A092194 Primes p that divide A001008(k), the numerator of the k-th harmonic number H(k), for some k < p-1.

11, 29, 37, 43, 53, 61, 97, 109, 137, 173, 199, 227, 257, 269, 271, 313, 347, 353, 379, 397, 401, 409, 421, 433, 439, 509, 521, 577, 599, 601, 617, 641, 643, 647, 659, 677, 733, 761, 773, 809, 821, 827, 839, 863, 911, 919, 929, 937, 941, 947, 953, 971, 1009

Offset: 1

Author

T. D. Noe, Feb 24 2004

Keywords

• nonn

Comments

These primes are a subset of the non-harmonic primes A092102. Because these primes are analogous to the irregular primes A000928 that divide the numerators of Bernoulli numbers, they might be called H-irregular primes. The density of these primes is about 0.4 -- very close to the density of irregular primes.

These primes are called Harmonic irregular primes in the Wikipedia entry for "Regular prime" (see links). It may be noted that if p is known to be of this type and H(k) is the smallest Harmonic number divisible by p, then not only does k < p-1 hold, but k <= (p-1)/2. This is because, by symmetry, H(p-1-n) == H(n) (mod p), so that any eligible k lying between (p+1)/2 and p-1 would have a counterpart in the range between 1 and (p-1)/2. Furthermore, the minimal k cannot be exactly equal to (p-1)/2, because then p would be a Wieferich prime (A001220) and would also divide H(Int(p/4)). Thus k <= (p-3)/2, and this inequality is sharp because exact equality holds for p = 29, 37, 3373 (see A072984). - John Blythe Dobson, Apr 09 2015

Links

• Robert Israel, Table of n, a(n) for n = 1..2500
• Eric Weisstein's World of Mathematics, Harmonic Number
• Eric Weisstein's World of Mathematics, Irregular Prime
• Wikipedia, Regular prime

Crossrefs

Cf. A072984 (least k such that prime(n) divides A001008(k)).

Programs

Maple

A1008:= 1: h:= 1: for n from 2 to 10000 do h:= h+1/n; A1008:= A1008, numer(h) od:
A1008:= [A1008]:
filter:= proc(p) ormap(i -> A1008[i] mod p = 0, [$1..p-2]) end proc:
select(filter, [seq(ithprime(i),i=1..1000)]); # Robert Israel, Jan 20 2025

Mathematica

n=2; Table[cnt=0; While[cnt==0, p=Prime[n]; k=1; h=0; While[cnt==0 && k<=(p-1)/2, h=h+1/k; If[Mod[Numerator[h], p]==0, cnt++ ]; k++ ]; n++ ]; p, {100}]