cp's OEIS Frontend

Equivalently, (binomial(2p,p)-2)/(2*p^3) where p runs through the primes >=5.

The values of this sequence's terms are replicated by conjectured general formula, given in A223886 (and also added to the formula section here) for k=2, j=1 and n>=3. - Alexander R. Povolotsky, Apr 18 2013

For p = prime(n), Boyd defines J_p to be the set of numbers k such that p divides A001008(k). This sequence gives the size of J_p. A072984 and A177734 give the smallest and largest elements of J_p, respectively.

A092101 gives primes prime(n) such that a(n) = 3 (i.e., a(A000720(A092101(m))) = 3 for all m). A092102 gives primes prime(n) such that a(n) > 3.

From Carlo Sanna, Apr 06 2016: (Start)

Eswarathasan and Levine conjectured that for any prime number p the set J_p is finite.

I proved that if J_p(x) is the number of integers in J_p that are less than x > 1, then J_p(x) < 129 p^(2/3) x^0.765 for any prime p. In particular, J_p has asymptotic density zero. (End)

Bing-Ling Wu and Yong-Gao Chen improved Sanna's (see previous comment) result showing that J_p(x) <= 3 x^(2/3 + 1/(25 log p)) for any prime p and any x > 1. - Carlo Sanna, Jan 12 2017

For any prime p, generation m consists of the numbers p^(m-1) <= k < p^m. The zeroth generation consists of just the number 0. When there is a k in generation m such that p divides A001008(k), then that k may generate solutions in generation m+1. It is conjectured that for all primes there are solutions for only a finite number of generations.

Boyd's table 3 states incorrectly that harmonic primes have 2 generations; harmonic primes have 3 generations.

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

For p = prime(n), Boyd defines J_p to be the set of numbers k such that p divides A001008(k). This sequence gives the largest element of J_p. The smallest element of J_p is given by A072984. The size of J_p is given by A092103.

Term a(23) is too large to include, see b-file. - Max Alekseyev, Apr 04 2025

a(n) = 0 iff A000040(n) is a Wolstenholme prime (given by A088164).

For n>2 and p=A000040(n), H(p^2-p) == H(p^2-1) == a(n)*p (mod p^2).

Wolstenholme's theorem states that prime p > 3 divides A007406(p-1). It is not difficult to show that this implies p also divides A007406((p-1)/2). In most instances, a(n) = (prime(n)-1)/2. Exceptions occur for primes in A093690, which have a smaller a(n).

Note that if p divides A007406(k) for k < (p-1)/2, then p divides A007406(p-k-1).

Another interesting observation: it appears that p=7 is the only prime that divides A007406(k) for some k > p-1; 7 divides A007406(26) = 23507608254234781649. Also note that when p > 3 and 2p-1 are both prime, they divide A007406(p-1).

Harmonic primes A092101 are a subset of these primes. Because these primes are analogous to the regular primes A007703 that divide the numerators of Bernoulli numbers, they might be called H-regular primes. The density of these primes is about 0.6 -- very close to the density of regular primes.