A334921 Expansion of Phi(x) = (1/(1+x))*Product_{k>=0} (1-(x/(1+x))^2^k).

1, -2, 2, 0, -6, 20, -48, 96, -166, 252, -340, 416, -480, 544, -544, 0, 2906, -13396, 44100, -121792, 296860, -652808, 1306560, -2377280, 3879136, -5461952, 5892512, -2171520, -11699616, 45871040, -114213888, 228427776, -377994406, 478195212, -252252460, -1013309824

Offset: 0

Author

Michel Marcus, May 16 2020

Keywords

• sign

Comments

The Hankel transforms of Phi(x) and Phi(x^2) are identical. See Theorem 2.8 in Han paper.

Links

• Michel Marcus, Table of n, a(n) for n = 0..1000
• Guo-Niu Han, Jacobi continued fraction and Hankel determinants of the Thue-Morse sequence, Quaestiones Mathematicae, 2016, 39 (7), pp.895-909. hal-02125285.

Crossrefs

Cf. A000120 (Hamming weight of n), A024036 (4^n - 1).

Programs

Mathematica

a[n_] := Sum[(-1)^(n - k) * Binomial[n, k] * (-1)^(2*n - 2*k + DigitCount[k, 2, 1]), {k, 0, n}]; Array[a, 36, 0] (* Amiram Eldar, May 16 2020 *)

PARI

a(n) = sum(k=0, n, (-1)^(n-k)*binomial(n, k)*(-1)^(2*n-2*k+hammingweight(k)));

Formula

a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n, k)*(-1)^(2*n-2*k+A000120(k)). See Theorem 2.8 in Han paper.

a(n) = 0 for positive terms of A024036.