A095112 a(n) is the sum of n/k over all prime powers k > 1 which divide n.
0, 1, 1, 3, 1, 5, 1, 7, 4, 7, 1, 13, 1, 9, 8, 15, 1, 17, 1, 19, 10, 13, 1, 29, 6, 15, 13, 25, 1, 31, 1, 31, 14, 19, 12, 43, 1, 21, 16, 43, 1, 41, 1, 37, 29, 25, 1, 61, 8, 37, 20, 43, 1, 53, 16, 57, 22, 31, 1, 77, 1, 33, 37, 63, 18, 61, 1, 55, 26, 59, 1, 95, 1, 39, 43, 61, 18, 71, 1, 91, 40
Offset: 1
Keywords
Examples
The prime power divisors of 24 are 2, 4, 8 and 3, so a(24) = 24/2 + 24/4 + 24/8 + 24/3 = 29.
Links
- Antti Karttunen, Table of n, a(n) for n = 1..65537
- Jon Maiga, Computer-generated formulas for A095112, Sequence Machine.
Crossrefs
Programs
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Maple
with(numtheory): seq(add(bigomega(d)*phi(n/d),d in divisors(n)), n=1..60); # Ridouane Oudra, Oct 30 2023
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Mathematica
a[n_]:=Plus@@(n/Flatten[ #[[1]]^Range[ #[[2]]]&/@FactorInteger[n]])
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PARI
A095112(n) = sumdiv(n,d,(1==omega(d))*(n/d)); \\ Antti Karttunen, Feb 25 2018
Formula
a(n) = Sum_{k=1..n} bigomega(gcd(n,k)). - Lechoslaw Ratajczak, Jun 18 2017
Sum_{k=1..n} a(k) ~ A154945 * n*(n+1)/2. - Daniel Suteu, Apr 01 2019
a(n) = Sum_{d|n} bigomega(d)*phi(n/d). - Ridouane Oudra, Oct 30 2023
a(n) = Sum_{d|n} A116512(d). [From Sequence Machine] - Antti Karttunen, Nov 22 2023
Comments