A141020 Pascal-like triangle with index of asymmetry y = 4 and index of obliqueness z = 0.

1, 1, 1, 1, 2, 1, 1, 4, 2, 1, 1, 8, 4, 2, 1, 1, 16, 8, 4, 2, 1, 1, 32, 16, 8, 4, 2, 1, 1, 63, 33, 16, 8, 4, 2, 1, 1, 124, 67, 33, 16, 8, 4, 2, 1, 1, 244, 136, 67, 33, 16, 8, 4, 2, 1, 1, 480, 276, 136, 67, 33, 16, 8, 4, 2, 1, 1, 944, 560, 276, 136, 67, 33, 16, 8, 4, 2, 1

Offset: 0

Author

Juri-Stepan Gerasimov, Jul 11 2008

Keywords

• nonn
• tabl

Comments

The left column is set to 1. The four rightmost columns start with powers of 2:

T(n, 0) = T(n, n)=1; T(n, n-1)=2; T(n, n-2)=4; T(n, n-3)=8; T(n, n-4)=16.

Recurrence: T(n, k) = T(n-1, k) + T(n-2, k) + T(n-3, k) + T(n-4, k) + T(n-5, k) + T(n-5,k-1), k = 1..n-5.

From Petros Hadjicostas, Jun 14 2019: (Start)

In the attached photograph we see that the index of asymmetry is denoted by s (rather than y) and the index of obliqueness by e (rather than z).

The general recurrence is G(n+s+2, k) = G(n+1, k-e*s+e-1) + Sum_{1 <= m <= s+1} G(n+m, k-e*s+m*e-2*e) for n >= 0 with k = 1..(n+1) when e = 0 and k = (s+1)..(n+s+1) when e = 1. The initial conditions are G(n+x+1, n-e*n+e*x-e+1) = 2^x for x=0..s and n >= 0. There is one more initial condition, namely, G(n, e*n) = 1 for n >= 0.

For s = 0, we get Pascal's triangle A007318. For s = 1, we get A140998 (e = 0) and A140993 (e = 1). For s = 2, we get A140997 (e = 0) and A140994 (e = 1). For s = 3, we get A140996 (e = 0) and A140995 (e = 1). For s = 4, we have the current array (with e = 0) and array A141021 (with e = 1). In some of these arrays, the indices n and k are sometimes shifted.

(End)

Examples

			Pascal-like triangle with y = 4 and z = 0 begins as follows:
  1
  1   1
  1   2   1
  1   4   2   1
  1   8   4   2   1
  1  16   8   4   2  1
  1  32  16   8   4  2  1
  1  63  33  16   8  4  2  1
  1 124  67  33  16  8  4  2 1
  1 244 136  67  33 16  8  4 2 1
  1 480 276 136  67 33 16  8 4 2 1
  1 944 560 276 136 67 33 16 8 4 2 1
  ...
		

Links

• Juri-Stepan Gerasimov, Stepan's triangles and Pascal's triangle are connected by the recurrence relation ...

Crossrefs

Cf. A001949, A007318, A140993, A140994, A140995, A140996, A140997, A140998, A141021, A141031, A141064, A141065, A141066, A141067, A141069, A141070, A141072, A141073.

Programs

Maple

A141020 := proc(n,k) option remember ; if k<0 or k>n then 0 ; elif k=0 or k=n then 1 ; elif k=n-1 then 2 ; elif k=n-2 then 4 ; elif k=n-3 then 8 ; elif k=n-4 then 16 ; else procname(n-1,k) +procname(n-2,k)+procname(n-3,k)+procname(n-4,k) +procname(n-5,k)+procname(n-5,k-1) ; fi; end:
for n from 0 to 20 do for k from 0 to n do printf("%d,",A141020(n,k)) ; od: od: # R. J. Mathar, Sep 19 2008

Mathematica

T[n_, k_] := T[n, k] = Which[k < 0 || k > n, 0, k == 0 || k == n, 1, k == n-1, 2, k == n-2, 4, k == n-3, 8, k == n-4, 16, True, T[n-1, k] + T[n-2, k] + T[n-3, k] + T[n-4, k] + T[n-5, k] + T[n-5, k-1]];
Table[T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Dec 18 2019, after R. J. Mathar *)

Formula

From Petros Hadjicostas, Jun 14 2019: (Start)

T(n, k) = A141021(n, n-k) for 0 <= k <= n.

Bivariate g.f.: Sum_{n,k >= 0} T(n, k)*x^n*y^k = (1 - x - x^2 - x^3 - x^4 - x^5 + y*x^2*(1 + x + x^2 + x^4)) / ((1 - x) * (1 - x*y) * (1 - x - x^2 - x^3 - x^4 - x^5 - x^5*y)).

Differentiating the bivariate w.r.t. y and setting y = 0, we get the g.f. of the column k = 1: x/((-1 + x)*(x^5 + x^4 + x^3 + x^2 + x - 1)). This is the g.f. of a shifted version of sequence A001949.

(End)

Extensions

Partially edited by N. J. A. Sloane, Jul 18 2008

Recurrence rewritten by R. J. Mathar, Sep 19 2008