A309696 Numbers m such that the numerator of Sum_{k=1..m, gcd(k,m) = 1} 1/k^2 is divisible by m^2.

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

Offset: 1

Author

Amiram Eldar and Thomas Ordowski, Aug 13 2019

Keywords

• nonn

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 ?

Crossrefs

Cf. A290815.

Programs

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

Mathematica

aQ[n_] := Divisible[Numerator[Plus @@ ((1/Select[Range[n], CoprimeQ[n, #] &])^2)], n^2]; Select[Range[10^4], aQ]

PARI

isok(m) = !(numerator(sum(k=1, m, if (gcd(k,m) == 1, 1/k^2))) % m^2); \\ Michel Marcus, Aug 13 2019