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.

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, 0, 0, 128, 2730, 12400, 11776, 0, 0, 0, 0, 256, 10922, 99696, 206336, 65536, 0, 0, 0, 0, 512, 43690, 798576, 3380736, 3080192, 0, 0, 0, 0, 0, 1024, 174762, 6390640, 54425088, 108724224, 33554432, 0, 0, 0, 0, 0

Offset

0,2

Comments

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).

Links

Table of n, a(n) for n=0..65.

Formula

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)).

Example

Triangle begins:
n\k  0 |  1 |  2 | 3 | 4 | 5
[0]  1,
[1]  2,   0
[2]  4,   2,   0
[3]  8,  10,   0,  0
[4] 16,  42,  16,  0,  0
[5] 32, 170, 176,  0,  0,  0

Prog

(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))))

Crossrefs

Cf. A002412, A015083, A016061, A377132.

Sequence in context: A185879 A356918 A081880 * A144289 A293815 A373691

Adjacent sequences: A376920 A376921 A376922 * A376924 A376925 A376926

Keyword

nonn,tabl

Author

Thomas Scheuerle, Oct 17 2024

Status

approved