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A081115
(p^2 - 1)/12 where p > 3 runs through the primes.
5
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
Hojoo Lee, Problems in Elementary Number Theory, p. 14, problem 10.
George PĆ³lya 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
KEYWORD
nonn
AUTHOR
Benoit Cloitre, Apr 16 2003
EXTENSIONS
Offset set to 3 and edited by Michel Marcus, Nov 11 2013
STATUS
approved