(2,1)-Pascal triangle

The (2,1)-Pascal triangle has its rightmost nonzero entries initialized to 1 and its leftmost nonzero entries (except the first row for n = 0) initialized to 2. Thus the rows of the (2,1)-Pascal triangle are the left-right reversal of the rows of the (1,2)-Pascal triangle, with the exception of the first row (for n = 0) which is now 1 instead of 2.

The (2,1)-Pascal triangle is a geometric arrangement of numbers produced recursively which generates (in its columns, for the rectangular version) the square gnomonic numbers (the odd numbers,) the square numbers, the square pyramidal numbers and then the square hyperpyramidal numbers for dimension greather than 3 (The (1,2)-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 (2,1)-Pascal triangle
(Figurate Number Triangle)

n = 0	1
1	2	1
2	2	3	1
3	2	5	4	1
4	2	7	9	5	1
5	2	9	16	14	6	1
6	2	11	25	30	20	7	1
7	2	13	36	55	50	27	8	1
8	2	15	49	91	105	77	35	9	1
9	2	17	64	140	196	182	112	44	10	1
10	2	19	81	204	336	378	294	156	54	11	1
11	2	21	100	285	540	714	672	450	210	65	12	1
12	2	23	121	385	825	1254	1386	1122	660	275	77	13	1
	d = 0	1	2	3	4	5	6	7	8	9	10	11	12

(2+x) (1+x)^(n-1)
or
"The rectangular version of the (2,1)-Pascal polynomials triangle"

n = 0	1
1	2 +	1x
2	2 +	3x +	1x 2
3	2 +	5x +	4x 2 +	1x 3
4	2 +	7x +	9x 2 +	5x 3 +	1x 4
5	2 +	9x +	16x 2 +	14x 3 +	6x 4 +	1x 5
6	2 +	11x +	25x 2 +	30x 3 +	20x 4 +	7x 5 +	1x 6
7	2 +	13x +	36x 2 +	55x 3 +	50x 4 +	27x 5 +	8x 6 +	1x 7
8	2 +	15x +	49x 2 +	91x 3 +	105x 4 +	77x 5 +	35x 6 +	9x 7 +	1x 8
9	2 +	17x +	64x 2 +	140x 3 +	196x 4 +	182x 5 +	112x 6 +	44x 7 +	10x 8 +	1x 9
10	2 +	19x +	81x 2 +	204x 3 +	336x 4 +	378x 5 +	294x 6 +	156x 7 +	54x 8 +	11x 9 +	1x 10
	d = 0	1	2	3	4	5	6	7	8	9	10

In the equilateral version of the (2,1)-Pascal triangle, we start with a cell (row 0) initialized to 1, with all the leftmost nonzero cells in the rows below initialized to 2, 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 (2,1)-Pascal triangle, we start with a cell (row 0) initialized to 1, with all the cells below it initialized to 2, 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 2. All the interior cells are necessarily greater than 2. The number of cells from rows 0 to n which are equal to 1 is n+1. (Cf. A001477(n+1),) the number of cells from rows 0 to n which are equal to 2 is n (Cf. A001477(n),) and the number of cells from rows 0 to n which are greather than or equal to 3 is $\scriptstyle P_{3}^{(2)}(n-1)\,$, the (n-1)^th triangular number.

Recursion rule

The (2,1)-Pascal triangle recursion rule is:

$T_{(2,1)}(n,n)=1,\ n\geq 0,\,$

$T_{(2,1)}(n,0)=2,\ n\geq 1,\,$

$T_{(2,1)}(n,d)=T_{(2,1)}(n-1,d-1)+T_{(2,1)}(n-1,d),\ 0<d<n.\,$

Formulae

$T_{(2,1)}(0,0)=1,\,$

$T_{(2,1)}(n,j)={\binom {n-1}{j-1}}+2{\binom {n-1}{j}}=T_{(1,1)}(n-1,j-1)+2T_{(1,1)}(n-1,j),\ n\geq 1,\,$

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.

