A236358 -id:A236358 - cp's OEIS frontend

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.

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

Zhi-Wei Sun, Dec 26 2013

nonn

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.

			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.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Cf. A000010, A000040, A000079, A050415, A057733, A234309, A234310, A234337, A234344, A234346, A234347, A234359, A234360, A234361, A234388, A234399, A234470, A236358.

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}]

A234503 Number of ways to write n = k + m with k > 0 and m > 0 such that 3^(phi(k)/2 + phi(m)/12) + 2 is prime, where phi(.) is Euler's totient function.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 2, 1, 1, 2, 3, 2, 1, 1, 1, 2, 1, 2, 2, 3, 2, 4, 4, 4, 2, 3, 2, 1, 3, 4, 8, 3, 4, 4, 4, 6, 3, 4, 6, 3, 5, 5, 3, 2, 2, 6, 5, 3, 2, 3, 7, 4, 3, 4, 4, 3, 4, 4, 4, 5, 2, 5, 2, 6, 5, 7, 3, 5, 7, 6, 13, 5, 7, 7, 10, 6, 8, 8, 9, 6, 7, 8, 6, 6, 5, 7, 9, 6, 7, 8, 10

Offset: 1

Zhi-Wei Sun, Dec 26 2013

nonn

It might seem that a(n) > 0 for all n > 14, but a(43905) = 0. If a(n) > 0 infinitely often, then there are infinitely many primes of the form 3^m + 2.
Similarly, it might seem that for n > 26 there is a positive integer k < n such that m = phi(k)/2 + phi(n-k)/12 is an integer with 3^m - 2 prime, but n = 41213 is a counterexample.
See also A234451 and A236358 for similar sequences.

			a(15) = 1 since 15 = 1 + 14 with 3^(phi(1)/2 + phi(14)/12) + 2 = 3 + 2 = 5 prime.
a(23) = 1 since 23 = 10 + 13 with 3^(phi(10)/2 + phi(13)/12) + 2 = 3^3 + 2 = 29 prime.
a(24) = 1 since 24 = 3 + 21 with 3^(phi(3)/2 + phi(21)/12) + 2 = 3^2 + 2 = 11 prime.
a(37) = 1 since 37 = 9 + 28 with 3^(phi(9)/2 + phi(28)/12) + 2 = 3^4 + 2 = 83 prime.

Zhi-Wei Sun, Table of n, a(n) for n = 1..4000

Cf. A000010, A000040, A000244, A014232, A057735, A079363, A111974, A234344, A234346, A234347, A234361, A234451, A234470, A234475, A234504, A236358.

f[n_,k_]:=3^(EulerPhi[k]/2+EulerPhi[n-k]/12)+2
a[n_]:=Sum[If[PrimeQ[f[n,k]],1,0],{k,1,n-1}]
Table[a[n],{n,1,100}]

A234504 Number of ways to write n = k + m with k > 0 and m > 0 such that 2^(phi(k) + phi(m)/4) - 5 is prime, where phi(.) is Euler's totient function.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 2, 1, 1, 3, 2, 3, 2, 3, 2, 3, 4, 5, 5, 4, 5, 6, 7, 4, 5, 6, 7, 6, 5, 7, 8, 5, 7, 9, 8, 8, 6, 8, 7, 10, 7, 10, 10, 9, 9, 8, 9, 10, 5, 10, 10, 9, 10, 10, 9, 10, 9, 7, 12, 14, 10, 9, 5, 11, 7, 13, 8, 13, 6, 9, 11, 11, 14, 15, 9, 13

Offset: 1

Zhi-Wei Sun, Dec 26 2013

nonn

Conjecture: a(n) > 0 for all n > 10.

We have verified this for n up to 50000. The conjecture implies that there are infinitely many primes of the form 2^n - 5.

			a(15) = 2 since 2^(phi(2) + phi(13)/4) - 5 = 2^4 - 5 = 11 and 2^(phi(3) + phi(12)/4) - 5 = 2^3 - 5 = 3 are both prime.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Cf. A000010, A000040, A000079, A050522, A156560, A234309, A234451, A234470, A234475, A234503, A236358.

f[n_,k_]:=2^(EulerPhi[k]+EulerPhi[n-k]/4)-5
a[n_]:=Sum[If[PrimeQ[f[n,k]],1,0],{k,1,n-1}]
Table[a[n],{n,1,100}]

cp's OEIS Frontend

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.

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A234503 Number of ways to write n = k + m with k > 0 and m > 0 such that 3^(phi(k)/2 + phi(m)/12) + 2 is prime, where phi(.) is Euler's totient function.

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A234504 Number of ways to write n = k + m with k > 0 and m > 0 such that 2^(phi(k) + phi(m)/4) - 5 is prime, where phi(.) is Euler's totient function.

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