A237168 Number of ways to write 2*n - 1 = 2*p + q with p, q, phi(p+1) - 1 and phi(p+1) + 1 all prime, where phi(.) is Euler's totient function.
0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 2, 2, 2, 2, 3, 1, 3, 3, 2, 2, 4, 1, 1, 3, 2, 2, 3, 1, 1, 3, 2, 2, 2, 1, 2, 3, 2, 2, 4, 1, 4, 5, 2, 1, 6, 3, 3, 2, 3, 2, 5, 1, 2, 5, 3, 3, 4, 3, 2, 6, 4, 4, 5, 2, 3, 7, 2, 2
Offset: 1
Keywords
Examples
a(9) = 1 since 2*9 - 1 = 2*7 + 3 with 7, 3, phi(7+1) - 1 = 3 and phi(7+1) + 1 = 5 all prime. a(934) = 1 since 2*934 - 1 = 2*457 + 953 with 457, 953, phi(457+1) - 1 = 227 and phi(457+1) + 1 = 229 all prime.
Links
- Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
- Zhi-Wei Sun, Unification of Goldbach's conjecture and the twin prime conjecture, a message to Number Theory List, Jan. 29, 2014.
Crossrefs
Programs
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Mathematica
PQ[n_]:=PrimeQ[EulerPhi[n]-1]&&PrimeQ[EulerPhi[n]+1] a[n_]:=Sum[If[PQ[Prime[k]+1]&&PrimeQ[2n-1-2*Prime[k]],1,0],{k,1,PrimePi[n-1]}] Table[a[n],{n,1,70}]
Comments