A240130 Least prime of the form prime(n)^2 + k^2, or 0 if none.
5, 13, 29, 53, 137, 173, 293, 397, 593, 857, 977, 1373, 1697, 1913, 2213, 2909, 3517, 3821, 4493, 5077, 5333, 6257, 7213, 7937, 9413, 10301, 10613, 11549, 11897, 13093, 16193, 17417, 18773, 19421, 22397, 22817, 24749, 26573, 27893, 30029
Offset: 1
Keywords
Examples
Prime(2) = 3 and 3^2 + 1^2 = 10 is not prime but 3^2 + 2^2 = 13 is prime, so a(2) = 13.
Links
- Robert Israel, Table of n, a(n) for n = 1..10000
- Stephan Baier and Liangyi Zhao, On Primes Represented by Quadratic Polynomials, Anatomy of Integers, CRM Proc. & Lecture Notes, Vol. 46, Amer. Math. Soc. 2008, pp. 169 - 166.
- Étienne Fouvry and Henryk Iwaniec, Gaussian primes, Acta Arithmetica 79:3 (1997), pp. 249-287.
- E.W. Weisstein, Fermat's 4n+1 Theorem, MathWorld.
- Wikipedia, Bunyakovsky's conjecture
Programs
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Maple
g:= proc(p) local k; for k from 2 by 2 do if isprime(p^2 + k^2) then return p^2+k^2 fi od end proc: g(2):= 5: seq(g(ithprime(i)),i=1..1000); # Robert Israel, Nov 04 2015
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Mathematica
Table[First[Select[Prime[n]^2 + Range[20]^2, PrimeQ]], {n, 40}]
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PARI
a(n) = {p = prime(n); k = 1 - p%2; inc = 2; while (!isprime(q=p^2+k^2), k += inc); q;} \\ Michel Marcus, Nov 04 2015
Formula
a(n) == 1 (mod 4) if a(n) > 0.
a(n) > 0 if Bunyakovsky's conjecture is true.
a(n) <> a(m) if n <> m and a(n) > 0, by uniqueness in Fermat's 4n+1 Theorem.
a(n) = prime(n)^2 + A240131(n)^2 if a(n) > 0.
Comments