A219185 Number of prime pairs {p,q} (p>q) with 3(p-q)-1 and 3(p-q)+1 both prime such that p+(1+(n mod 2))q=n.
0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 0, 0, 2, 1, 1, 1, 2, 1, 0, 1, 3, 1, 1, 1, 1, 2, 1, 1, 1, 0, 1, 1, 2, 0, 1, 2, 2, 0, 2, 2, 0, 2, 1, 0, 3, 1, 2, 1, 1, 2, 1, 1, 1, 2, 2, 1, 4, 1, 1, 1, 0, 1, 1, 2, 1, 1, 3, 1, 5, 2, 1, 2, 1, 0, 2, 0, 2, 3, 4, 2, 3, 3, 2, 2, 1, 3, 2, 1, 1, 2, 0, 0, 2, 1, 3, 2, 3
Offset: 1
Keywords
Examples
a(11)=1 since 11=5+2*3, and both 3(5-3)-1=5 and 3(5-3)+1=7 are prime. a(16)=2 since 16=11+5=13+3, and 3(11-5)-1, 3(11-5)+1, 3(13-3)-1, 3(13-3)+1 are all prime.
Links
- Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
- Zhi-Wei Sun, Conjectures involving primes and quadratic forms, arXiv:1211.1588.
Crossrefs
Programs
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Mathematica
a[n_]:=a[n]=Sum[If[PrimeQ[n-(1+Mod[n,2])Prime[k]]==True&&PrimeQ[3(n-(2+Mod[n,2])Prime[k])-1]==True&&PrimeQ[3(n-(2+Mod[n,2])Prime[k])+1]==True,1,0], {k,1,PrimePi[(n-1)/(2+Mod[n,2])]}] Do[Print[n," ",a[n]],{n,1,100000}]
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PARI
a(n)=if(n%2, aOdd(n), aEven(n)) aOdd(n)=my(s); forprime(q=2,(n-1)\3, my(p=n-2*q); if(isprime(n-2*q) && isprime(3*n-9*q-1) && isprime(3*n-9*q+1), s++)); s aEven(n)=my(s); forprime(q=2,n/2, if(isprime(n-q) && isprime(3*n-6*q-1) && isprime(3*n-6*q+1), s++)); s \\ Charles R Greathouse IV, Jul 31 2016
Comments