponds to the j^th column, j ≥ 0, of the (1,k)-Pascal triangle. With this convention, n would indicate the number of nondegenerate subfigures of the figurate number.

The rectangular version of the (k,1)-Pascal triangle
(Figurate Number Triangle)

n = 0

1

1

k

1

2

k

$\scriptstyle Y_{k+1}^{(1)}(2)\,$

1

3

k

$\scriptstyle Y_{k+1}^{(1)}(3)\,$

$\scriptstyle Y_{k+2}^{(2)}(2)\,$

1

4

k

$\scriptstyle Y_{k+1}^{(1)}(4)\,$

$\scriptstyle Y_{k+2}^{(2)}(3)\,$

$\scriptstyle Y_{k+3}^{(3)}(2)\,$

1

5

k

$\scriptstyle Y_{k+1}^{(1)}(5)\,$

$\scriptstyle Y_{k+2}^{(2)}(4)\,$

$\scriptstyle Y_{k+3}^{(3)}(3)\,$

$\scriptstyle Y_{k+4}^{(4)}(2)\,$

1

6

k

$\scriptstyle Y_{k+1}^{(1)}(6)\,$

$\scriptstyle Y_{k+2}^{(2)}(5)\,$

$\scriptstyle Y_{k+3}^{(3)}(4)\,$

$\scriptstyle Y_{k+4}^{(4)}(3)\,$

$\scriptstyle Y_{k+5}^{(5)}(2)\,$

1

7

k

$\scriptstyle Y_{k+1}^{(1)}(7)\,$

$\scriptstyle Y_{k+2}^{(2)}(6)\,$

$\scriptstyle Y_{k+3}^{(3)}(5)\,$

$\scriptstyle Y_{k+4}^{(4)}(4)\,$

$\scriptstyle Y_{k+5}^{(5)}(3)\,$

$\scriptstyle Y_{k+6}^{(6)}(2)\,$

1

8

k

$\scriptstyle Y_{k+1}^{(1)}(8)\,$

$\scriptstyle Y_{k+2}^{(2)}(7)\,$

$\scriptstyle Y_{k+3}^{(3)}(6)\,$

$\scriptstyle Y_{k+4}^{(4)}(5)\,$

$\scriptstyle Y_{k+5}^{(5)}(4)\,$

$\scriptstyle Y_{k+6}^{(6)}(3)\,$

$\scriptstyle Y_{k+7}^{(7)}(2)\,$

1

9

k

$\scriptstyle Y_{k+1}^{(1)}(9)\,$

$\scriptstyle Y_{k+2}^{(2)}(8)\,$

$\scriptstyle Y_{k+3}^{(3)}(7)\,$

$\scriptstyle Y_{k+4}^{(4)}(6)\,$

$\scriptstyle Y_{k+5}^{(5)}(5)\,$

$\scriptstyle Y_{k+6}^{(6)}(4)\,$

$\scriptstyle Y_{k+7}^{(7)}(3)\,$

$\scriptstyle Y_{k+8}^{(8)}(2)\,$

1

10

k

$\scriptstyle Y_{k+1}^{(1)}(10)\,$

$\scriptstyle Y_{k+2}^{(2)}(9)\,$

$\scriptstyle Y_{k+3}^{(3)}(8)\,$

$\scriptstyle Y_{k+4}^{(4)}(7)\,$

$\scriptstyle Y_{k+5}^{(5)}(6)\,$

$\scriptstyle Y_{k+6}^{(6)}(5)\,$

$\scriptstyle Y_{k+7}^{(7)}(4)\,$

$\scriptstyle Y_{k+8}^{(8)}(3)\,$

$\scriptstyle Y_{k+9}^{(9)}(2)\,$

1

11

k

$\scriptstyle Y_{k+1}^{(1)}(11)\,$

$\scriptstyle Y_{k+2}^{(2)}(10)\,$

$\scriptstyle Y_{k+3}^{(3)}(9)\,$

$\scriptstyle Y_{k+4}^{(4)}(8)\,$

