|
|
A309696
|
|
Numbers m such that the numerator of Sum_{k=1..m, gcd(k,m) = 1} 1/k^2 is divisible by m^2.
|
|
0
|
|
|
1, 39, 42, 78, 155, 156, 266, 310, 465, 546, 620, 793, 798, 930, 1092, 1586, 1596, 1638, 1860, 2170, 2184, 2379, 2394, 3172, 3276, 3720, 3965, 4340, 4758, 4788, 4914, 5219, 6045, 6344, 6510, 6552, 7137, 7182, 7930, 8680, 9516, 9828, 10374, 10438, 11102, 11895, 12090, 13020, 14274, 14364, 15657, 15860, 16843, 16891, 18135
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
Probably A290815 is a subset of these numbers. How to prove it?
Conjecture: odd terms of this sequence are odd terms of A290815.
Problem: are there numbers m > 1 such that the numerator of Sum_{k=1..m, gcd(k,m)=1} 1/k^2 is divisible by m^3 ?
|
|
LINKS
|
|
|
MATHEMATICA
|
aQ[n_] := Divisible[Numerator[Plus @@ ((1/Select[Range[n], CoprimeQ[n, #] &])^2)], n^2]; Select[Range[10^4], aQ]
|
|
PROG
|
(PARI) isok(m) = !(numerator(sum(k=1, m, if (gcd(k, m) == 1, 1/k^2))) % m^2); \\ Michel Marcus, Aug 13 2019
(Magma) v:= [Numerator(&+[1/k^2:k in [1..n]|Gcd(k, n) eq 1]):n in [1..10000]]; [m:m in [1..#v]| v[m] mod m^2 eq 0]; // Marius A. Burtea, Aug 14 2019
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|