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Least prime p = prime(n)^2 + prime(n+1)^2 + prime(n+2)^2 + prime(n+3)^2 + q^2, where q > prime(n+3) is also prime.
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%I #12 Nov 04 2015 11:41:14

%S 373,653,1997,1901,2309,3389,4373,5381,6701,8069,10589,12269,18269,

%T 18461,19541,22973,24821,29021,32909,38261,46589,45869,50021,53549,

%U 56909,66029,69389,77261,87629,93581,102101,107741,118901,128981,131837,145517,152909,159869,170021,188261,184901,196661,214469,229781,237821,252509,277157,281429,291101,305933,317693,333029,344021,359981,370661,387341,395069,418349,460949

%N Least prime p = prime(n)^2 + prime(n+1)^2 + prime(n+2)^2 + prime(n+3)^2 + q^2, where q > prime(n+3) is also prime.

%C The corresponding prime q is in A263725.

%C The prime p exists for all n > 1 under Schinzel's Hypothesis H; see Sierpinski (1988), p. 221.

%C If q = prime(n+4), then p is in A133559 (prime sums of squares of 5 consecutive primes). The converse holds if a(n) != a(m) when n != m (which holds if a(n) < a(n+1), as appears to be true).

%D W. Sierpinski, Elementary Theory of Numbers, 2nd English edition, revised and enlarged by A. Schinzel, Elsevier, 1988; see p. 221.

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Schinzel%27s_hypothesis_H">Schinzel's Hypothesis H</a>

%e The primes 373 = 3^2 + 5^2 + 7^2 + 11^2 + 13^2, 653 = 5^2 + 7^2 + 11^2 + 13^2 + 17^2, and 1997 = 7^2 + 11^2 + 13^2 + 17^2 + 37^2 lead to a(1) = 373, a(2) = 653, and a(3) = 1997.

%t Table[k = 4;

%t While[p = Sum[Prime[n + j]^2, {j, 0, 3}] + Prime[n + k]^2; ! PrimeQ[p],

%t k++]; p, {n, 2, 60}]

%Y Cf. A133559, A263725.

%K nonn

%O 2,1

%A _Jonathan Sondow_, Oct 24 2015