A073642 -id:A073642 - cp's OEIS frontend

A326032 a(2^x + ... + 2^z) = w(x) + ... + w(z), where x...z are distinct nonnegative integers and w = A000120.

0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 3, 3, 4, 4, 4, 4, 5

Offset: 0

Gus Wiseman, Jul 22 2019

nonn

From Robert Israel, Jul 23 2019: (Start)
a(2*n+1)=a(2*n).
a(n)=1 if and only if n > 1 is in A283526. (End)

			For example, a(6) = a(2^2 + 2^1) = w(2) + w(1) = 2.

Other sequences that are built by replacing 2^k in the binary representation with other numbers: A022290 (Fibonacci), A059590 (factorials), A073642, A089625 (primes), A116549, A326031.

Cf. A000120, A029931, A035327, A048793, A070939, A283526, A305830, A326031, A326669, A326702.

Bwt:= proc(n) option remember; convert(convert(n,base,2),`+`) end proc:
f:= proc(n) local L,i;
  L:= convert(n,base,2);
  add(L[i]*Bwt(i-1),i=1..nops(L))
end proc:
map(f, [$0..100]); # Robert Israel, Jul 23 2019

bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
Table[Total[Length/@bpe/@(bpe[n]-1)],{n,0,100}]

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

Original entry on oeis.org

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

Ilya Gutkovskiy, Mar 11 2025

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

			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.

Cf. A029931, A033922, A073642.

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 *)

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

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

cp's OEIS Frontend

A326032 a(2^x + ... + 2^z) = w(x) + ... + w(z), where x...z are distinct nonnegative integers and w = A000120.

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