A084920 a(n) = (prime(n)-1)*(prime(n)+1).

3, 8, 24, 48, 120, 168, 288, 360, 528, 840, 960, 1368, 1680, 1848, 2208, 2808, 3480, 3720, 4488, 5040, 5328, 6240, 6888, 7920, 9408, 10200, 10608, 11448, 11880, 12768, 16128, 17160, 18768, 19320, 22200, 22800, 24648, 26568, 27888, 29928

Offset: 1

Author

Reinhard Zumkeller, Jun 11 2003

Keywords

• nonn
• easy

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.
• Lực Ta, Enumeration of virtual quandles up to isomorphism, arXiv:2506.16536 [math.GT], 2025. See p. 6.

Crossrefs

Cf. A000040, A005563, A049001, A154945, A166010, A182200, A182174.

Cf. A006093, A008864, A084921, A084922.

Cf. A066872, A001248, A024702, A013661, A065469.

Programs

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

Maple

A084920:=n->ithprime(n)^2-1; seq(A084920(k), k=1..50); # Wesley Ivan Hurt, Oct 11 2013

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 *)

PARI

a(n) = (prime(n)-1)*(prime(n)+1); \\ Michel Marcus, Jul 28 2016

Sage

[(p-1)*(p+1) for p in primes(200)] # Bruno Berselli, Mar 30 2015

Formula

a(n) = A006093(n) * A008864(n);

a(n) = A084921(n)*2, for n > 1; a(n) = A084922(n)*6, for n > 2.

Product_{n > 0} a(n)/A066872(n) = 2/5. a(n) = A001248(n) - 1. - R. J. Mathar, Feb 01 2009

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)