A322434 Numbers k that are the numerator of a harmonic number such that k is divisible by the square of a prime >= 5.

25, 49, 363, 7381, 86021, 2436559, 14274301, 19093197, 315404588903, 9227046511387, 9304682830147, 54801925434709, 2078178381193813, 12309312989335019, 5943339269060627227, 14063600165435720745359, 254381445831833111660789, 15117092380124150817026911

Offset: 1

Author

Bernard Schott, Dec 08 2018

Keywords

• nonn
• frac

Comments

From Bernard Schott, Dec 09 2018: (Start)

In the data, there are two families of numerators.

1) Numerators of harmonic numbers H_{p-1} which are divisible by p^2 for p >= 5, by Wolstenholme's theorem; these numerators are in A076637.

2) Numerators of harmonic numbers which are also divisible by squares of primes, but not as a result of Wolstenholme's theorem. E.g., the numbers 363, numerator of H_7, and 9227046511387, numerator of H_{29}, are respectively divisible by 11^2 and 43^2. Up through H_{60}, only the two numerators of H_7 and H_{29} belong to this second family.

(End)

The third term in the second family is the numerator of H_{297} ~ 1.153... * 10^129 which is divisible by 1153^2, and the fourth numerator of H_k has k > 335. - Amiram Eldar, Dec 09 2018

Numbers n <= 50000 such that 11^2 divides the numerator of the n-th harmonic number: 7, 10 (by Wolstenholme's theorem), 848, 9328, 9338, 10583, 10591, 102718, 102721, 102728, 116413, 116423. - Jon E. Schoenfield, Dec 09 2018

Examples

			49 is a term because the numerator of the harmonic number H_6 = 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 = 49/20 is divisible by the square of 7 (by Wolstenholme's Theorem).
363 is a term because the numerator of the harmonic number H_7 = 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 = 363/140 is divisible by the square of 11.
		

Links

• Hisanori Mishima, Wolstenholme Numbers, WIFC (World Integer Factorization Center).

Crossrefs

Cf. A001008, A076637.

Programs

Mathematica

Select[Numerator@ HarmonicNumber@ Range@ 60, AnyTrue[FactorInteger[#], And[First[#] > 2, Last[#] > 1] &] &] (* Michael De Vlieger, Dec 08 2018 *)

PARI

isok(nh) = {my(f = factor(nh)); for (i=1, #f~, if ((f[i, 1] >= 5) && (f[i, 2] >= 2), return (1)); ); return (0); }
lista(nn) = {my(h = 0, nh); for (n=1, nn, h += 1/n; nh = numerator(h); if (isok(nh), print1(nh, ", "); ); ); } \\ _Michel Marcus, Dec 08 2018