A075892 Average of squares of successive primes: a(n) = (prime(n+1)^2 + prime(n)^2)/2, with n >= 2.

17, 37, 85, 145, 229, 325, 445, 685, 901, 1165, 1525, 1765, 2029, 2509, 3145, 3601, 4105, 4765, 5185, 5785, 6565, 7405, 8665, 9805, 10405, 11029, 11665, 12325, 14449, 16645, 17965, 19045, 20761, 22501, 23725, 25609, 27229, 28909, 30985, 32401

Offset

2,1

Comments

a(n) is prime for n in A240749. - Robert Israel, Jul 06 2017

If p and q are primes such that p > q > 3, then ((p^2 - q^2)/2, p*q, (p^2 + q^2)/2) is a primitive Pythagorean triple. - César Aguilera, Jun 02 2022

Links

Zak Seidov, Table of n, a(n) for n = 2..10001

Formula

a(n)^2 = A124434(n)^2 + A006094(n)^2. - César Aguilera, Jun 02 2022

Example

a(2)=17 because (prime(3)^2 + prime(2)^2)/2 = (5^2 + 3^2)/2 = 17.

Maple

seq((ithprime(i)^2 + ithprime(i+1)^2)/2, i=2..100); # Robert Israel, Jul 06 2017

Mathematica

Table[(Prime[n + 1]^2 + Prime[n]^2)/2, {n, 2, 50}] (* Vincenzo Librandi, Mar 07 2015 *)
p=2; q=3; Table[p=q; q=NextPrime[q]; (q^2+p^2)/2, {100}] (* Zak Seidov, Jul 06 2017 *)

Prog

(PARI) a(n) = (prime(n+1)^2+prime(n)^2)/2; \\ Michel Marcus, Oct 03 2013
(Magma) [(NthPrime(n+1)^2+NthPrime(n)^2)/2: n in [2..50]]; // Vincenzo Librandi, Mar 07 2015

Crossrefs

Cf. A006094, A124434, A143850, A240749.

Sequence in context: A319245 A093930 A048880 * A155143 A261529 A141886

Adjacent sequences: A075889 A075890 A075891 * A075893 A075894 A075895

Keyword

nonn,easy

Author

Zak Seidov, Oct 17 2002

Status

approved