A118917 Number of inequivalent primes in ring of integers Z[sqrt(2)] with absolute value of norm = n.

0, 0, 1, 0, 0, 0, 0, 2, 0, 1, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 2, 0, 1, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 2, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 2, 0

Offset: 0

Author

Franklin T. Adams-Watters, May 05 2006

Keywords

• easy
• nice
• nonn

Comments

Since there are infinitely many units in Z[sqrt(2)], the total number of primes with a given norm is infinite (when there are any).

Links

• Antti Karttunen, Table of n, a(n) for n = 0..65537

Crossrefs

Cf. A055029, A118916.

Programs

Mathematica

a[n_] := Which[PrimeQ[n] && MatchQ[Mod[n, 8], 1|7], 2, p = Sqrt[n]; n == 2 || IntegerQ[p] && PrimeQ[p] && MatchQ[Mod[p, 8], 3|5], 1, True, 0]; Table[a[n], {n, 0, 104}] (* Jean-François Alcover, Nov 22 2016 *)

PARI

A118917(n) = { my(p); if(isprime(n)&&((1==(n%8))||(7==(n%8))), 2, if((2==n)||((issquare(n,&p)&&isprime(p))&&((3==(p%8))||(5==(p%8)))),1,0)); }; \\ Antti Karttunen, Aug 30 2017

Formula

a(n) = 2 if n is a prime = 1,7 (mod 8); a(n) = 1 if n is 2 or p^2 where p is a prime = 3,5 (mod 8); otherwise a(n) = 0.