OFFSET
1,2
COMMENTS
a(n) is multiplicative with a(p^e) = -1 + 2^(e+1).
If s is squarefree then a(s) = A048691(s).
More generally: Let a_q(n) be multiplicative with a_q(p^e) = (q^(e+1)-1)/ (q-1) for prime p, e >= 0 and some fixed integer q. Then a_q(n) is the inverse Moebius transform of the completely multiplicative sequence b_q(n) = q^bigomega(n) with b_q(p) = q and b_q(1) = 1. For q = 1 see a_q(n) = A000005(n) and b_q(n) = A000012(n), for q = 0 see a_q(n) = A000012(n) and b_q(n) = A000007(n) with offset 1, and for q = -1 see a_q(n) = A010052(n) with offset 1 and b_q(n) = A008836(n). - Werner Schulte, Feb 20 2019
LINKS
Enrique Pérez Herrero, Table of n, a(n) for n = 1..5000
FORMULA
EXAMPLE
a(12) = a(2^2 * 3^1) = (-1 + 2^(2+1)) * (-1 + 2^(1+1)) = 7 * 3 = 21; or, using the divisors set {1,2,3,4,6,12}: 2^0 + 2^1 + 2^1 + 2^2 + 2^2 + 2^3 = 21.
MATHEMATICA
t[n_] := DivisorSum[n, 2^PrimeOmega[#]&]; Table[t[n], {n, 100}]
PROG
(PARI) for(n=1, 100, print1(direuler(p=2, n, 1/(1 - X)/(1 - 2*X))[n], ", ")) \\ Vaclav Kotesovec, Mar 14 2023
CROSSREFS
KEYWORD
nonn,mult
AUTHOR
Enrique Pérez Herrero, Apr 14 2012
STATUS
approved