cp's OEIS Frontend

Note that p^2 - 1 is always divisible by 24 since p == 1 or 2 (mod 3), so p^2 == 1 (mod 3) and p == 1, 3, 5, or 7 (mod 8) so p^2 == 1 (mod 8). - Michael B. Porter, Sep 02 2016

For n > 3 and m > 1, a(n) = A000330(m)/(2*m + 1), where 2*m + 1 = prime(n). For example, for m = 8, 2*m + 1 = 17 = prime(7), A000330(8) = 204, 204/17 = 12 = a(7). - Richard R. Forberg, Aug 20 2013

For primes => 5, a(n) == 0 or 2 (mod 5). - Richard R. Forberg, Aug 28 2013

The only primes in this sequence are 2, 5 and 7 (checked up to n = 10^7). The set of prime factors, however, appears to include all primes. - Richard R. Forberg, Feb 28 2015

Subsequence of generalized pentagonal numbers (cf. A001318): a(n) = k_n*(3*k_n - 1)/2, for k_n in {1, -1, 2, -2, 3, -3, 4, 5, -5, -6, 7, -7, 8, 9, 10, -10, ...} = A024699(n-2)*((A000040(n) mod 6) - 3)/2, n >= 3. - Daniel Forgues, Aug 02 2016

The only primes in this sequence are indeed 2, 5 and 7. For a prime p >= 5, if both p + 1 and p - 1 contains a prime factor > 3, then (p^2 - 1)/24 = (p + 1)*(p - 1)/24 contains at least 2 prime factors, so at least one of p + 1 and p - 1 is 3-smooth. Let's call it s. Also, If (p^2 - 1)/24 is a prime, then A001222(p^2-1) = 5. Since A001222(p+1) and A001222(p-1) are both at least 2, A001222(s) <= 5 - 2 = 3. From these we can see the only possible cases are p = 7, 11 and 13. - Jianing Song, Dec 28 2018

The trivial solutions of the congruence x^2 == 1 (mod 3*prime(n+2)), n>=1, with the primes prime(n+2) = A000040(n+2) have positive representatives 1 and 3*prime(n+2)-1. There are all-together four incongruent solutions due to a general theorem (see, e.g., the Hardy-Wright reference, Theorem 122, p. 96, and also A060594) and the fact that the number of incongruent solutions of this congruence with odd prime modulus p is two, namely with positive representative p and p-1 (see, e.g., Hardy-Wright, Theorem 109, p. 85). a(n) is the smallest positive odd representative >1 which solves this congruence. The other nontrivial even representative solving this congruence is 3*prime(n+2) - a(n), i.e. 4, 8, 10, 14, 16, 20, ... See 2*A207336.

a(n) solves also the congruence x^2 == 1 (Modd A001748(n+2)), n>=1. For Modd n (not to be confused with mod n) see a comment on A203571. This follows from floor(a(n)^2/3*prime(n+2)) being even, in fact it is 8*A024699(n) (see a comment there), hence a(n)^2 (Modd 3*prime(n+2)) = a(n)^2 (mod 3*prime(n+2)) = 1. For those multiplicative groups Modd 3*p with p an odd prime which are cyclic (this is not possible in the mod case, see A033949), a(n) is the representative of the only other nontrivial solution of this congruence. The representative of the trivial solution is 1 (-1 belongs to the same Modd class). (The conjecture stated here earlier is wrong, that is, the multiplicative group Modd (91=7*13) is non-cyclic. It may still be true for 3*p. - Wolfdieter Lang, Mar 15 2012)

Offset is 3 because 5=prime(3) is the first prime of the given form. It is provable that if 6m-1 and 6m+1 are a pair of twin primes, then for all k, 0