cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A290815 Numbers m such that the numerator of Sum_{k=1..m, gcd(k,m) = 1} 1/k is divisible by m^3.

Original entry on oeis.org

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
Offset: 1

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Author

Amiram Eldar, Aug 11 2017

Keywords

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^(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

Examples

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

References

  • G. H. Hardy and E. M. Wright, Introduction to the theory of numbers, 5th edition, Oxford, England: Clarendon Press, 1979, pp. 100-102.

Crossrefs

Programs

  • 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]
  • 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