A096956 Pascal (1,6) triangle.

6, 1, 6, 1, 7, 6, 1, 8, 13, 6, 1, 9, 21, 19, 6, 1, 10, 30, 40, 25, 6, 1, 11, 40, 70, 65, 31, 6, 1, 12, 51, 110, 135, 96, 37, 6, 1, 13, 63, 161, 245, 231, 133, 43, 6, 1, 14, 76, 224, 406, 476, 364, 176, 49, 6, 1, 15, 90, 300, 630, 882, 840, 540, 225, 55, 6, 1, 16, 105, 390, 930

Offset: 0

Author

Wolfdieter Lang, Aug 13 2004

Keywords

• nonn
• easy
• tabl

Comments

Except for the first row this is the row reversed (6,1)-Pascal triangle A093563.

This is the sixth member, q=6, in the family of (1,q) Pascal triangles: A007318 (Pascal (q=1)), A029635 (q=2) (but with a(0,0)=2, not 1), A095660 (q=3), A095666 (q=4), A096940 (q=5).

This is an example of a Riordan triangle (see A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group) with o.g.f. of column no. m of the type g(x)*(x*f(x))^m with f(0)=1. Therefore the o.g.f. for the row polynomials p(n,x):=Sum_{m=0..n} a(n,m)*x^m is G(z,x)=g(z)/(1-x*z*f(z)). Here: g(x)=(6-5*x)/(1-x), f(x)=1/(1-x), hence G(z,x)=(6-5*z)/(1-(1+x)*z).

The SW-NE diagonals give Sum_{k=0..ceiling((n-1)/2)} a(n-1-k,k) = A022097(n-2), n >= 2, with n=1 value 6. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.

Examples

			Triangle begins:
  [0]  6;
  [1]  1,  6;
  [2]  1,  7,  6;
  [3]  1,  8, 13,  6;
  [4]  1,  9, 21, 19,  6;
  [5]  1, 10, 30, 40, 25,  6;
  ...
		

Links

• Paolo Xausa, Table of n, a(n) for n = 0..11475 (rows 0..150 of triangle, flattened).
• Wolfdieter Lang, First 10 rows.

Crossrefs

Row sums: A005009(n-1), n>=1, 6 if n=0; g.f.: (6-5*x)/(1-2*x). Alternating row sums are [6, -5, followed by 0's].

Column sequences (without leading zeros) give for m=1..9, with n >= 0: A000027(n+6), A056115, A096957-9, A097297-A097300.

Programs

Maple

a(n,k):=piecewise(n=0,6,0Mircea Merca, Apr 08 2012

Mathematica

A096956[n_, k_] := If[n == k, 6, (5*k/n + 1)*Binomial[n, k]];
Table[A096956[n, k], {n, 0, 12}, {k, 0, n}] (* Paolo Xausa, Apr 14 2025 *)

Formula

Recursion: a(n,m)=0 if m > n, a(0,0) = 6; a(n,0) = 1 if n >= 1; a(n,m) = a(n-1, m) + a(n-1, m-1).

G.f. column m (without leading zeros): (6-5*x)/(1-x)^(m+1), m >= 0.

a(n,k) = (1+5*k/n)*binomial(n,k), for n > 0. - Mircea Merca, Apr 08 2012