A380944 a(n) = b(n,A000120(n)) for n >= 0 where b(n,k) is defined in Comments.

1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 9, 3, 4, 1, 5, 4, 7, 3, 13, 5, 7, 2, 23, 9, 12, 3, 16, 4, 5, 1, 6, 5, 9, 4, 17, 7, 10, 3, 31, 13, 18, 5, 25, 7, 9, 2, 53, 23, 32, 9, 44, 12, 15, 3, 64, 16, 20, 4, 25, 5, 6, 1, 7, 6, 11, 5, 21, 9, 13, 4, 39, 17, 24, 7, 34, 10

Offset: 0

Author

Mikhail Kurkov, Feb 09 2025

Keywords

• nonn
• base

Comments

Here b(2n+1,k) = b(n,k) + b(n,k-1) for n >= 0, k > 0, b(2n,k) = (A000120(n)-k+1)*b(2n+1,k) for n > 0, k > 0 with T(n, 0) = 1 for n >= 0, T(0, k) = 0 for k > 0 (see A379817, A379819 for similar recurrence).

Crossrefs

Cf. A000120, A379817, A379819.

Programs

PARI

b(n,k) = if(k==0, 1, if(n==0, 0, if(n%2, b((n-1)/2,k) + b((n-1)/2,k-1), (hammingweight(n/2)-k+1)*b(n+1,k))))
a(n) = b(n, hammingweight(n))

Formula

Conjecture: a(2^m*(2^n-1)) = (n+1)^m - n*n!*c(m,n+1) for n >= 0, m >= 0 where c(n,k) = Sum_{i=0..n-k} Stirling2(k+i,k) for n >= 0, k >= 0.