A235343 a(n) = |{0 < k < n: f(n,k) - 1, f(n,k) + 1 and q(f(n,k)) + 1 are all prime with f(n,k) = phi(k) + phi(n-k)/4}|, where phi(.) is Euler's totient function, and q(.) is the strict partition function (A000009).
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 2, 3, 3, 2, 4, 2, 2, 3, 4, 4, 2, 3, 0, 3, 2, 3, 3, 3, 3, 4, 0, 2, 1, 1, 2, 2, 1, 2, 2, 2, 1, 1, 2, 4, 0, 2, 1, 5, 2, 2, 0, 2, 3, 2, 3, 4, 4, 2, 2, 2, 1, 3, 6, 3, 3, 1, 5, 2, 2, 2, 4, 2, 2, 2, 2, 2, 3, 2, 2
Offset: 1
Keywords
Examples
a(50) = 1 since phi(34) + phi(16)/4 = 18 with 18 - 1, 18 + 1 and q(18) + 1 = 47 all prime. a(215) = 1 since phi(87) + phi(128)/4 = 72 with 72 - 1, 72 + 1 and q(72) + 1 = 36353 all prime. a(645) = 1 since phi(365) + phi(280)/4 = 312 with 312 - 1, 312 + 1 and q(312) + 1 = 207839472391 all prime.
Links
- Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
- Z.-W. Sun, Problems on combinatorial properties of primes, arXiv:1402.6641, 2014
Crossrefs
Programs
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Mathematica
f[n_,k_]:=EulerPhi[k]+EulerPhi[n-k]/4 p[n_,k_]:=PrimeQ[f[n,k]-1]&&PrimeQ[f[n,k]+1]&&PrimeQ[PartitionsQ[f[n,k]]+1] a[n_]:=Sum[If[p[n,k],1,0],{k,1,n-1}] Table[a[n],{n,1,100}]
Comments