A125193 Smallest prime p such that p^2 divides the numerator of generalized harmonic number H((p-1)/2,2n) = Sum[ 1/k^(2n), {k,1,(p-1)/2} ].

7, 31, 127, 7, 5, 8191, 7, 2591, 149, 7, 11, 31, 7, 7, 5, 7, 17, 223, 7, 37, 431, 7, 23, 127, 5, 13, 23, 7, 29, 547, 7, 31, 11, 7, 5, 59, 7, 19, 13, 7, 41, 31, 7, 11, 5, 7, 31, 2371, 7

Offset: 1

Author

Alexander Adamchuk, Jan 13 2007

Keywords

• hard
• more
• nonn

Comments

Generalized harmonic number is H(n,m)= Sum[ 1/k^m, {k,1,n} ].

For prime p>3, p^2 divides H((p-1)/2,2p), implying that a(p)<=p. a(p)=p for prime p in {5,7,11,17,23,29,41,53,59,83,89,101,113,131,...}.

Note that many a(n) are of the form 2^m - 1 (for example, a(1) = 7, a(2) = 31, a(3) = 127, a(6) = 8191, etc.). a(n) = 5 for n = 5 + 10k, where k = {1,2,3,4,5,6,7,...}. a(n) = 7 for n = 1 + 3k, where k = {1,2,3,4,5,6,7,9,10,11,12,13,14,15,16,17,19,20,...}. a(n) = 31 for n = 2 + 5k, where k = {2,6,8,9,12,14,...}.

a(50) > 3*10^6.

a(51)-a(62) = {17,7,53,131,5,7,19,7,59,23,7,31}. a(64)-a(77) = {7,5,11,7,17,23,7,23,31,7,37,5,7,7}. a(79)-a(119) = {7,47,263,7,83,2543,5,43,29,7,89,103,7,23,23,7,5,16193,7,7,11,7,101,17,7,13,5,7,31,127,7,37,37,7,113,19,5,29,13,7,7}. a(121)-a(150) = {7,31,41,7,5,23,7,37,43,7,131,11,7,67,5,7,23,23,7,7,47,7,11,1847,5,37,31,7,47,127}.

Currently a(n) is unknown for n = {50,63,78,120,...}.

Links

• Eric Weisstein's World of Mathematics, Harmonic Number
• Eric Weisstein's World of Mathematics, Wolstenholme's Theorem

Crossrefs

Cf. A120290.

Extensions

a(48), a(84), a(96), a(144) from Max Alekseyev, Sep 12 2009