|
|
A263326
|
|
Denominator of the rational number Sum_{d|n}1/(d+1).
|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
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
|
|
|
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))):
|
|
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
|
|
|
STATUS
|
approved
|
|
|
|