$\scriptstyle Y_{k+5}^{(5)}(7)\,$

$\scriptstyle Y_{k+6}^{(6)}(6)\,$

$\scriptstyle Y_{k+7}^{(7)}(5)\,$

$\scriptstyle Y_{k+8}^{(8)}(4)\,$

$\scriptstyle Y_{k+9}^{(9)}(3)\,$

$\scriptstyle Y_{k+10}^{(10)}(2)\,$

1

12

k

$\scriptstyle Y_{k+1}^{(1)}(12)\,$

$\scriptstyle Y_{k+2}^{(2)}(11)\,$

$\scriptstyle Y_{k+3}^{(3)}(10)\,$

$\scriptstyle Y_{k+4}^{(4)}(9)\,$

$\scriptstyle Y_{k+5}^{(5)}(8)\,$

$\scriptstyle Y_{k+6}^{(6)}(7)\,$

$\scriptstyle Y_{k+7}^{(7)}(6)\,$

$\scriptstyle Y_{k+8}^{(8)}(5)\,$

$\scriptstyle Y_{k+9}^{(9)}(4)\,$

$\scriptstyle Y_{k+10}^{(10)}(3)\,$

$\scriptstyle Y_{k+11}^{(11)}(2)\,$

1

d = 0

1

2

3

4

5

6

7

8

9

10

11

12

(k,1)-Pascal (rectangular) triangle falling diagonals and (Chebyshev polynomials?)

Table of falling diagonals sequences

The index j, j ≥ 0, refers to the to the j ^th falling diagonal starting from the right. The i ^th member of the j ^th falling diagonal is in row j+i and column i of the triangle.

(k,1)-Pascal triangle falling diagonals sequences

j	$\displaystyle T_{(k,1)}(j+i,i),\ i\geq 0\,$ sequences	OEIS number
0	{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	{$\scriptstyle \ Y_{k+i}^{(i)}(2)|_{i=0}^{\infty }$}
2	{$\scriptstyle \ Y_{k+i}^{(i)}(3)|_{i=0}^{\infty }$}
3	{$\scriptstyle \ Y_{k+i}^{(i)}(4)|_{i=0}^{\infty }$}
4	{$\scriptstyle \ Y_{k+i}^{(i)}(5)|_{i=0}^{\infty }$}
5	{$\scriptstyle \ Y_{k+i}^{(i)}(6)|_{i=0}^{\infty }$}
6	{$\scriptstyle \ Y_{k+i}^{(i)}(7)|_{i=0}^{\infty }$}
7	{$\scriptstyle \ Y_{k+i}^{(i)}(8)|_{i=0}^{\infty }$}
8	{$\scriptstyle \ Y_{k+i}^{(i)}(9)|_{i=0}^{\infty }$}
9	{$\scriptstyle \ Y_{k+i}^{(i)}(10)|_{i=0}^{\infty }$}
10	{$\scriptstyle \ Y_{k+i}^{(i)}(11)|_{i=0}^{\infty }$}
11	{$\scriptstyle \ Y_{k+i}^{(i)}(12)|_{i=0}^{\infty }$}
12	{$\scriptstyle \ Y_{k+i}^{(i)}(13)|_{i=0}^{\infty }$}

Table of falling diagonals sequences related formulae

The index j, j ≥ 0, refers to the to the j ^th falling diagonal starting from the right. The i ^th member of the j ^th falling diagonal is in row j+i and column i of the triangle.

