A081115 (p^2 - 1)/12 where p > 3 runs through the primes.

2, 4, 10, 14, 24, 30, 44, 70, 80, 114, 140, 154, 184, 234, 290, 310, 374, 420, 444, 520, 574, 660, 784, 850, 884, 954, 990, 1064, 1344, 1430, 1564, 1610, 1850, 1900, 2054, 2214, 2324, 2494, 2670, 2730, 3040, 3104, 3234, 3300, 3710, 4144, 4294, 4370, 4524

Offset

3,1

Comments

If p=4k+1, (p^2 - 1)/12 = Sum_{i=1..k} floor(sqrt(i*k)) (see links). - R. J. Mathar, Jul 07 2006

For n=1 and 2, the corresponding primes being 2 and 3, and a(n) is a fraction, not entered here. - Michel Marcus, Nov 11 2013

For prime p > 3, (p^2 - 1)/12 = (1/p)*Sum_{k=0..floor(p/2)} (p - k)*k. - Joseph Wheat, Feb 03 2018

Links

Muniru A Asiru, Table of n, a(n) for n = 3..5000

Hojoo Lee, Problems in Elementary Number Theory, p. 14, problem 10.

George Pólya and Gabor Szego, Problems and Theorems in Analysis II, p. 113, problem 20.

S. A. Shirali, A family portrait of primes -- a case study in discrimination, Math. Mag.. Vol. 70, No. 4 (Oct. 1997), pp. 263-272.

Formula

a(n) = j*(j+1)/3 where A000040(n)=2*j+1. - R. J. Mathar, Jul 07 2006

a(n) = (A001248(n) - 1)/12. - Vicente Izquierdo Gomez, May 25 2013

a(n) = 2*A024702(n). - R. J. Mathar, Jan 09 2017

a(n) = (prime(n)^2 - 1)/12 for n >= 3. - Jon E. Schoenfield, Dec 25 2019

Maple

seq((ithprime(p)^2-1)/12, p=3..20); # Muniru A Asiru, Feb 04 2018

Mathematica

(Prime[Range[3, 51]]^2 - 1)/12 (* Giovanni Resta, May 25 2013 *)

Prog

(PARI) a(n) = p = prime(n); (p^2-1)/12; \\ Michel Marcus, Nov 11 2013
(GAP) List(Filtered([5..20], IsPrime), p->(p^2-1)/12); # Muniru A Asiru, Feb 04 2018

Crossrefs

Sequence in context: A139480 A227388 A152749 * A053417 A082230 A236547

Adjacent sequences: A081112 A081113 A081114 * A081116 A081117 A081118

Keyword

nonn

Author

Benoit Cloitre, Apr 16 2003

Extensions

Offset set to 3 and edited by Michel Marcus, Nov 11 2013

Status

approved