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A376923 T(n, k) = T(n - 1, k) + 2^(n - 1)*T(n - 2, k - 1), if k > 0 and T(n, 0) = 2^n.

%I A376923 #108 Oct 19 2024 22:15:40
%S A376923 1,2,0,4,2,0,8,10,0,0,16,42,16,0,0,32,170,176,0,0,0,64,682,1520,512,0,
%T A376923 0,0,128,2730,12400,11776,0,0,0,0,256,10922,99696,206336,65536,0,0,0,
%U A376923 0,512,43690,798576,3380736,3080192,0,0,0,0,0,1024,174762,6390640,54425088,108724224,33554432,0,0,0,0,0
%N A376923 T(n, k) = T(n - 1, k) + 2^(n - 1)*T(n - 2, k - 1), if k > 0 and T(n, 0) = 2^n.
%C A376923 This is the case r = 2 of the more general recurrence: T(n, k, r) = T(n-1, k, r) + r^(n-1)*T(n - 2, k - 1, r), if k > 0 and T(n, 0, r) = 1 + (r^n - 1)/(r - 1) if r > 1. Consider the sequence b(n) = Sum_{k=0..n-1} b(n - k - 1)*T(n - 1, k, r)*(-1)^k, with b(0) = 1. The sequence b(n) will have an ordinary generating function which can be represented as the continued fraction expansion: 1/(1 - x/(1 - r^0*x/(1 - r^1*x/(1 - r^2*x/(1 - r^3*x/(...)))))). In short b(n) will have the ordinary generating function 1/(1-G(x)*x), where G(x) is the generating function of the Carlitz-Riordan q-Catalan numbers for q = r. The Hankel determinant of b(0)..b(2*n) will be r^A016061(n). The Hankel determinant of b(1)..b(2*n+1) will be r^A002412(n).
%F A376923 Column k has o.g.f.: x^(2*k)*2^(k^2)/((1 - 2^(k+1)*x)*Product_{m=1..k}(1 - 2^(m-1)*x)).
%e A376923 Triangle begins:
%e A376923 n\k  0 |  1 |  2 | 3 | 4 | 5
%e A376923 [0]  1,
%e A376923 [1]  2,   0
%e A376923 [2]  4,   2,   0
%e A376923 [3]  8,  10,   0,  0
%e A376923 [4] 16,  42,  16,  0,  0
%e A376923 [5] 32, 170, 176,  0,  0,  0
%o A376923 (PARI) T(n, k) = if(n < 0, return(0), return(if(k == 0, return(2^n), T(n-1,k) + 2^(n-1)*T(n-2,k-1))))
%Y A376923 Cf. A002412, A015083, A016061, A377132.
%K A376923 nonn,tabl
%O A376923 0,2
%A A376923 _Thomas Scheuerle_, Oct 17 2024