A201509 - OEIS

%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