A263326 Denominator of the rational number Sum_{d|n}1/(d+1).

2, 6, 4, 30, 3, 84, 8, 90, 20, 11, 12, 5460, 7, 40, 48, 1530, 9, 7980, 20, 1155, 88, 276, 24, 81900, 78, 189, 35, 1160, 15, 38192, 32, 16830, 51, 315, 72, 3838380, 19, 780, 280, 142065, 21, 132440, 44, 828, 5520, 376, 48, 9746100, 200, 14586

Offset: 1

Author

Zhi-Wei Sun, Oct 14 2015

Keywords

• nonn

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.

Examples

			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.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000203, A001157, A263317, A263319, A263325.

Programs

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}]

PARI

a(n) = denominator(sumdiv(n, d, 1/(d+1))); \\ Michel Marcus, Oct 15 2015