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(PARI) a304903(n) = forprime(p=3, , if(ispseudoprime(2*n^2-p), return(p)))
a(n) = 2*n^2-a304903(n) \\ Felix Fröhlich, May 20 2018
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Each square > 1 can be written as the average of 2 primes p1 < p2. a(n) gives the greatest prime p2 such that n^2 = (p1 + p2) / 2. The corresponding p1 is provided in A304903.
allocated for Hugo PfoertnerGreatest prime p such that 2*n^2 - p is prime.
5, 13, 29, 47, 67, 79, 109, 157, 197, 239, 283, 331, 389, 443, 509, 571, 643, 719, 797, 877, 937, 1051, 1129, 1237, 1321, 1453, 1549, 1669, 1789, 1879, 2029, 2161, 2309, 2447, 2579, 2731, 2857, 3037, 3187, 3359, 3517
2,1
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Hugo Pfoertner, May 20 2018
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allocated for Hugo Pfoertner
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