(a,b)-Pascal triangle

The (a,b)-Pascal triangle is a geometric arrangement of numbers produced recursively in the same way as for the original Pascal triangle, except for its boundary conditions (for the leftmost and rightmost entries on each rows.) Assuming its boundary conditions to be positive integers, the rightmost nonzero entries are initialized to b, b > 0, and its leftmost nonzero entries (except the first row for n = 0) are initialized to a, a > 0. The (a,b)-Pascal triangle can be built by superimposing the (k,1)-Pascal triangle with k = a-1 and the (1,k)-Pascal triangle with k = b-1, i.e.:

$T_{(a,b)}(n,j)=T_{(a-1,1)}(n,j)+T_{(1,b-1)}(n,j)\,$.

The rectangular version of the (a,b)-Pascal triangle

n = 0

b

1

a

b

2

a

$\scriptstyle a+b\,$

b

3

a

$\scriptstyle 2a+b\,$

$\scriptstyle a+2b\,$

b

4

a

$\scriptstyle 3a+b\,$

$\scriptstyle 3a+3b\,$

$\scriptstyle a+3b\,$

b

5

a

$\scriptstyle 4a+b\,$

$\scriptstyle 6a+4b\,$

$\scriptstyle 4a+6b\,$

$\scriptstyle a+4b\,$

b

6

a

$\scriptstyle 5a+b\,$

$\scriptstyle 10a+5b\,$

$\scriptstyle 10a+10b\,$

$\scriptstyle 5a+10b\,$

$\scriptstyle a+5b\,$

b

7

a

$\scriptstyle 6a+b\,$

$\scriptstyle 15a+6b\,$

$\scriptstyle 20a+15b\,$

$\scriptstyle 15a+20b\,$

$\scriptstyle 6a+15b\,$

$\scriptstyle a+6b\,$

b

8

a

$\scriptstyle 7a+b\,$

$\scriptstyle 21a+7b\,$

$\scriptstyle 35a+21b\,$

$\scriptstyle 35a+35b\,$

$\scriptstyle 21a+35b\,$

$\scriptstyle 7a+21b\,$

$\scriptstyle a+7b\,$

b

9

a

$\scriptstyle 8a+b\,$

$\scriptstyle 28a+8b\,$

$\scriptstyle 56a+28b\,$

$\scriptstyle 70a+56b\,$

$\scriptstyle 56a+70b\,$

$\scriptstyle 28a+56b\,$

$\scriptstyle 8a+28b\,$

$\scriptstyle a+8b\,$

b

10

a

$\scriptstyle 9a+b\,$

$\scriptstyle 36a+9b\,$

$\scriptstyle 84a+36b\,$

$\scriptstyle 126a+84b\,$

$\scriptstyle 126a+126b\,$

$\scriptstyle 84a+126b\,$

$\scriptstyle 36a+84b\,$

$\scriptstyle 9a+36b\,$

$\scriptstyle a+9b\,$

b

11

a

$\scriptstyle 10a+b\,$

$\scriptstyle 45a+10b\,$

$\scriptstyle 120a+45b\,$

$\scriptstyle 210a+120b\,$

$\scriptstyle 252a+210b\,$

$\scriptstyle 210a+252b\,$

$\scriptstyle 120a+210b\,$

$\scriptstyle 45a+120b\,$

$\scriptstyle 10a+45b\,$

$\scriptstyle a+10b\,$

b

12

a

$\scriptstyle 11a+b\,$

$\scriptstyle 55a+11b\,$

$\scriptstyle 165a+55b\,$

$\scriptstyle 330a+165b\,$

$\scriptstyle 462a+330b\,$

$\scriptstyle 462a+462b\,$

$\scriptstyle 330a+462b\,$

$\scriptstyle 165a+330b\,$

$\scriptstyle 55a+165b\,$

$\scriptstyle 11a+55b\,$

$\scriptstyle a+11b\,$

b

j = 0

1

2

3

4

5

6

7

8

9

10

11

12

In the equilateral version of the (a,b)-Pascal triangle, we start with a cell (row 0) initialized to b, with the leftmost nonzero cell in the row below initialized to a, 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 (a,b)-Pascal triangle, we start with a cell (row 0) initialized to b, with the cell below it initialized to a, 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.

If a and b are positive numbers, then all the interior cells are necessarily greater than or equal to a + b. The number of cells from rows 0 to n which are equal to a is thus n. (Cf. A000027(n),) the number of cells from rows 0 to n which are equal to b is thus n+1 (Cf. A000027(n+1),) and the number of cells from rows 0 to n which are greater than or equal to a + b is $\scriptstyle P_{3}^{(2)}(n-1)\,$, the (n-1)^th triangular number.

