OFFSET
1,1
COMMENTS
Squares of primes minus 1. - Wesley Ivan Hurt, Oct 11 2013
Integers k for which there exist exactly two positive integers b such that (k+1)/(b+1) is an integer. - Benedict W. J. Irwin, Jul 26 2016
LINKS
Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Barry Brent, On the constant terms of certain meromorphic modular forms for Hecke groups, arXiv:2212.12515 [math.NT], 2022.
Barry Brent, On the Constant Terms of Certain Laurent Series, Preprints (2023) 2023061164.
Nik Lygeros and Olivier Rozier, A new solution to the equation tau(p) == 0 (mod p), J. Int. Seq. 13 (2010), Article 10.7.4.
FORMULA
a(n) = prime(n)^2 - 1 = A001248(n) - 1. - Vladimir Joseph Stephan Orlovsky, Oct 17 2009
a(n) ~ n^2*log(n)^2. - Ilya Gutkovskiy, Jul 28 2016
a(n) = (1/2) * Sum_{|k|<=2*sqrt(p)} k^2*H(4*p-k^2) where H() is the Hurwitz class number and p is n-th prime. - Seiichi Manyama, Dec 31 2017
a(n) = 24 * A024702(n) for n > 2. - Jianing Song, Apr 28 2019
Sum_{n>=1} 1/a(n) = A154945. - Amiram Eldar, Nov 09 2020
From Amiram Eldar, Nov 07 2022: (Start)
Product_{n>=1} (1 + 1/a(n)) = Pi^2/6 (A013661).
Product_{n>=1} (1 - 1/a(n)) = A065469. (End)
MAPLE
MATHEMATICA
Table[Prime[n]^2 - 1, {n, 50}] (* Wesley Ivan Hurt, Oct 11 2013 *)
Prime[Range[50]]^2-1 (* Harvey P. Dale, Oct 02 2021 *)
PROG
(Haskell)
a084920 n = (p - 1) * (p + 1) where p = a000040 n
-- Reinhard Zumkeller, Aug 27 2013
(Magma) [p^2-1: p in PrimesUpTo(200)]; // Vincenzo Librandi, Mar 30 2015
(Sage) [(p-1)*(p+1) for p in primes(200)] # Bruno Berselli, Mar 30 2015
(PARI) a(n) = (prime(n)-1)*(prime(n)+1); \\ Michel Marcus, Jul 28 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Reinhard Zumkeller, Jun 11 2003
STATUS
approved