

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 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, 1979, pp. 100102.


LINKS



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



KEYWORD

nonn


AUTHOR



STATUS

approved



