OFFSET
1,2
COMMENTS
It is easy to show that a(n) is odd iff n is a square.
a(n) = sigma(n) for odd n, since any divisor of an odd number is odd.
LINKS
Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
FORMULA
Multiplicative with a(2^k) = k + 1, a(p^k) = sigma(p^k) = (p^(k+1)-1) / (p-1) for p > 2.
Sum_{k=1..n} a(k) ~ c * n^2, where c = Pi^2/18 = 0.548311... (A086463). - Amiram Eldar, Nov 04 2022
a(n) = Sum_{d|n} A000265(d). - Ridouane Oudra, Oct 30 2023
EXAMPLE
Divisors of 10 are 1, 2, 5, 10. The odd parts of these are 1, 1, 5, 5, so a(10) = 1+1+5+5 = 12.
MAPLE
with(numtheory): with(padic): seq(add(d/2^ordp(d, 2), d in divisors(n)), n=1..80); # Ridouane Oudra, Oct 30 2023
MATHEMATICA
f[p_, e_] := If[p == 2, e + 1, (p^(e + 1) - 1)/(p - 1)]; a[1] = 1; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Jun 30 2020 *)
PROG
(PARI) a(n)=my(k=valuation(n, 2)); sigma(n)\(2^(k+1)-1)*(k+1)
(Haskell)
a264740 = sum . map a000265 . a027750_row'
-- Reinhard Zumkeller, Nov 23 2015
CROSSREFS
KEYWORD
AUTHOR
Franklin T. Adams-Watters, Nov 22 2015
STATUS
approved