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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A234451 Number of ways to write n = k + m with k > 0 and m > 0 such that 2^(phi(k)/2 + phi(m)/6) + 3 is prime, where phi(.) is Euler's totient function.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 2, 2, 2, 3, 4, 4, 4, 3, 5, 4, 5, 4, 6, 4, 4, 5, 5, 5, 6, 6, 6, 5, 6, 8, 7, 6, 5, 7, 8, 7, 10, 6, 7, 9, 7, 5, 5, 8, 6, 6, 7, 9, 3, 7, 10, 9, 3, 8, 6, 8, 6, 9, 9, 12, 5, 8, 8, 10, 9, 10, 9, 8, 8, 8, 10, 9, 12, 10, 13, 11, 9, 10
Offset: 1

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Author

Zhi-Wei Sun, Dec 26 2013

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 9. Also, any integer n > 13 can be written as k + m with k > 0 and m > 0 such that 2^(phi(k)/2 + phi(m)/6) - 3 is prime.
(ii) Each integer n > 25 can be written as k + m with k > 0 and m > 0 such that 3*2^(phi(k)/2 + phi(m)/8) + 1 (or 3*2^(phi(k)/2 + phi(m)/12) + 1 when n > 38) is prime. Also, any integer n > 14 can be written as k + m with k > 0 and m > 0 such that 3*2^(phi(k)/2 + phi(m)/12) - 1 is prime.
This conjecture implies that there are infinitely many primes in any of the four forms 2^n + 3, 2^n - 3, 3*2^n + 1, 3*2^n - 1.
We have verified the conjecture for n up to 50000.

Examples

			a(10) = 1 since 10 = 3 + 7 with 2^(phi(3)/2 + phi(7)/6) + 3 = 7 prime.
a(11) = 1 since 11 = 4 + 7 with 2^(phi(4)/2 + phi(7)/6) + 3 = 7 prime.
a(12) = 2 since 12 = 3 + 9 = 5 + 7 with 2^(phi(3)/2 + phi(9)/6) + 3 = 7 and 2^(phi(5)/2 + phi(7)/6) + 3 = 11 both prime.
a(769) = 1 since 769 = 31 + 738 with 2^(phi(31)/2 + phi(738)/6) + 3 = 2^(55) + 3 prime.
a(787) = 1 since 787 = 112 + 675 with 2^(phi(112)/2 + phi(675)/6) + 3 = 2^(84) + 3 prime.
a(867) = 1 since 867 = 90 + 777 with 2^(phi(90)/2 + phi(777)/6) + 3 = 2^(84) + 3 prime.
a(869) = 1 since 869 = 51 + 818 with 2^(phi(51)/2 + phi(818)/6) + 3 = 2^(84) + 3 prime.
a(913) = 1 since 913 = 409 + 504 with 2^(phi(409)/2 + phi(504)/6) + 3 = 2^(228) + 3 prime.
a(1085) = 1 since 1085 = 515 + 570 with 2^(phi(515)/2 + phi(570)/6) + 3 = 2^(228) + 3 prime.
		

Crossrefs

Programs

  • Mathematica
    f[n_,k_]:=2^(EulerPhi[k]/2+EulerPhi[n-k]/6)+3
    a[n_]:=Sum[If[PrimeQ[f[n,k]],1,0],{k,1,n-1}]
    Table[a[n],{n,1,100}]