a(n) = 2^[n(n+1) - A000120(n)] * [x^n] 1/(1-x)^(1/2^n) for n>=0, where A000120(n) = number of 1's in binary expansion of n.

A134097

a(n) = 2^[n(n+1) - A000120(n)] * [x^n] 1/(1-x)^(1/2^n) for n>=0, where A000120(n) = number of 1's in binary expansion of n.

Terms

    a(0) =1a(1) =1a(2) =5a(3) =51a(4) =9163a(5) =1789359a(6) =2966784613

External references