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Numbers k that are the numerator of a harmonic number such that k is divisible by the square of a prime >= 5.
2

%I #43 Dec 15 2024 07:20:31

%S 25,49,363,7381,86021,2436559,14274301,19093197,315404588903,

%T 9227046511387,9304682830147,54801925434709,2078178381193813,

%U 12309312989335019,5943339269060627227,14063600165435720745359,254381445831833111660789,15117092380124150817026911

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

%C From _Bernard Schott_, Dec 09 2018: (Start)

%C In the data, there are two families of numerators.

%C 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.

%C 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.

%C (End)

%C 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

%C 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

%H Hisanori Mishima, Wolstenholme Numbers, <a href="http://www.asahi-net.or.jp/~KC2H-MSM/mathland/matha1/index.htm">WIFC (World Integer Factorization Center)</a>.

%e 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).

%e 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.

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

%o (PARI) isok(nh) = {my(f = factor(nh)); for (i=1, #f~, if ((f[i, 1] >= 5) && (f[i, 2] >= 2), return (1)); ); return (0); }

%o 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

%Y Cf. A001008, A076637.

%K nonn,frac

%O 1,1

%A _Bernard Schott_, Dec 08 2018