A141021 Pascal-like triangle with index of asymmetry y = 4 and index of obliqueness z = 1.

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

Offset: 0

Author

Juri-Stepan Gerasimov, Jul 11 2008

Keywords

• nonn
• tabl

Comments

The triangle here is A141020 with each row reversed.

From Petros Hadjicostas, Jun 16 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 array A141020 (with e = 0) and the current array (with e = 1). In some of these arrays, the indices n and k are sometimes shifted.

Putting k = 1 in Stepan's triangles with index of asymmetry s and index of obliqueness e = 0, we get G(n + s + 2, 1) = 1 + Sum_{1 <= m <= s+1} G(n+m, 1) for n >= 0 and k = 1..(n+1) with initial conditions G(x+1, 1) = 2^x for x = 0..s. Thus, we get a shifted version of column s+1 in array A172119. These sequences were first studied by Dunkel (1925).

Thus, the second main diagonal of Stepan's triangles with index of asymmetry s and index of obliqueness e = 1 is equal to a shifted version of column s + 1 in array A172119.

It follows from Eq. (20) on p. 360 in Dunkel (1925) that, for Stepan's triangles with index of asymmetry s and index of obliqueness e = 0, we have G(n, 1) = Sum_{t = 1..floor((n + s + 1)/(s + 2))} (-1)^(t + 1) * binomial(n + s - t*(s + 1), t - 1) * 2^(n + s - t*(s + 2) + 1) for n >= 0.

In a similar way, for Stepan's triangles with index of asymmetry s and index of obliqueness e = 1, we have G(n, n - 1) = Sum_{t = 1..floor((n + s + 1)/(s + 2))} (-1)^(t + 1) * binomial(n + s - t*(s + 1), t - 1) * 2^(n + s - t*(s + 2) + 1) for n >= 1.

Let A_s(x, y) be the bivariate g.f. of G(n, k) with index of asymmetry s and index of obliqueness e = 0 and let B_s(x, y) be the bivariate g.f. of the other G(n, k) with index of asymmetry s and index of obliqueness e = 1. Because the two triangular arrays are mirror images of each other, we have B_s(x, y) = A_s(x*y, y^(-1)).

(End)

Examples

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

Links

• O. Dunkel, Solutions of a probability difference equation, Amer. Math. Monthly, 32 (1925), 354-370; see p. 356 with r = s + 1 (where s is the index of asymmetry).
• Juri-Stepan Gerasimov, Stepan's triangles and Pascal's triangle are connected by the recurrence relation ...

Crossrefs

Cf. A007318, A140993, A140994, A140995, A140996, A140997, A140998, A141020, A141031, A172119, A308808.

Programs

Maple

# This is a slight modification of R. J. Mathar's Maple program from array A141020:
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:
A141021 := proc(n, k) A141020(n, n-k): end:
for n1 from 0 to 20 do for k1 from 0 to n1 do printf("%d, ", A141021(n1, k1)) ; od: od: # Petros Hadjicostas, Jun 16 2019

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]];
T[n_, k_] := t[n, n - k];
Table[Table[T[n, k], {k, 0, n}], {n, 0, 10}] // Flatten (* Jean-François Alcover, Apr 24 2020 *)

Formula

T(n, k) = A141020(n, n-k). - R. J. Mathar, Sep 19 2008

From Petros Hadjicostas, Jun 16 2019: (Start)

Recurrence: G(n+6, k) = G(n+1, k-4) + G(n+1, k-5) + G(n+2, k-4) + G(n+3, k-3) + G(n+4, k-2) + G(n+5, k-1) for n >= 0 and k = 5..(n+5) with G(n+x+1, x) = 2^x for x = 0..4 and n >= 0.

Bivariate g.f.: Sum_{n,k >=0} T(n, k)*x^n*y^k = (x^6*y^5 - x^5*y^5 - x^4*y^4 + x^4*y^3 - x^3*y^3 + x^3*y^2 - x^2*y^2 + x^2*y - x*y + 1)/((1 - x*y) * (1 - x) * (1 - x*y - x^2*y^2 - x^3*y^3 - x^4*y^4 - x^5*y^4 - x^5*y^5)).

Second main diagonal: G(n, n - 1) = Sum_{t = 1..floor((n + 5)/6)} (-1)^(t + 1) * binomial(n + 4 - 5*t, t - 1) * 2^(n + 5 - 6*t) for n >= 1.

Limiting row: Let b(k) = lim_{n -> infinity} G(n, k) for k >= 0. Then b(k) = b(k-5) + 2*b(k-4) + b(k-3) + b(k-2) + b(k-1) for k >= 5 with b(x) = 2^x for x = 0..4. This is the sequence 1, 2, 4, 8, 16, 33, 67, 136, 276, 561, 1140, 2316, 4705, 9559, 19421, 39457, 80163, 162864, 330885, 672247, ..., which is A308808.

(End)

Extensions

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

Comment simplified by R. J. Mathar, Sep 19 2008

Data corrected by Jean-François Alcover, Apr 24 2020