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%I #48 Apr 24 2020 07:52:51
%S 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,
%T 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,
%U 16,33,67,136,276,480,1
%N Pascal-like triangle with index of asymmetry y = 4 and index of obliqueness z = 1.
%C The triangle here is A141020 with each row reversed.
%C From _Petros Hadjicostas_, Jun 16 2019: (Start)
%C 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).
%C 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.
%C 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.
%C 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).
%C 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.
%C 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.
%C 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.
%C 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)).
%C (End)
%H O. Dunkel, <a href="http://www.jstor.org/stable/2298801">Solutions of a probability difference equation</a>, Amer. Math. Monthly, 32 (1925), 354-370; see p. 356 with r = s + 1 (where s is the index of asymmetry).
%H Juri-Stepan Gerasimov, <a href="/A140998/a140998.jpg">Stepan's triangles and Pascal's triangle are connected by the recurrence relation ...</a>
%F T(n, k) = A141020(n, n-k). - _R. J. Mathar_, Sep 19 2008
%F From _Petros Hadjicostas_, Jun 16 2019: (Start)
%F 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.
%F 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)).
%F 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.
%F 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.
%F (End)
%e Pascal-like triangle with y = 4 and z = 1 (with rows n >= 0 and columns k >= 0) begins as follows:
%e 1
%e 1 1
%e 1 2 1
%e 1 2 4 1
%e 1 2 4 8 1
%e 1 2 4 8 16 1
%e 1 2 4 8 16 32 1
%e 1 2 4 8 16 33 63 1
%e 1 2 4 8 16 33 67 124 1
%e 1 2 4 8 16 33 67 136 244 1
%e 1 2 4 8 16 33 67 136 276 480 1
%e 1 2 4 8 16 33 67 136 276 560 944 1
%e ...
%p # This is a slight modification of _R. J. Mathar_'s Maple program from array A141020:
%p 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:
%p A141021 := proc(n, k) A141020(n, n-k): end:
%p 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
%t 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 T[n_, k_] := t[n, n - k];
%t Table[Table[T[n, k], {k, 0, n}], {n, 0, 10}] // Flatten (* _Jean-François Alcover_, Apr 24 2020 *)
%Y Cf. A007318, A140993, A140994, A140995, A140996, A140997, A140998, A141020, A141031, A172119, A308808.
%K nonn
%O 0,5
%A _Juri-Stepan Gerasimov_, Jul 11 2008
%E Partially edited by _N. J. A. Sloane_, Jul 18 2008
%E Comment simplified by _R. J. Mathar_, Sep 19 2008
%E Data corrected by _Jean-François Alcover_, Apr 24 2020
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