(2,1)-Pascal triangle rows

n = 0	1
1	2	1
2	2	3	1
3	2	5	4	1
4	2	7	9	5	1
5	2	9	16	14	6	1
6	2	11	25	30	20	7	1
7	2	13	36	55	50	27	8	1
8	2	15	49	91	105	77	35	9	1
9	2	17	64	140	196	182	112	44	10	1
10	2	19	81	204	336	378	294	156	54	11	1
11	2	21	100	285	540	714	672	450	210	65	12	1
12	2	23	121	385	825	1254	1386	1122	660	275	77	13	1
	d = 0	1	2	3	4	5	6	7	8	9	10	11	12

The (2,1)-Pascal triangle rows give an infinite sequence of finite sequences:

{{1}, {2, 1}, {2, 3, 1}, {2, 5, 4, 1}, {2, 7, 9, 5, 1}, {2, 9, 16, 14, 6, 1}, {2, 11, 25, 30, 20, 7, 1}, {2, 13, 36, 55, 50, 27, 8, 1}, {2, 15, 49, 91, 105, 77, 35, 9, 1}...}

The generating function for the d^th, d ≥ 0, member of the n^th, n ≥ 1, subsequence is:

$G_{T_{(2,1)}(n,d)}(x)=(1+x)^{n}+(1+x)^{n-1}=(2+x)(1+x)^{n-1}\,$

The concatenation of the infinite sequence of finite sequences gives the infinite sequence (cf. A029653(n)):

{1, 2, 1, 2, 3, 1, 2, 5, 4, 1, 2, 7, 9, 5, 1, 2, 9, 16, 14, 6, 1, 2, 11, 25, 30, 20, 7, 1, 2, 13, 36, 55, 50, 27, 8, 1, 2, 15, 49, 91, 105, 77, 35, 9, 1, ...}

The generating function for the i^th, i ≥ 0, member is:

$G_{T_{(2,1)}(i={\frac {n(n+1)}{2}}+d)}(x)=1+\sum _{n=1}^{\infty }x^{\tfrac {n(n+1)}{2}}(2+x)(1+x)^{n-1}\,$

(2,1)-Pascal triangle rows sums

The sums of the respective finite sequences give the infinite sequence (Cf. A003945(n)):

{1, 3, 6, 12, 24, 48, 96, 192, 384, 768, 1536, 3072, 6144, 12288, 24576, 49152, 98304, 196608, 393216, 786432, 1572864, 3145728, ...}

with members given by the formula:

$\sum _{d=0}^{n}T_{(2,1)}(n,d)={\frac {(3\cdot 2^{n}-0^{n})}{2}},\,$

where:

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

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

The generating function is:

$G_{\{\sum _{d=0}^{n}T_{(2,1)}(n,d)\}}={\frac {3\ G_{2^{n}}(x)-G_{0^{n}}(x)}{2}}={\frac {3({\frac {1}{1-2x}})-(1)}{2}}={\frac {1+x}{1-2x}}\,$

(2,1)-Pascal triangle rows alternating sign sums

The alternating sign sums of the respective finite sequences give the infinite sequence (coinciding with A019590(n+1):

{1, 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, 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:

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

where:

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

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

The generating function is:

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

(2,1)-Pascal (rectangular) triangle columns and square (hyper)pyramidal numbers

(Square Hyperpyramidal) Figurate Number Triangle

n = 0	1
1	2	1
2	2	3	1
3	2	5	4	1
4	2	7	9	5	1
5	2	9	16	14	6	1
6	2	11	25	30	20	7	1
7	2	13	36	55	50	27	8	1
8	2	15	49	91	105	77	35	9	1
9	2	17	64	140	196	182	112	44	10	1
10	2	19	81	204	336	378	294	156	54	11	1
11	2	21	100	285	540	714	672	450	210	65	12	1
12	2	23	121	385	825	1254	1386	1122	660	275	77	13	1
	d = 0	1	2	3	4	5	6	7	8	9	10	11	12

The partial sums of the d^th column build the entries of the (d+1)^th column, thus begetting the (d+1)-dimensional square hyperpyramidal numbers from the d-dimensional ones:

$\sum _{i=0}^{n}T_{(2,1)}(i,d)=\sum _{i=d}^{n}T_{(2,1)}(i,d)=T_{(2,1)}(n+1,d+1)\,$

