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%I #43 Dec 23 2023 14:33:36
%S 1,1,2,2,4,5,1,8,12,4,16,28,13,1,32,64,38,6,64,144,104,25,1,128,320,
%T 272,88,8,256,704,688,280,41,1,512,1536,1696,832,170,10,1024,3328,
%U 4096,2352,620,61,1,2048,7168
%N Irregular triangle read by rows: T(n,k) = 2*T(n-1,k) + T(n-2,k-1) with T(0,0) = 0, T(n,0) = T(1,1) = 1 and T(n,k) = 0 if k < 0 or if n < k.
%C This is the pseudo-triangle whose successive lines are of the type T(n,0), T(n,1)+T(n-1,0), T(n,2)+T(n-1,1), ... T(n,k)+T(n-1,k-1), without 0's, with T=A201701. [e-mail, _Philippe Deléham_, Dec 04 2011]
%F T(n,k) = 2*T(n-1,k) + T(n-2,k-1) with T(0,0) = 0, T(n,0) = T(1,1) = 1 and T(n,k) = 0 if k < 0 or if n < k. - _Philippe Deléham_, Dec 05 2011
%F The n-th row polynomial appears to equal Sum_{k = 1..floor((n+1)/2)} binomial(n,2*k-1)*(1+t)^k. Cf. A034867. - _Peter Bala_, Sep 10 2012
%F Aside from the first two rows below, the signed coefficients appear in the expansion (b*x - 1)^2 / (a*b*x^2 - 2a*x + 1) = 1 + (2 a - 2 b)x + (4 a^2 - 5 a b + b^2)x^2 + (8 a^3 - 12 a^2b + 4 ab^2)x^3 + ..., the reciprocal of the derivative of x*(1-a*x) / (1-b*x). This is related to A263633 via the expansion (a*b*x^2 - 2a*x + 1) / (b*x - 1)^2 = 1 + (b - a) (2x + 3b x^2 + 4b^2 x^3 + ...). See also A201780. - _Tom Copeland_, Oct 30 2023
%e Triangle starts:
%e 1 1
%e 2 2
%e 4 5 1
%e 8 12 4
%e 16 28 13 1
%e 32 64 38 6
%e 64 144 104 25 1
%e 128 320 272 88 8
%e ...
%e Triangle begins (full version):
%e 0
%e 1, 1
%e 2, 2, 0
%e 4, 5, 1, 0
%e 8, 12, 4, 0, 0
%e 16, 28, 13, 1, 0, 0
%e 32, 64, 38, 6, 0, 0, 0
%e 64, 144, 104, 25, 1, 0, 0, 0
%e 128, 320, 272, 88, 8, 0, 0, 0, 0
%Y Cf. A052542 (row sums).
%Y Cf. A039991, A034867.
%Y Cf. A201780, A263633.
%K nonn,tabf
%O 0,3
%A _Paul Curtz_, Dec 02 2011
%E Edited and new name using _Philippe Deléham_'s formula, _Joerg Arndt_, Dec 13 2023
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