

A290815


Numbers m such that the numerator of Sum_{k=1..m, gcd(k,m) = 1} 1/k is divisible by m^3.


2



1, 39, 78, 155, 310, 465, 546, 793, 798, 930, 1092, 1586, 1638, 1860, 2170, 2379, 2394, 3172, 3276, 3965, 4340, 4758, 4914, 5219, 6045, 6510, 7137, 7182, 7930, 9516, 9828, 10374, 10438, 11102, 11895, 12090, 13020, 14274, 15657, 15860, 16843, 16891, 18135
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OFFSET

1,2


COMMENTS

A generalization of Wolstenholme primes (A088164) for composite number.
Leudesdorf proved in 1888 that the numerator of Sum_{k=1..n, gcd(k,n)=1} 1/k is divisible by n^2 for all (but not only) numbers n with gcd(n,6)=1, which is a generalization of Wolstenholme's theorem.
Terms that are coprime to 6: 1, 155, 793, 3965, 5219, 16843, 16891, 51305, ...
a(41) = A088164(1) = 16843.
A general conjecture: if, for some e > 0, m^e  Numerator(Sum_{k=1..m, gcd(k,m)=1} 1/k), then m^(e1)  Numerator(Sum_{k=1..m, gcd(k,m)=1} 1/k^2). Note: in this case, the exponent e = 3. Problem: are there numbers m > 1 such that m^4  Numerator(Sum_{k=1..m, gcd(k,m)=1} 1/k)?  Thomas Ordowski, Aug 10 2019
This general conjecture was checked up to 10^4. This problem has no solution up to 10^5.  Amiram Eldar, Aug 10 2019
It appears that all odd terms of this sequence are odd numbers m such that the numerator of Sum_{k=1..m, gcd(k,m)=1} 1/k^2 is divisible by m^2.  Thomas Ordowski, Aug 12 2019


REFERENCES

G. H. Hardy and E. M. Wright, Introduction to the theory of numbers, 5th edition, Oxford, England: Clarendon Press, pp. 100102, 1979.


LINKS

David A. Corneth, Table of n, a(n) for n = 1..174 (first 100 terms from Amiram Eldar, terms <= 4*10^5)
C. Leudesdorf, Some results in the elementary theory of numbers, Proceedings of the London Mathematical Society, Vol. 20 (1888), pp. 199212.
Eric Weisstein's World of Mathematics, Leudesdorf Theorem
Wikipedia, Wolstenholme's theorem.


EXAMPLE

Sum_{k=1..39, gcd(k,39)=1} 1/k = 46855131783993/15222026943200, and 46855131783993 = 39^3 * 789884047, thus 39 is in the sequence.


MATHEMATICA

seqQ[n_] := Module[{}, g[m_] := GCD[n, m] == 1; Divisible[Numerator[Plus @@ (1/Select[Range[n], g])], n^3]]; Select[Range[10^5], seqQ]


PROG

(PARI) isok(n) = numerator(sum(k=1, n, if (gcd(n, k)==1, 1/k))) % n^3 == 0; \\ Michel Marcus, Aug 11 2017
(PARI) upto(n) = {my(v = vector(n), d = divisors(n), res = List(), squarefreepart(n) = factorback(factorint(n)[, 1])); v[1] = 1; for(i = 2, n, v[i] = v[i1] + 1/i; ); for(j = 1, n, fr = v[j]; d = divisors(squarefreepart(j)); for(i = 2, #d, fr += 1/d[i] * v[j/d[i]] * (1)^omega(d[i]) ); if(numerator(fr) % j^3 == 0, listput(res, j); ) ); res } \\ David A. Corneth, Aug 23 2019


CROSSREFS

Cf. A088164, A093600.
Sequence in context: A044486 A072122 A355852 * A355857 A354227 A063335
Adjacent sequences: A290812 A290813 A290814 * A290816 A290817 A290818


KEYWORD

nonn


AUTHOR

Amiram Eldar, Aug 11 2017


STATUS

approved



