A233864 -id:A233864 - cp's OEIS frontend

A233867 a(n) = |{0 < m < 2n: m is a square with 2n - 1 - phi(m) prime}|, where phi(.) is Euler's totient function (A000010).

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

Offset: 1

Zhi-Wei Sun, Dec 17 2013

nonn

Conjecture: (i) a(n) > 0 for all n > 1.
(ii) For any odd number 2*n - 1 > 4, there is a positive integer k < 2*n such that 2*n - 1 - phi(k) and 2*n - 1 + phi(k) are both prime.
By Goldbach's conjecture, 2*n > 2 could be written as p + q with p and q both prime, and hence 2*n - 1 = p + (q - 1) = p + phi(q).
By induction, phi(k^2) (k = 1,2,3,...) are pairwise distinct.

			a(29) = 1 since 2*29 - 1 = 37 + phi(5^2) with 37 prime.
a(39) = 1 since 2*39 - 1 = 71 + phi(3^2) with 71 prime.
a(66) = 1 since 2*66 - 1 = 89 + phi(7^2) with 89 prime.
a(128) = 1 since 2*128 - 1 = 223 + phi(8^2) with 223 prime.
a(182) = 1 since 2*182 - 1 = 331 + phi(8^2) with 331 prime.
a(413) = 1 since 2*413 - 1 = 823 + phi(2^2) with 823 prime.
a(171) = 3 since 2*171 - 1 = 233 + phi(18^2) = 257 + phi(14^2) = 293 + phi(12^2) with 233, 257, 293 all prime.

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

Cf. A000010, A000040, A002372, A002375, A233542, A233544, A233547, A233654, A233793, A233864.

a[n_]:=Sum[If[PrimeQ[2n-1-EulerPhi[k^2]],1,0],{k,1,Sqrt[2n-1]}]
Table[a[n],{n,1,100}]

A236765 Number of ways to write n = k^2 + m with k > 1 and m > 1 such that sigma(k^2) + prime(m) - 1 is prime, where sigma(j) denotes the sum of all positive divisors of j.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 2, 0, 2, 3, 1, 3, 2, 1, 2, 2, 3, 2, 4, 3, 2, 3, 2, 2, 3, 3, 2, 2, 2, 3, 4, 4, 3, 4, 4, 1, 3, 4, 2, 2, 5, 3, 3, 4, 4, 3, 1, 5, 3, 4, 3, 4, 5, 4, 3, 1, 5, 2, 6, 4, 3, 4, 2, 1, 5, 4, 7, 4, 4, 3, 1, 3, 1, 4, 4, 4, 2, 5, 6, 3, 6, 5, 5, 1, 4, 5, 5, 4, 3, 6

Offset: 1

Zhi-Wei Sun, Jan 30 2014

nonn

Conjecture: (i) If n > 6 is not equal to 18, then a(n) > 0.

(ii) Any integer n > 14 can be written as p + q with q > 0 such that p, p + 6 and prime(p) + sigma(q) are all prime.

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

Cf. A000040, A000203, A233544, A233654, A233793, A233864, A236548.

a(10) = 1 since 10 = 2^2 + 6 with sigma(2^2) + prime(6) - 1 = 7 + 13 - 1 = 19 prime.
a(253) = 1 since 253 = 15^2 + 28 with sigma(15^2) + prime(28) - 1 = 403 + 107 - 1 = 509 prime.

p[n_,k_]:=PrimeQ[DivisorSigma[1,k^2]+Prime[n-k^2]-1]
a[n_]:=If[n<6,0,Sum[If[p[n,k],1,0],{k,2,Sqrt[n-2]}]]
Table[a[n],{n,1,100}]

cp's OEIS Frontend

A233867 a(n) = |{0 < m < 2n: m is a square with 2n - 1 - phi(m) prime}|, where phi(.) is Euler's totient function (A000010).

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A236765 Number of ways to write n = k^2 + m with k > 1 and m > 1 such that sigma(k^2) + prime(m) - 1 is prime, where sigma(j) denotes the sum of all positive divisors of j.

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cp's OEIS Frontend

A233867 a(n) = |{0 < m < 2*n: m is a square with 2*n - 1 - phi(m) prime}|, where phi(.) is Euler's totient function (A000010).

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Author

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Mathematica

A236765 Number of ways to write n = k^2 + m with k > 1 and m > 1 such that sigma(k^2) + prime(m) - 1 is prime, where sigma(j) denotes the sum of all positive divisors of j.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

A233867 a(n) = |{0 < m < 2n: m is a square with 2n - 1 - phi(m) prime}|, where phi(.) is Euler's totient function (A000010).