The OEIS is supported by the many generous donors to the OEIS Foundation.
%I #68 Oct 03 2021 15:28:43
%S 1,1,1,1,2,1,1,4,4,1,1,8,12,8,1,1,16,32,32,16,1,1,32,80,104,80,32,1,1,
%T 64,192,304,304,192,64,1,1,128,448,832,1008,832,448,128,1,1,256,1024,
%U 2176,3072,3072,2176,1024,256,1,1,512,2304,5504,8832,10272,8832,5504,2304,512,1
%N Triangle read by rows: T(n,k) = 2*(T(n-1,k-1) - T(n-2,k-1) + T(n-1,k)) for 0 < k < n, T(n,0) = T(n,n) = 1.
%C From _Petros Hadjicostas_, Feb 09 2021: (Start)
%C The rectangular version (R(n,k): n,k >= 0) of this symmetric triangular array (T(n,k): 0 <= k <= n) is given by R(n,k) = T(n+k,k) for n,k >= 0. Conversely, T(n,k) = R(n-k, k) for 0 <= k <= n.
%C Note that [o.g.f of R](x,y) = [o.g.f. of T](x, y/x) and [o.g.f of T](x,y) = [o.g.f of R](x,x*y). (End)
%C From _Petros Hadjicostas_, Feb 10 2021: (Start)
%C All the conjectures below are true because one has to prove only one of them, and the rest follow from the proved one.
%C As _Peter Luschny_ pointed out, one has to show only that the function S(n,k) = 2^n*hypergeom([-k + 1, n], [1], -1) satisfies the recurrence S(n,k) = 2*(S(n,k-1) - S(n-1,k-1) + S(n-1,k)) for n, k > 0 and the initial conditions S(n,0) = S(0,n) = 1 for n >= 0.
%C This is quite easy to achieve because S(n,k) = 2^n*Sum_{s=0}^{k-1} binomial(k-1,s)*binomial(n+s-1,s) for n >= 0 and k >= 1. The proof of the recurrence relies on the identity binomial(m,n) = binomial(m-1, n) + binomial(m-1,n-1).
%C Note that without the 2^n in the formula R(n,k) = 2^n*hypergeom([-k + 1, n], [1], -1), we essentially get array A049600.
%C In addition, note that without the 2^(n-k-1) in the formula T(n,k+1) = 2^(n-k-1)*hypergeom([-k, n-k+1], [1], -1), we essentially get A208341 (without the first column and the main diagonal of T). (End)
%H <a href="/index/Pas#Pascal">Index entries for triangles and arrays related to Pascal's triangle</a>
%F From _Petros Hadjicostas_, Feb 09 2021: (Start)
%F Formulas for the triangular array (T(n,k): 0 <= k <= n):
%F T(n,k) = T(n,n-k) for 0 <= k <= n.
%F Sum_{k=0..n} T(n,k) = A087161(n+1).
%F T(n,1) = T(n,n-1) = 2^(n-1) = A000079(n-1) for n >= 1.
%F T(n,2) = T(n,n-2) = (n-1)*2^(n-2) = A001787(n-1) for n >= 2.
%F T(n,3) = T(n,n-3) = (n^2-n-4)*2^(n-4) = A100312(n-3) for n >= 3.
%F T(n,floor(n/2)) = T(n,ceiling(n/2)) = A341344(n).
%F Bivariate o.g.f.: Sum_{n,k >= 0} T(n,k)*x^n*y^k = (3*x^2*y - 2*x*y - 2*x + 1)/((1 - x)*(-x*y + 1)*(2*x^2*y - 2*x*y - 2*x + 1)).
%F Conjecture based on _Peter Luschny_'s formulas in other sequences: T(n,k) = 2^(n-k)*hypergeom([-k + 1, n-k], [1], -1) = 2^k*hypergeom([-(n-k) + 1, k], [1], -1).
%F Formulas for the rectangular array (R(n,k): n,k >= 0):
%F R(n,k) = 2*(R(n,k-1) - R(n-1,k-1) + R(n-1,k)) for n,k > 0 with R(n,0) = R(0,n) = 1 for n >= 0.
%F R(n,k) = R(k, n) for n,k >= 0.
%F R(1,n) = R(n,1) = 2^n = A000079(n).
%F R(2,n) = R(n,2) = (n+1)*2^n = A001787(n+1).
%F R(3,n) = R(n,3) = (n^2+5*n+2)*2^(n-1) = A100312(n).
%F R(n,n) = A152254(n-1) = 2*A084773(n-1) for n >= 1.
%F Bivariate o.g.f.: Sum_{n,k >= 0} R(n,k)*x^n*y^k = (3*x*y - 2*x - 2*y - 1)/((1 - x)*(1 - y)*(2*x*y - 2*x - 2*y - 1)).
%F Conjecture based on _Peter Luschny_'s formulas in other sequences: R(n,k) = 2^n*hypergeom([-k + 1, n], [1], -1) = 2^k*hypergeom([-n + 1, k], [1], -1). (End)
%F From _Petros Hadjicostas_, Feb 10 2021: (Start)
%F The above conjecture is true (see the comments).
%F R(n,k) = 2^k*Sum_{s=0}^{n-1} binomial(n-1,s)*binomial(k+s-1,s) = 2^n*Sum_{s=0}^{k-1} binomial(k-1,s)*binomial(n+s-1,s) for n, k >= 1.
%F To get two binomial formulas for T(n,k), use the equation T(n,k) = R(n-k, k) for 1 <= k <= n and the above formulas for R(n,k). (End)
%e From _Petros Hadjicostas_, Feb 09 2021: (Start)
%e Triangle T(n,k) (with rows n >= 0 and columns 0 <= k <= n) begins:
%e 1,
%e 1, 1,
%e 1, 2, 1,
%e 1, 4, 4, 1,
%e 1, 8, 12, 8, 1,
%e 1, 16, 32, 32, 16, 1,
%e 1, 32, 80, 104, 80, 32, 1,
%e 1, 64, 192, 304, 304, 192, 64, 1,
%e 1, 128, 448, 832, 1008, 832, 448, 128, 1,
%e 1, 256, 1024, 2176, 3072, 3072, 2176, 1024, 256, 1,
%e ...
%e Rectangular array R(n,k) (with rows n >= 0 and columns k >= 0) begins:
%e 1, 1, 1, 1, 1, 1, 1, 1, ...
%e 1, 2, 4, 8, 16, 32, 64, 128, ...
%e 1, 4, 12, 32, 80, 192, 448, 1024, ...
%e 1, 8, 32, 104, 304, 832, 2176, 5504, ...
%e 1, 16, 80, 304, 1008, 3072, 8832, 24320, ...
%e 1, 32, 192, 832, 3072, 10272, 32064, 95104, ...
%e 1, 64, 448, 2176, 8832, 32064, 107712, 341504, ...
%e 1, 128, 1024, 5504, 24320, 95104, 341504, 1150592, ...
%e ... (End)
%t T[_, 0] = T[n_, n_] = 1;
%t T[n_, k_] /; 0<k<n := T[n, k] = 2(T[n-1, k-1]-T[n-2, k-1]+T[n-1, k]);
%t T[_, _] = 0;
%t Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Oct 03 2021 *)
%Y Cf. A000079, A001787, A049600, A084773, A087161, A100312, A152254, A208341, A341344.
%K nonn,tabl
%O 0,5
%A _Reinhard Zumkeller_, Dec 03 2004
%E Offset changed by _Petros Hadjicostas_, Feb 09 2021