A253108 Numbers k such that (sum of k^2 through (k+2)^2) + (k+1)^2 is prime.

2, 4, 6, 9, 14, 17, 20, 21, 25, 32, 34, 35, 40, 45, 49, 51, 52, 56, 60, 62, 65, 76, 80, 82, 86, 87, 89, 94, 95, 96, 100, 104, 105, 107, 112, 114, 115, 116, 117, 124, 126, 135, 137, 140, 145, 147, 151, 164, 167, 172, 174, 179, 180, 181, 182, 199, 200, 202, 205, 206, 207

Offset: 1

Author

César Aguilera, Dec 26 2014

Keywords

• nonn

Comments

Sequence is related to the Legendre conjecture.

No terms == 3 mod 5 or == 1 mod 7 or 0 mod 11. - Robert Israel, Jun 24 2015

Examples

			For n=2, n+1=3, n+2=4: we have
Sum(n^2,(n+1)^2)=Sum(2^2,3^2)=Sum(4,9)=Sum(4+5+6+7+8+9)=39,
Sum((n+1)^2,(n+2)^2)=Sum(3^2,4^2)=Sum(9,16)=Sum(9+10+11+12+13+14+15+16)=100,
39+100=139,
139 is prime; hence 2 is a term.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Programs

Maple

select(n -> isprime(4*n^3+14*n^2+20*n+11), [$1..1000]); # Robert Israel, Dec 28 2014

Mathematica

Select[Range[250],PrimeQ[Total[Range[#^2,(#+2)^2]]+(#+1)^2]&] (* Harvey P. Dale, Aug 04 2022 *)

PARI

for (n=1,1000,if(isprime(4*n^3+14*n^2+20*n+11),print1(n",")))

Extensions

a(47) corrected by Robert Israel, Jun 24 2015