(k,1)-Pascal triangle falling diagonals sequences related formulae

j	Formulae $T_{(k,1)}(j+i,i)=\,$ ${\binom {i+j-1}{j-1}}{\frac {(i+kj)}{j}}\,$	Generating function for i th (i ≥ 0) member of falling diagonal $G_{T_{(k,1)}}(j,x)=\,$ ${\frac {k-(k-1)x}{(1-x)^{j+1}}}\,$	Order of basis $g_{T_{(k,1)}}(j)=\,$	Differences $T_{(k,1)}(j+i,i)-\,$ $T_{(k,1)}(j+i-1,i)=\,$	Partial sums $\sum _{i=0}^{m}T_{(k,1)}(j+i,i)=\,$	Partial sums of reciprocals $\sum _{i=0}^{m}{1 \over {T_{(k,1)}(j+i,i)}}=$	Sum of Reciprocals $\sum _{i=0}^{\infty }{1 \over {T_{(k,1)}(j+i,i)}}=$
0	$\,$	$\displaystyle {\frac {k-(k-1)x}{(1-x)}}\,$	$\,$	$\,$	$\,$	$\,$	$\,$
1	$\displaystyle {\binom {i+0}{0}}{\frac {(i+k)}{1}}\,$ $\displaystyle {\frac {(i+k)}{1}}\,$	$\displaystyle {\frac {k-(k-1)x}{(1-x)^{2}}}\,$	$\,$	$\,$	$\,$	$\,$	$\,$
2	$\displaystyle {\binom {i+1}{1}}{\frac {(i+2k)}{2}}\,$ $\displaystyle {\frac {(i+1)(i+2k)}{2}}\,$	$\displaystyle {\frac {k-(k-1)x}{(1-x)^{3}}}\,$	$\,$	$\,$	$\,$	$\,$	$\,$
3	$\displaystyle {\binom {i+2}{2}}{\frac {(i+3k)}{3}}\,$ $\textstyle {\frac {(i+1)(i+2)(i+3k)}{6}}\,$	$\displaystyle {\frac {k-(k-1)x}{(1-x)^{4}}}\,$	$\,$	$\,$	$\,$	$\,$	$\,$
4	$\displaystyle {\binom {i+3}{3}}{\frac {(i+4k)}{4}}\,$	$\displaystyle {\frac {k-(k-1)x}{(1-x)^{5}}}\,$	$\,$	$\,$	$\,$	$\,$	$\,$
5	$\displaystyle {\binom {i+4}{4}}{\frac {(i+5k)}{5}}\,$	$\displaystyle {\frac {k-(k-1)x}{(1-x)^{6}}}\,$	$\,$	$\,$	$\,$	$\,$	$\,$
6	$\displaystyle {\binom {i+5}{5}}{\frac {(i+6k)}{6}}\,$	$\displaystyle {\frac {k-(k-1)x}{(1-x)^{7}}}\,$	$\,$	$\,$	$\,$	$\,$	$\,$
7	$\displaystyle {\binom {i+6}{6}}{\frac {(i+7k)}{7}}\,$	$\displaystyle {\frac {k-(k-1)x}{(1-x)^{8}}}\,$	$\,$	$\,$	$\,$	$\,$	$\,$
8	$\displaystyle {\binom {i+7}{7}}{\frac {(i+8k)}{8}}\,$	$\displaystyle {\frac {k-(k-1)x}{(1-x)^{9}}}\,$	$\,$	$\,$	$\,$	$\,$	$\,$
9	$\displaystyle {\binom {i+8}{8}}{\frac {(i+9k)}{9}}\,$	$\displaystyle {\frac {k-(k-1)x}{(1-x)^{10}}}\,$	$\,$	$\,$	$\,$	$\,$	$\,$
10	$\displaystyle {\binom {i+9}{9}}{\frac {(i+10k)}{10}}\,$	$\displaystyle {\frac {k-(k-1)x}{(1-x)^{11}}}\,$	$\,$	$\,$	$\,$	$\,$	$\,$
11	$\displaystyle {\binom {i+10}{10}}{\frac {(i+11k)}{11}}\,$	$\displaystyle {\frac {k-(k-1)x}{(1-x)^{12}}}\,$	$\,$	$\,$	$\,$	$\,$	$\,$
12	$\displaystyle {\binom {i+11}{11}}{\frac {(i+12k)}{12}}\,$	$\displaystyle {\frac {k-(k-1)x}{(1-x)^{13}}}\,$	$\,$	$\,$	$\,$	$\,$	$\,$

(k,1)-Pascal triangle central elements

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

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

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:

$T_{(k,1)}(0,0)=1\,$

$T_{(k,1)}(2m,m)=(k+1){\frac {T_{(1,1)}(2m,m)}{2}}=(k+1)(m+1){\frac {C_{m}}{2}},\ m\geq 1,\,$

where:

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

is the m^th, 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:

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

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

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

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

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

See also

• (1,1)-Pascal triangle or Pascal's triangle
• (1,2)-Pascal triangle or Lucas triangle
• (1,k)-Pascal triangle
• (2,1)-Pascal triangle
• (k,1)-Pascal triangle
• (a,b)-Pascal triangle
• (a(n),b(n))-Pascal triangle

External links

• S. Plouffe, Approximations de Séries Génératrices et Quelques Conjectures, Dissertation, Université du Québec à Montréal, 1992.
• S. Plouffe, 1031 Generating Functions and Conjectures, Université du Québec à Montréal, 1992.
• Herbert S. Wilf, generatingfunctionology, 2^nd ed., 1994.