A234200 - OEIS

A234200

a(n) = |{0 < k < n/2: k*phi(n-k) - 1 and k*phi(n-k) + 1 are both prime}|, where phi(.) is Euler's totient function.

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

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OFFSET

1,6

COMMENTS

Conjecture: (i) a(n) > 0 for all n > 4.

(ii) If n > 3 is different from 9 and 29, then k*sigma(n-k) - 1 and k*sigma(n-k) + 1 are both prime for some 0 < k < n.

Obviously, either of the two parts implies the twin prime conjecture. We have verified part (i) for n up to 10^8.

LINKS

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

Zhi-Wei Sun, New representation problems involving Euler's totient function, a message to Number Theory List, Dec. 18, 2013.

EXAMPLE

a(5) = 1 since 2*phi(3) - 1 = 3 and 2*phi(3) + 1 = 5 are both prime.