OFFSET
1,1
COMMENTS
Conjecture: For any positive integers k and s, all the numbers Sum_{d|n}1/(d+k)^s (n = 1,2,3,...) have pairwise distinct fractional parts, and none of them is an integer.
This implies that a(n) > 1 for all n > 0.
See also A001157 for a similar conjecture involving Sum_{d|n}1/d^s.
I have verified that Sum_{d|n}1/(d+1) (n = 1..2*10^5) indeed have pairwise distinct fractional parts and none of them is an integer. For each k = 2,3,4,5,6 I have verified that Sum_{d|n}1/(d+k) (n = 1..10^5) have pairwise distinct fractional parts and none of them is integral. - Zhi-Wei Sun, Oct 20 2015.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
EXAMPLE
a(1) = 2 since sum_{d|1}1/(d+1) = 1/2.
a(2) = 6 since sum_{d|2}1/(d+1) = 1/2 + 1/3 = 5/6.
a(3) = 4 since sum_{d|3}1/(d+1) = 1/2 + 1/4 = 3/4.
MAPLE
f:= n -> denom(add(1/(d+1), d=numtheory:-divisors(n))):
map(f, [$1..100]); # Robert Israel, Oct 20 2015
MATHEMATICA
Dv[n_]:=Dv[n]=Divisors[n]
a[n_]:=a[n]=Denominator[Sum[1/(Part[Dv[n], i]+1), {i, 1, Length[Dv[n]]}]]
Do[Print[n, " ", a[n]], {n, 1, 50}]
PROG
(PARI) a(n) = denominator(sumdiv(n, d, 1/(d+1))); \\ Michel Marcus, Oct 15 2015
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Oct 14 2015
STATUS
approved