login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

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
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 ?
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
Cf. A290815.
Sequence in context: A041759 A257442 A261374 * A061756 A236673 A119028
KEYWORD
nonn
AUTHOR
Amiram Eldar and Thomas Ordowski, Aug 13 2019
STATUS
approved