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%I #33 Dec 18 2019 08:56:28
%S 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,
%T 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,
%U 276,136,67,33,16,8,4,2,1,1,944,560,276,136,67,33,16,8,4,2,1
%N Pascal-like triangle with index of asymmetry y = 4 and index of obliqueness z = 0.
%C The left column is set to 1. The four rightmost columns start with powers of 2:
%C 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.
%C 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.
%C From _Petros Hadjicostas_, Jun 14 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 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.
%C (End)
%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 14 2019: (Start)
%F T(n, k) = A141021(n, n-k) for 0 <= k <= n.
%F 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)).
%F 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.
%F (End)
%e Pascal-like triangle with y = 4 and z = 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 32 16 8 4 2 1
%e 1 63 33 16 8 4 2 1
%e 1 124 67 33 16 8 4 2 1
%e 1 244 136 67 33 16 8 4 2 1
%e 1 480 276 136 67 33 16 8 4 2 1
%e 1 944 560 276 136 67 33 16 8 4 2 1
%e ...
%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 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
%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 Table[T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Dec 18 2019, after _R. J. Mathar_ *)
%Y Cf. A001949, A007318, A140993, A140994, A140995, A140996, A140997, A140998, A141021, A141031, A141064, A141065, A141066, A141067, A141069, A141070, A141072, A141073.
%K nonn,tabl
%O 0,5
%A _Juri-Stepan Gerasimov_, Jul 11 2008
%E Partially edited by _N. J. A. Sloane_, Jul 18 2008
%E Recurrence rewritten by _R. J. Mathar_, Sep 19 2008
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