A328892 If n = Product (p_j^k_j) then a(n) = Sum (2^(k_j - 1)).

0, 1, 1, 2, 1, 2, 1, 4, 2, 2, 1, 3, 1, 2, 2, 8, 1, 3, 1, 3, 2, 2, 1, 5, 2, 2, 4, 3, 1, 3, 1, 16, 2, 2, 2, 4, 1, 2, 2, 5, 1, 3, 1, 3, 3, 2, 1, 9, 2, 3, 2, 3, 1, 5, 2, 5, 2, 2, 1, 4, 1, 2, 3, 32, 2, 3, 1, 3, 2, 3, 1, 6, 1, 2, 3, 3, 2, 3, 1, 9, 8, 2, 1, 4, 2, 2, 2, 5, 1, 4

Offset: 1

Author

Ilya Gutkovskiy, Oct 29 2019

Keywords

• nonn

Examples

			a(72) = 6 because 72 = 2^3 * 3^2 and 2^(3 - 1) + 2^(2 - 1) = 6.
		

Links

• Index entries for sequences computed from exponents in factorization of n

Crossrefs

Cf. A000040 (positions of 1's), A008481, A011782, A162510, A324910.

Programs

Maple

a:= n-> add(2^(i[2]-1), i=ifactors(n)[2]):
seq(a(n), n=1..100);  # Alois P. Heinz, Oct 29 2019

Mathematica

a[1] = 0; a[n_] := Plus @@ (2^(#[[2]] - 1) & /@ FactorInteger[n]); Table[a[n], {n, 1, 90}]

PARI

a(n)={vecsum([2^(k-1) | k<-factor(n)[,2]])} \\ Andrew Howroyd, Oct 29 2019

Formula

If n = Product (p_j^k_j) then a(n) = Sum ordered partition(k_j).

Additive with a(p^e) = 2^(e-1).