For comparison, the (1,1)-Pascal triangle (Pascal's triangle) is shown below:

The rectangular version of Pascal's triangle
(Figurate Number Triangle)

n = 0	1
1	1	1
2	1	2	1
3	1	3	3	1
4	1	4	6	4	1
5	1	5	10	10	5	1
6	1	6	15	20	15	6	1
7	1	7	21	35	35	21	7	1
8	1	8	28	56	70	56	28	8	1
9	1	9	36	84	126	126	84	36	9	1
10	1	10	45	120	210	252	210	120	45	10	1
11	1	11	55	165	330	462	462	330	165	55	11	1
12	1	12	66	220	495	792	924	792	495	220	66	12	1
	j = 0	1	2	3	4	5	6	7	8	9	10	11	12

Recursion rule

The (a,b)-Pascal triangle recursion rule is:

$T_{(a,b)}(n,n)=b,\ n\geq 0,\,$

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

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

Formulae

$T_{(a,b)}(0,0)=b,\,$

$T_{(a,b)}(n,j)=T_{(a-1,1)}(n,j)+T_{(1,b-1)}(n,j)={\bigg \{}{\binom {n-1}{j-1}}+(a-1)\ {\binom {n-1}{j}}{\bigg \}}+{\bigg \{}(b-1)\ {\binom {n-1}{j-1}}+{\binom {n-1}{j}}{\bigg \}}\,$

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

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.

(a,b)-Pascal triangle formulae and additive commutative semigroups

The rectangular version of the (a,b)-Pascal triangle

n = 0

b

1

a

b

2

a

$\scriptstyle a+b\,$

b

3

a

$\scriptstyle a+(a+b)\,$

$\scriptstyle (a+b)+b\,$

b

4

a

$\scriptstyle a+(a+(a+b))\,$

$\scriptstyle (a+(a+b))+((a+b)+b)\,$

$\scriptstyle ((a+b)+b)+b\,$

b

5

a

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

b

6

a

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

b

7

a

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

b

8

a

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

b

9

a

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

b

10

a

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

b

11

a

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

b

12

a

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

$\scriptstyle \,$

b

j = 0

1

2

3

4

5

6

7

8

9

10

11

12

Since the binomial coefficients are counting devices (telling how many times that something was added,) we only need a and b to belong to an additive commutative semigroup (a semigroup being a set with an operation having the associativity and closure properties.)

Both associativity and commutativity are required to allow for the grouping of the a's and b's:

$(a+(a+b))+((a+b)+b)=3a+3b\,$

Therefore a and b could both belong to, among others:

• Natural numbers $\mathbb {N} \,$
• Rational integers $\mathbb {Z} \,$
• Rational integers (mod n) $\mathbb {Z} /{n\mathbb {Z} }\,$
• Rational numbers $\mathbb {Q} \,$
• Algebraic integers $\mathbb {Z} (\rho _{1},...,\rho _{n})\,$ (e.g. Gaussian integers $\mathbb {Z} (i)\,$)
• Algebraic numbers $\mathbb {Q} (\rho _{1},...,\rho _{n})\,$
• Real numbers $\mathbb {R} \,$
• Complex numbers $\mathbb {C} \,$ (equivalent to two (a,b)-Pascal triangles: one for the real parts, one for the imaginary parts)
• Quaternions (equivalent to four (a,b)-Pascal triangles)
• Octonions (equivalent to eight (a,b)-Pascal triangles)
• Sedenions (equivalent to sixteen (a,b)-Pascal triangles)
• Vectors (tensors of rank 1) in $\mathbb {R} ^{n}\,$ or over any given field (equivalent to n (a,b)-Pascal triangles)
• Matrices (tensors of rank 2) in $\mathbb {R} ^{m}\times \mathbb {R} ^{n}\,$ or over any given field (equivalent to m × n (a,b)-Pascal triangles)
• Tensors with same rank r in $\scriptstyle \prod _{i=1}^{r}\mathbb {R} ^{m_{i}}\,$ or over any given field (equivalent to $\scriptstyle \prod _{i=1}^{r}m_{i}\,$ (a,b)-Pascal triangles)
• ...

(a,b)-Pascal triangle central elements

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

{b, a+b, 3a+3b, 10a+10b, 35a+35b, 126a+126b, 462a+462b, ...}

which is given by the formulae:

$T_{(a,b)}(0,0)=b\,$

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

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

The generating function is:

$G_{\{T_{(a,b)}(2m,m)\}}(x)=\,$

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

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

Generalizations of the (a,b)-Pascal triangle

Keeping the original Pascal triangle recurrence rule unchanged for the interior cells of the triangle, then one might want to consider a generalized (a(n),b(n))-Pascal triangle, n ≥ 0, where a_n ≡ a(n) and b_n ≡ b(n) are integer sequences.

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.