The d^th column, d ≥ 1, gives the d-dimensional square hyperpyramidal numbers, forming square (hyper)pyramids, e.g.:

d=1	1-dimensional square hyperpyramidal numbers	Square gnomon numbers	(2 0D-cells facets)	(Square gnomons)
d=2	2-dimensional square hyperpyramidal numbers	Square numbers	(4 1D-cells facets)	(Squares)
d=3	3-dimensional square hyperpyramidal numbers	Square pyramidal numbers	(5 2D-cells facets)	(Square pyramids)
d=4	4-dimensional square hyperpyramidal numbers	Square 4D-hyperpyramidal numbers	(? 3D-cells facets)	(Square 4D-hyperpyramids)
d=5	5-dimensional square hyperpyramidal numbers	Square 5D-hyperpyramidal numbers	(? 4D-cells facets)	(Square 5D-hyperpyramids)
d=6	6-dimensional square hyperpyramidal numbers	Square 6D-hyperpyramidal numbers	(? 5D-cells facets)	(Square 6D-hyperpyramids)
d=7	7-dimensional square hyperpyramidal numbers	Square 7D-hyperpyramidal numbers	(? 6D-cells facets)	(Square 7D-hyperpyramids)
d=8	8-dimensional square hyperpyramidal numbers	Square 8D-hyperpyramidal numbers	(? 7D-cells facets)	(Square 8D-hyperpyramids)

where (-1D)-cells correspond to the empty set, 0D-cells are vertices, 1D-cells are edges, 2D-cells are faces, and so on...

Table of columns sequences

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

(2,1)-Pascal triangle columns sequences

d	Sequence	A-number
0	{1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, ...}	A040000(i)
1	{1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, ...}	A005408(i)
2	{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, ...}	A000290(i+1)
3	{1, 5, 14, 30, 55, 91, 140, 204, 285, 385, 506, 650, 819, 1015, 1240, 1496, 1785, 2109, 2470, 2870, 3311, 3795, 4324, 4900, 5525, 6201, 6930, 7714, 8555, 9455, ...}	A000330(i+1)
4	{1, 6, 20, 50, 105, 196, 336, 540, 825, 1210, 1716, 2366, 3185, 4200, 5440, 6936, 8721, 10830, 13300, 16170, 19481, 23276, 27600, 32500, 38025, 44226, 51156, ...}	A002415(i+2)
5	{1, 7, 27, 77, 182, 378, 714, 1254, 2079, 3289, 5005, 7371, 10556, 14756, 20196, 27132, 35853, 46683, 59983, 76153, 95634, 118910, 146510, 179010, 217035, ...}	A005585(i+1)
6	{1, 8, 35, 112, 294, 672, 1386, 2640, 4719, 8008, 13013, 20384, 30940, 45696, 65892, 93024, 128877, 175560, 235543, 311696, 407330, 526240, 672750, ...}	A040977(i)
7	{1, 9, 44, 156, 450, 1122, 2508, 5148, 9867, 17875, 30888, 51272, 82212, 127908, 193800, 286824, 415701, 591261, 826804, 1138500, 1545830, 2072070, ...}	A050486(i)
8	{1, 10, 54, 210, 660, 1782, 4290, 9438, 19305, 37180, 68068, 119340, 201552, 329460, 523260, 810084, 1225785, 1817046, 2643850, 3782350, 5328180, ...}	A053347(i)
9	{1, 11, 65, 275, 935, 2717, 7007, 16445, 35750, 72930, 140998, 260338, 461890, 791350, 1314610, 2124694, 3350479, 5167525, 7811375, 11593725, ...}	A054333(i)
10	{1, 12, 77, 352, 1287, 4004, 11011, 27456, 63206, 136136, 277134, 537472, 999362, 1790712, 3105322, 5230016, 8580495, 13748020, 21559395, 33153120, ...}	A054334(i)
11	{1, 13, 90, 442, 1729, 5733, 16744, 44200, 107406, 243542, 520676, 1058148, 2057510, 3848222, 6953544, 12183560, 20764055, 34512075, 56071470, ...}	A057788(i)
12	{1, 14, 104, 546, 2275, 8008, 24752, 68952, 176358, 419900, 940576, 1998724, 4056234, 7904456, 14858000, 27041560, 47805615, 82317690, 138389160, ...}	A??????

