A323355 a(1)=1; for n >= 2, a(n) = Sum_{i=1..A000120(n)} a(z(i)), where z(i) are the positions of 1's in the binary expansion of n, counted from left to right.

1, 1, 2, 1, 3, 2, 4, 1, 2, 3, 4, 2, 3, 4, 5, 1, 4, 2, 5, 3, 6, 4, 7, 2, 5, 3, 6, 4, 7, 5, 8, 1, 3, 4, 6, 2, 4, 5, 7, 3, 5, 6, 8, 4, 6, 7, 9, 2, 4, 5, 7, 3, 5, 6, 8, 4, 6, 7, 9, 5, 7, 8, 10, 1, 5, 3, 7, 4, 8, 6, 10, 2, 6, 4, 8, 5, 9, 7, 11, 3, 7, 5, 9, 6, 10, 8, 12, 4, 8, 6, 10, 7, 11, 9, 13, 2, 6, 4

Offset: 1

Author

Ctibor O. Zizka, Jan 12 2019

Keywords

• base
• nonn

Examples

			n=13, decimal 13 is 1101 in binary, the 1's are at positions 1,2,4. So a(13) = a(1) + a(2) + a(4).
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A007088, A000120.

Programs

Mathematica

a[1] = 1; a[n_] := a[n] = Total[a /@ (Position[IntegerDigits[n, 2], 1] // Flatten)]; Array[a, 100] (* Amiram Eldar, Jul 24 2023 *)

PARI

lista(nn) = {my(va = vector(nn), vb); va[1] = 1; for (n=2, nn, vb = binary(n); va[n] = sum(k=1, #vb, vb[k]*va[k]);); va;} \\ Michel Marcus, Jan 12 2019