A234503 -id:A234503 - cp's OEIS frontend

A234514 Number of ways to write n = k + m with k > 0 and m > 0 such that p = k + phi(m)/2 and q(p) + 1 are both prime, where phi(.) is Euler's totient function, and q(.) is the strict partition function (A000009).

0, 0, 0, 1, 2, 2, 1, 1, 1, 0, 1, 0, 2, 2, 2, 3, 4, 2, 4, 2, 3, 3, 3, 2, 2, 3, 1, 4, 2, 1, 4, 2, 4, 2, 5, 3, 4, 1, 5, 6, 4, 2, 5, 5, 5, 3, 5, 4, 6, 3, 5, 7, 10, 2, 4, 5, 6, 5, 5, 2, 3, 5, 6, 6, 4, 2, 5, 3, 7, 4, 5, 3, 8, 7, 2, 5, 9, 3, 3, 2, 9, 9, 6, 6, 7, 6, 9, 4, 7, 4, 10, 8, 6, 11, 11, 4, 6, 4, 9, 7

Offset: 1

Zhi-Wei Sun, Dec 27 2013

nonn

Conjecture: (i) a(n) > 0 for all n > 12.
(ii) For any integer n > 4, there is a prime p < n - 2 such that q(p + phi(n-p)/2) + 1 is prime.
Clearly, part (i) of the conjecture implies that there are infinitely many primes p with q(p) + 1 prime (cf. A234530).
We have verified part (i) for n up to 10^5.

			a(11) = 1 since 11 = 1 + 10 with 1 + phi(10)/2 = 3 and q(3) + 1 = 3 both prime.
a(27) = 1 since 27 = 7 + 20 with 7 + phi(20)/2 = 11 and q(11) + 1 = 13 both prime.
a(30) = 1 since 30 = 8 + 22 with 8 + phi(22)/2 = 13 and q(13) + 1 = 19 both prime.
a(38) = 1 since 38 = 21 + 17 with 21 + phi(17)/2 = 29 and q(29) + 1 = 257 both prime.
a(572) = 1 since 572 = 77 + 495 with 77 + phi(495)/2 = 197 and q(197) + 1 = 406072423 both prime.
a(860) = 1 since 860 = 523 + 337 with 523 + phi(337)/2 = 691 and q(691) + 1 = 712827068077888961 both prime.

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

Cf. A000009, A000010, A000040, A229835, A233307, A233390, A233417, A234451, A234470, A234475, A234503, A234504, A234530

f[n_,k_]:=k+EulerPhi[n-k]/2
q[n_,k_]:=PrimeQ[f[n,k]]&&PrimeQ[PartitionsQ[f[n,k]]+1]
a[n_]:=Sum[If[q[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}]

A236358 a(n) = |{0 < k < n: m = phi(k)/2 + phi(n-k)/12 is an integer with 2*3^m + 1 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, 0, 2, 3, 1, 1, 0, 2, 1, 0, 2, 3, 3, 3, 2, 3, 2, 1, 4, 1, 4, 1, 4, 5, 3, 5, 7, 7, 8, 5, 5, 4, 4, 7, 7, 4, 7, 3, 6, 4, 5, 5, 6, 7, 6, 4, 5, 7, 6, 9, 5, 8, 7, 7, 4, 6, 5, 4, 6, 9, 8, 3, 6, 8, 9, 8, 8, 7, 8, 8, 9, 8, 4, 7, 4, 7, 7, 5, 4, 8, 6, 6, 7, 11

Offset: 1

Zhi-Wei Sun, Jan 23 2014

nonn

Conjecture: (i) a(n) > 0 for all n > 26.
(ii) For any integer n > 37, there is a positive integer k < n such that m = phi(k)/2 + phi(n-k)/12 is an integer with 2*3^m - 1 prime.
We have verified both parts for n up to 50000. Clearly, part (i) implies that there are infinitely many positive integers m with 2*3^m + 1 prime, while part (ii) implies that there are infinitely many positive integers m with 2*3^m - 1 prime.

			 a(36) = 1 since phi(15)/2 + phi(21)/12 = 4 + 1 = 5 with 2*3^5 + 1 = 487 prime.

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

Cf. A000010, A000040, A000244, A079363, A111974, A234451, A234503, A234504.

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

A234963 Number of ways to write n = k + m with k > 0 and m > 2 such that C(2*sigma(k) + phi(m), sigma(k) + phi(m)/2) - 1 is prime, where sigma(k) is the sum of all positive divisors of k and phi(.) is Euler's totient function.

Original entry on oeis.org

0, 0, 0, 1, 1, 2, 3, 0, 3, 2, 2, 3, 3, 5, 3, 4, 3, 3, 3, 2, 3, 0, 3, 3, 4, 3, 0, 1, 2, 3, 1, 2, 3, 3, 1, 3, 3, 4, 1, 2, 3, 3, 2, 6, 4, 1, 4, 2, 3, 2, 2, 2, 4, 3, 2, 3, 3, 2, 4, 3, 3, 0, 2, 3, 1, 3, 1, 2, 0, 3, 1, 4, 4, 4, 1, 0, 5, 2, 1, 3, 2, 2, 1, 2, 1

Offset: 1

Zhi-Wei Sun, Jan 01 2014

nonn

Conjecture: a(n) > 0 for all n >= 180.
Clearly, this implies that there are infinitely many primes of the form C(2*n,n) - 1. We have verified the conjecture for n up to 10000.
Note that every n = 400, ..., 9123 can be written as k + m with k > 0 and m > 0 such that f(k, m) = sigma(k) + phi(m) is even and C(f(k, m) + 2, f(k, m)/2 + 1) + 1 is prime, but this fails for n = 9124.

			a(5) = 1 since 5 = 1 + 4 with C(2*sigma(1) + phi(4), sigma(1) + phi(4)/2) - 1 = C(4, 2) - 1 = 5 prime.
a(28) = 1 since 28 = 2 + 26 with C(2*sigma(2) + phi(26), sigma(2) + phi(26)/2) - 1 = C(18, 9) - 1 = 48619 prime.

Zhi-Wei Sun, Table of n, a(n) for n = 1..2000

Cf. A000010, A000040, A000203, A000984, A066726, A066699, A232270, A233544, A234451, A234470, A234503, A234514, A234567, A234615.

sigma[n_] := DivisorSigma[1, n];
f[n_,k_] := Binomial[2*sigma[k] + EulerPhi[n-k], sigma[k] + EulerPhi[n-k]/2] - 1;
a[n_] := Sum[If[PrimeQ[f[n,k]], 1, 0], {k, 1, n-3}];
Table[a[n], {n, 1, 100}]

cp's OEIS Frontend

A234514 Number of ways to write n = k + m with k > 0 and m > 0 such that p = k + phi(m)/2 and q(p) + 1 are both prime, where phi(.) is Euler's totient function, and q(.) is the strict partition function (A000009).

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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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A236358 a(n) = |{0 < k < n: m = phi(k)/2 + phi(n-k)/12 is an integer with 2*3^m + 1 prime}|, where phi(.) is Euler's totient function.

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A234963 Number of ways to write n = k + m with k > 0 and m > 2 such that C(2*sigma(k) + phi(m), sigma(k) + phi(m)/2) - 1 is prime, where sigma(k) is the sum of all positive divisors of k and phi(.) is Euler's totient function.

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