Table of columns sequences related formulae

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

(2,1)-Pascal triangle columns sequences related formulae

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

(2,1)-Pascal (rectangular) triangle falling diagonals and (Chebyshev polynomials?)

n = 0	1
1	2	1
2	2	3	1
3	2	5	4	1
4	2	7	9	5	1
5	2	9	16	14	6	1
6	2	11	25	30	20	7	1
7	2	13	36	55	50	27	8	1
8	2	15	49	91	105	77	35	9	1
9	2	17	64	140	196	182	112	44	10	1
10	2	19	81	204	336	378	294	156	54	11	1
11	2	21	100	285	540	714	672	450	210	65	12	1
12	2	23	121	385	825	1254	1386	1122	660	275	77	13	1
	d = 0	1	2	3	4	5	6	7	8	9	10	11	12

Table of falling diagonals sequences

The falling diagonals sequences of the (2,1)-Pascal triangle correspond to the columns sequences of the (1,2)-Pascal triangle, except that the 0^th member of the 0^th falling diagonal is 1 instead of 2 for the 0^th member of the 0^th column of the (1,2)-Pascal triangle.

The i ^th, i ≥ 0, member of falling diagonal j appears in row j+i.

(2,1)-Pascal triangle falling diagonals sequences

j	$T_{(2,1)}(j+i,j),\ i\geq 0\ \,$ sequences	A-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, ...}	A000012(i)

The first falling diagonal from the right (for j = 0) happens to be the continued fraction expansion of $\scriptstyle \phi ={\frac {1+{\sqrt {5}}}{2}},\,$ where $\scriptstyle \phi \,$ is the Golden ratio.

Table of falling diagonals sequences related formulae

The falling diagonals sequences related formulae of the (2,1)-Pascal triangle correspond to the columns sequences related formulae of the (1,2)-Pascal triangle, except that the 0^th member of the 0^th falling diagonal is 1 instead of 2 for the 0^th member of the 0^th column of the (1,2)-Pascal triangle.

The i ^th, i ≥ 0, member of falling diagonal j appears in row j+i.

(2,1)-Pascal triangle falling diagonals sequences related formulae

j	Formulae $T_{(2,1)}(j+i,j)=\,$ ${\binom {i+j-1}{j-1}}{\frac {(i+2j)}{j}},\,$ $j\geq 1\,$	Generating function for i th (i ≥ 0) falling diagonal member $G_{T_{(2,1)}}(x,j)=\,$ ${\frac {(2-x)}{(1-x)^{j+1}}},\,$ $j\geq 1\,$	Order of basis	Differences $T_{(2,1)}(j+i+1,j)-\,$ $T_{(2,1)}(j+i,j)\,$	Partial sums $\sum _{i=0}^{m}T_{(2,1)}(j+i,j)\,$	Partial sums of reciprocals $\sum _{i=0}^{m}{1 \over {T_{(2,1)}(j+i,j)}}$	Sum of Reciprocals $\sum _{i=0}^{\infty }{1 \over {T_{(2,1)}(j+i,j)}}$
0	$1\,$	${\frac {1}{(1-x)}}\,$	$\infty \,$	$0\,$	$m+1\,$	$m+1\,$	$\infty \,$

(2,1)-Pascal (rectangular) triangle rising diagonals

n = 0	1
1	2	1
2	2	3	1
3	2	5	4	1
4	2	7	9	5	1
5	2	9	16	14	6	1
6	2	11	25	30	20	7	1
7	2	13	36	55	50	27	8	1
8	2	15	49	91	105	77	35	9	1
9	2	17	64	140	196	182	112	44	10	1
10	2	19	81	204	336	378	294	156	54	11	1
11	2	21	100	285	540	714	672	450	210	65	12	1
12	2	23	121	385	825	1254	1386	1122	660	275	77	13	1
	d = 0	1	2	3	4	5	6	7	8	9	10	11	12

The rising diagonals (starting with the 0^th diagonal) give an infinite sequence of finite sequences:

{{1}, {2}, {2, 1}, {2, 3}, {2, 5, 1}, {2, 7, 4}, {2, 9, 9, 1}, {2, 11, 16, 5}, {2, 13, 25, 14, 1}, {2, 15, 36, 30, 6}, {2, 17, 49, 55, 20, 1}, {2, 19, 64, 91, 50, 7}, {2, 21, 81, 140, 105, 27, 1}, ... }

The concatenated infinite sequence of finite sequences give the infinite sequence (Cf. A129710):

{1, 2, 2, 1, 2, 3, 2, 5, 1, 2, 7, 4, 2, 9, 9, 1, 2, 11, 16, 5, 2, 13, 25, 14, 1, 2, 15, 36, 30, 6, 2, 17, 49, 55, 20, 1, 2, 19, 64, 91, 50, 7, 2, 21, 81, 140, 105, 27, 1, 2, 23, 100, 204, 196, 77, 8, ... }

