A381957 If n = Sum 2^e(k), then a(n) = Sum 2^a(e(k)), with a(0) = 1.

1, 2, 4, 6, 16, 18, 20, 22, 64, 66, 68, 70, 80, 82, 84, 86, 65536, 65538, 65540, 65542, 65552, 65554, 65556, 65558, 65600, 65602, 65604, 65606, 65616, 65618, 65620, 65622, 262144, 262146, 262148, 262150, 262160, 262162, 262164, 262166, 262208, 262210, 262212, 262214, 262224, 262226

Offset: 0

Author

Ilya Gutkovskiy, Mar 11 2025

Keywords

• nonn
• base

Comments

Replace 2^k in the binary representation of n with 2^a(k).

Examples

			25 = 2^4 + 2^3 + 2^0, hence a(25) = 2^a(4) + 2^a(3) + 2^a(0) = 2^16 + 2^6 + 2^1 = 65602.
		

Crossrefs

Cf. A029931, A033922, A073642.

Programs

Mathematica

e[n_] := -1 + Position[Reverse[IntegerDigits[n, 2]], 1] // Flatten; a[0] = 1; a[n_] := a[n] = Total[2^a /@ e[n]]; Array[a, 50, 0] (* Amiram Eldar, Mar 11 2025 *)

PARI

a(n) = if (n==0, 1, my(v=Vecrev(binary(n))); sum(k=1, #v, if (v[k], 2^a(k-1)))); \\ Michel Marcus, Mar 11 2025

Formula

G.f.: 1 + (1/(1 - x)) * Sum_{k>=0} 2^a(k) * x^(2^k) / (1 + x^(2^k)).