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A290815
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Numbers m such that the numerator of Sum_{k=1..m, gcd(k,m) = 1} 1/k is divisible by m^3.
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2
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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
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1,2
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COMMENTS
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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^(e-1) | 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
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REFERENCES
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G. H. Hardy and E. M. Wright, Introduction to the theory of numbers, 5th edition, Oxford, England: Clarendon Press, 1979, pp. 100-102.
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LINKS
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EXAMPLE
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Sum_{k=1..39, gcd(k,39)=1} 1/k = 46855131783993/15222026943200, and 46855131783993 = 39^3 * 789884047, thus 39 is in the sequence.
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MATHEMATICA
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seqQ[n_] := Module[{}, g[m_] := GCD[n, m] == 1; Divisible[Numerator[Plus @@ (1/Select[Range[n], g])], n^3]]; Select[Range[10^5], seqQ]
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PROG
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(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[i-1] + 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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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