(2,1)-Pascal (rectangular) triangle rising diagonals sums and Fibonacci numbers

The sums of the respective rising diagonals give:

{1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578, ...}

which are the (m+2)^th Fibonacci numbers (Cf. A000045(m+2),) and may be obtained using the Binet formula:

$\sum _{n+d=m}^{}T_{(2,1)}(n,d)=F_{m+2}={\frac {{\big (}{\tfrac {1+{\sqrt {5}}}{2}}{\big )}^{m+2}-{\big (}{\tfrac {1-{\sqrt {5}}}{2}}{\big )}^{m+2}}{\sqrt {5}}}={\frac {(\phi _{+})^{m+2}-(\phi _{-})^{m+2}}{(\phi _{+})-(\phi _{-})}},\ m\geq 0,\,$

where $\scriptstyle \phi _{+}={\frac {1+{\sqrt {5}}}{2}}$ and $\scriptstyle \phi _{-}={\frac {1-{\sqrt {5}}}{2}}$ are the roots of $\scriptstyle 1+x-x^{2}=0,\,$ and where $\scriptstyle \phi \ \equiv \ \phi _{+}\,$ is the Golden ratio.

The generating function is:

$G_{\{\sum _{n+d=m}^{}T_{(2,1)}(n,d)\}}=G_{F_{m+2}}={\frac {G_{F_{m}}}{x^{2}}}={\frac {1}{x^{2}}}{\bigg (}{\frac {x}{1-x-x^{2}}}{\bigg )}={\frac {1}{x(1-x-x^{2})}}\,$

Compare with the rising diagonals sums of the (1,2)-Pascal triangle, which give the Lucas numbers.

(2,1)-Pascal (rectangular) triangle rising diagonals alternating sign sums

The alternating sign sums of the respective rising diagonals give the periodic sequence (Cf. A057079):

{1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, ...}

which is given by the formulae (note the Binet like formula):

$\sum _{n+d=m}^{}(-1)^{d}\ T_{(2,1)}(n,d)={\big (}{\tfrac {1+i{\sqrt {3}}}{2}}{\big )}^{m-1}+{\big (}{\tfrac {1-i{\sqrt {3}}}{2}}{\big )}^{m-1}={\big (}e^{i{\tfrac {\pi }{3}}}{\big )}^{m-1}+{\big (}e^{-i{\tfrac {\pi }{3}}}{\big )}^{m-1}={\frac {{(\rho _{+})}^{m-1}+{(\rho _{-})}^{m-1}}{(\rho _{+})+(\rho _{-})}},\,$

where $\scriptstyle \rho _{+}={\frac {1+i{\sqrt {3}}}{2}}=e^{i{\tfrac {\pi }{3}}}\,$ and $\scriptstyle \rho _{-}={\frac {1-i{\sqrt {3}}}{2}}=e^{-i{\tfrac {\pi }{3}}}\,$ are the roots of $\scriptstyle 1-x+x^{2}=0.\,$

The generating function is:

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

(2,1)-Pascal triangle central elements

The central elements (for row 2m, m ≥ 0) of the (2,1)-Pascal triangle give the sequence (Cf. A029651(m):

{1, 3, 9, 30, 105, 378, 1386, 5148, 19305, 72930, 277134, 1058148, 4056234, 15600900, 60174900, 232676280, 901620585, 3500409330, 13612702950, 53017895700, ...}

which is given by the formulae:

$T_{(2,1)}(0,0)=1,\,$

$T_{(2,1)}(2m,m)=3\ {\binom {2m-1}{m}}=3\ {\frac {(2m-1)!}{m!(m-1)!}},\ m\geq 1,\,$

or:

$T_{(2,1)}(2m,m)={\frac {3}{2}}(m+1)C_{m}-{\frac {1}{2}}0^{m},\ m\geq 0,\,$

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 Catalan number (also called Segner numbers) (Cf. A000108(m).)

The generating function is:

$G_{\{T_{(2,1)}(2m,m)\}}(x)={\frac {1+xC(x)}{1-2xC(x)}}={\frac {1+x({\frac {1-{\sqrt {1-4x}}}{2x}})}{1-2x({\frac {1-{\sqrt {1-4x}}}{2x}})}},\,$

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

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

Compare with the (1,2)-Pascal triangle central elements where the triangle apex (the first central element) is 2 instead of 1.

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.