A140994 Triangle G(n, k), for 0 <= k <= n, read by rows, where G(n, n) = G(n+1, 0) = 1, G(n+2, 1) = 2, G(n+3, 2) = 4, G(n+4, m) = G(n+1, m-2) + G(n+1, m-3) + G(n+2, m-2) + G(n+3, m-1) for n >= 0 and m = 3..(n+3).

1, 1, 1, 1, 2, 1, 1, 2, 4, 1, 1, 2, 4, 8, 1, 1, 2, 4, 9, 15, 1, 1, 2, 4, 9, 19, 28, 1, 1, 2, 4, 9, 19, 40, 52, 1, 1, 2, 4, 9, 19, 41, 83, 96, 1, 1, 2, 4, 9, 19, 41, 88, 170, 177, 1, 1, 2, 4, 9, 19, 41, 88, 188, 345, 326, 1, 1, 2, 4, 9, 19, 41, 88, 189, 400, 694, 600, 1, 1, 2, 4, 9, 19, 41, 88, 189, 406, 846, 1386, 1104, 1

Offset: 0

Author

Juri-Stepan Gerasimov, Jul 08 2008

Keywords

• nonn
• tabl

Comments

From Petros Hadjicostas, Jun 12 2019: (Start)

This is a mirror image of the triangular array A140997. The current array has index of asymmetry s = 2 and index of obliqueness (obliquity) e = 1. Array A140997 has the same index of asymmetry, but has index of obliqueness e = 0. (In other related sequences, the author uses the letter y for the index of asymmetry and the letter z for the index of obliqueness, but on the stone slab that appears over a tomb in a picture that he posted in those sequences, the letters s and e are used instead. See, for example, the documentation for sequences A140998, A141065, A141066, and A141067.)

In general, if the index of asymmetry (from the Pascal triangle A007318) is s, then the order of the recurrence is s + 2 (because the recurrence of the Pascal triangle has order 2). There are also s + 2 infinite sets of initial conditions (as opposed to the Pascal triangle, which has only 2 infinite sets of initial conditions, namely, G(n, 0) = G(n+1, n+1) = 1 for n >= 0).

Pascal's triangle A007318 has s = 0 and is symmetric, arrays A140998 and A140993 have s = 1 (with e = 0 and e = 1, respectively), and arrays A140996 and A140995 have s = 3 (with e = 0 and e = 1, respectively).

If A(x,y) = Sum_{n,k >= 0} G(n, k)*x^n*y^k is the bivariate g.f. for this array (with G(n, k) = 0 for 0 <= n < k) and B(x, y) = Sum_{n, k} A140997(n, k)*x^n*y^k, then A(x, y) = B(x*y, y^(-1)). This can be proved using formal manipulation of double series expansions and the fact G(n, k) = A140997(n, n-k) for 0 <= k <= n.

If we let b(k) = lim_{n -> infinity} G(n, k) for k >= 0, then b(0) = 1, b(1) = 2, b(2) = 4, and b(k) = b(k-1) + 2*b(k-2) + b(k-3) for k >= 3. (The existence of the limit can be proved by induction on k.) It follows that b(k) = A141015(k) for k >= 0.

(End)

Examples

			Triangle begins:
  1
  1 1
  1 2 1
  1 2 4 1
  1 2 4 8 1
  1 2 4 9 15  1
  1 2 4 9 19 28  1
  1 2 4 9 19 40 52   1
  1 2 4 9 19 41 83  96   1
  1 2 4 9 19 41 88 170 177    1
  1 2 4 9 19 41 88 188 345  326    1
  1 2 4 9 19 41 88 189 400  694  600    1
  1 2 4 9 19 41 88 189 406  846 1386 1104 1
... [corrected by _Petros Hadjicostas_, Jun 12 2019]
E.g., G(12, 9) = G(9, 7) + G(9, 6) + G(10, 7) + G(11, 8) = 170 + 88 + 188 + 400 = 846.
		

Links

• Robert Price, Table of n, a(n) for n = 0..1325
• Juri-Stepan Gerasimov, Stepan's triangles and Pascal's triangle are connected by the recurrence relation ...

Crossrefs

Cf. A007318, A008937, A140993, A140995, A140996, A140997, A140998, A141015, A141018, A141020, A141021, A141031, A141065, A141066, A141067.

Programs

Maple

G := proc(n,k) if k=0 or n =k then 1; elif k= 1 then 2 ; elif k =2 then 4; elif k > n or k < 0 then 0 ; else procname(n-3,k-2)+procname(n-3,k-3)+procname(n-2,k-2)+procname(n-1,k-1) ; end if; end proc: seq(seq(G(n,k),k=0..n),n=0..15) ; # R. J. Mathar, Apr 14 2010

Mathematica

nlim = 50;
Do[G[n, 0] = 1, {n, 0, nlim}];
Do[G[n, n] = 1, {n, 1, nlim}];
Do[G[n + 2, 1] = 2, {n, 0, nlim}];
Do[G[n + 3, 2] = 4, {n, 0, nlim}];
Do[G[n + 4, m] =
   G[n + 1, m - 2] + G[n + 1, m - 3] + G[n + 2, m - 2] +
    G[n + 3, m - 1], {n, 0, nlim}, {m, 3, n + 3}];
A140994 = {}; For[n = 0, n <= nlim, n++,
 For[k = 0, k <= n, k++, AppendTo[A140994, G[n, k]]]];
A140994 (* Robert Price, Aug 19 2019 *)

Formula

From Petros Hadjicostas, Jun 12 2019: (Start)

G(n, k) = A140997(n, n-k) for 0 <= k <= n.

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

(End)

Extensions

Entries checked by R. J. Mathar, Apr 14 2010