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%I #61 Mar 17 2024 05:53:49
%S 1,1,1,1,2,1,1,4,2,1,1,8,4,2,1,1,16,8,4,2,1,1,31,17,8,4,2,1,1,60,35,
%T 17,8,4,2,1,1,116,72,35,17,8,4,2,1,1,224,148,72,35,17,8,4,2,1,1,432,
%U 303,149,72,35,17,8,4,2,1,1,833,618,308,149,72,35,17,8,4,2,1,1,1606,1257,636,308,149,72,35,17,8,4,2,1
%N Triangle G(n, k) read by rows for 0 <= k <= n, where G(n, 0) = G(n+1, n+1) = 1, G(n+2, n+1) = 2, G(n+3, n+1) = 4, G(n+4, n+1) = 8, and G(n+5, m) = G(n+1, m-1) + G(n+1, m) + G(n+2, m) + G(n+3, m) + G(n+4, m) for n >= 0 for m = 1..(n+1).
%C From _Petros Hadjicostas_, Jun 12 2019: (Start)
%C This is a mirror image of the triangular array A140995. The current array has index of asymmetry s = 3 and index of obliqueness (obliquity) e = 0. Array A140995 has the same index of asymmetry, but has index of obliqueness e = 1. (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.)
%C 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).
%C Pascal's triangle A007318 has s = 0 and is symmetric, arrays A140998 and A140993 have s = 1 (with e = 0 and e = 1, respectively), arrays A140997 and A140994 have s = 2 (with e = 0 and e = 1, respectively), and arrays A141020 and A141021 have s = 4 (with e = 0 and e = 1, respectively).
%C (End)
%H Robert Price, <a href="/A140996/b140996.txt">Table of n, a(n) for n = 0..5150</a>
%H Juri-Stepan Gerasimov, <a href="/A140998/a140998.jpg">Stepan's triangles and Pascal's triangle are connected by the recurrence relation ...</a>
%F From _Petros Hadjicostas_, Jun 12 2019: (Start)
%F G(n, k) = A140995(n, n - k) for 0 <= k <= n.
%F Bivariate g.f.: Sum_{n,k >= 0} G(n, k)*x^n*y^k = (1 - x - x^2 - x^3 - x^4 + x^2*y + x^3*y + x^5*y)/((1 - x) * (1 - x*y) * (1 - x - x^2 - x^3 - x^4 - x^4*y)).
%F If we take the first derivative of the bivariate g.f. w.r.t. y and set y = 0, we get the g.f. of column k = 1: x/((1 - x) * (1 - x - x^2 - x^3 - x^4)). This is the g.f. of a shifted version of sequence A107066.
%F Substituting y = 1 in the above bivariate function and simplifying, we get the g.f. of row sums: 1/(1 - 2*x). Hence, the row sums are powers of 2; i.e., A000079.
%F (End)
%e Triangle (with rows n >= 0 and columns k >= 0) begins as follows:
%e 1
%e 1 1
%e 1 2 1
%e 1 4 2 1
%e 1 8 4 2 1
%e 1 16 8 4 2 1
%e 1 31 17 8 4 2 1
%e 1 60 35 17 8 4 2 1
%e 1 116 72 35 17 8 4 2 1
%e 1 224 148 72 35 17 8 4 2 1
%e 1 432 303 149 72 35 17 8 4 2 1
%e 1 833 618 308 149 72 35 17 8 4 2 1
%e ...
%t nlim = 100;
%t For[n = 0, n <= nlim, n++, G[n, 0] = 1];
%t For[n = 1, n <= nlim, n++, G[n, n] = 1];
%t For[n = 2, n <= nlim, n++, G[n, n-1] = 2];
%t For[n = 3, n <= nlim, n++, G[n, n-2] = 4];
%t For[n = 4, n <= nlim, n++, G[n, n-3] = 8];
%t For[n = 5, n <= nlim, n++, For[k = 1, k < n - 3, k++,
%t G[n, k] = G[n-4, k-1] + G[n-4, k] + G[n-3, k] + G[n-2, k] + G[n-1, k]]];
%t A140996 = {}; For[n = 0, n <= nlim, n++,
%t For[k = 0, k <= n, k++, AppendTo[A140996, G[n, k]]]];
%t A140996 (* _Robert Price_, Jul 03 2019 *)
%t G[n_, k_] := G[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, True, G[n - 1, k] + G[n - 2, k] + G[n - 3, k] + G[n - 4, k] + G[n - 4, k - 1]];
%t Table[G[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Jan 28 2024 *)
%Y Cf. A007318, A107066, A140993, A140994, A140995, A140997, A140998, A141020, A141021, A141031, A141065, A141066, A141067, A141068, A141069, A141070, A141072, A141073, A309462.
%K nonn,tabl
%O 0,5
%A _Juri-Stepan Gerasimov_, Jul 08 2008
%E Name edited by _Petros Hadjicostas_, Jun 12 2019
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