A278922 Largest p such that n = p + q + r where p < q < r are all prime, or 0 if no such primes p, q, r exist.
0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 2, 0, 2, 3, 2, 0, 2, 3, 2, 3, 2, 5, 2, 5, 2, 3, 2, 5, 2, 7, 2, 5, 2, 7, 2, 7, 2, 7, 2, 11, 2, 11, 2, 5, 2, 11, 2, 13, 2, 11, 2, 13, 2, 13, 2, 11, 2, 17, 2, 13, 2, 13, 2, 17, 2, 17, 2, 17, 2, 19, 2, 19, 2, 13, 2, 17, 2, 19, 2, 17, 2, 23, 2, 19, 2, 19, 2, 23, 2, 23, 2, 23, 2, 23, 2, 29, 2, 23, 2, 29, 2, 29, 2, 23, 2, 29, 2, 31, 2, 31, 2, 29, 2, 31, 2, 29, 2, 31, 2, 37
Offset: 1
Keywords
Links
Programs
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Mathematica
f[n_] := If[OddQ@n || n < 18, Block[{p = 0, q = 3, r = 5}, While[q < r, r = NextPrime@ q; While[r < n - q - 1, If[n < 2q + r && PrimeQ[n - r - q], p = Max[p, n - r - q]; Break[]]; r = NextPrime@ r]; q = NextPrime@ q]; p], 2]; Array[f, 121] (* Robert G. Wilson v, Dec 02 2016 *)
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PARI
a(n,p=if(bittest(n,0),n\3-1,3))=while(p=precprime(p-1),forprime(q=p+1,(n-p-1)\2,isprime(n-p-q)&&return(p)))
Formula
a(2n) = 2 (for n > 4), since one of the three primes must necessarily be even, and that can only be p = 2.
a(n) = 0 for n < 2 + 3 + 5 = 10, and for odd n < 3 + 5 + 7 = 15.
Comments