A273457 Even numbers 2n that do not have a Goldbach partition 2n = p + q (p < q; p, q prime) satisfying sqrt(n) < p <= sqrt(2n).
2, 4, 6, 8, 12, 18, 20, 22, 24, 26, 30, 32, 38, 40, 44, 52, 56, 58, 62, 64, 70, 72, 76, 82, 84, 88, 92, 94, 98, 100, 102, 104, 106, 108, 110, 112, 114, 116, 118, 120, 122, 126, 128, 130, 132, 134, 136, 140, 144, 146, 152, 154, 156, 158, 164, 166, 172, 182, 188, 196, 198, 200, 214
Offset: 1
Keywords
Examples
32 is in the sequence because 32 has two Goldbach partitions: 32 = 3 + 29 with 3 < sqrt(16) and 32 = 13 + 19 with 13 > sqrt(32).
Links
- Corinna Regina Böger, Table of n, a(n) for n = 1..12831
Programs
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Mathematica
noGoldbatSqrQ[n_] := Block[{p = NextPrime[Sqrt[n/2]]}, While[2p < n && !PrimeQ[n - p], p = NextPrime@ p]; p > Sqrt[n]]; noGoldbatSqrQ[4] = True; Select[2Range[107], noGoldbatSqrQ] (* Robert G. Wilson v, Dec 15 2016 *)
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PARI
noSpecialGoldbach(n) = forprime(p=sqrtint(n/2-1) + 1, sqrtint(n), if(p<(n-p) && isprime(n-p), return(0))); 1 is(n) = n%2 == 0 && noSpecialGoldbach(n)
Comments