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

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.
a(7) = 1 since 3*phi(4) - 1 = 5 and 3*phi(4) + 1 = 7 are both prime.
a(18) = 1 since 5*phi(13) - 1 = 59 and 5*phi(13) + 1 = 61 are both prime.
a(91) = 1 since 13*phi(78) - 1 = 311 and 13*phi(78) + 1 = 313 are both prime.
a(101) = 1 since 6*phi(95) - 1 = 431 and 6*phi(95) + 1 = 433 are both prime.

Mathematica

TQ[n_]:=PrimeQ[n-1]&&PrimeQ[n+1]
a[n_]:=Sum[If[TQ[k*EulerPhi[n-k]], 1, 0], {k, 1, (n-1)/2}]
Table[a[n], {n, 1, 100}]

Crossrefs

Cf. A000010, A000040, A001359, A006512, A014574, A233542, A233544, A233547, A233566, A233567, A233867, A233918.

Sequence in context: A089216 A351599 A350743 * A102746 A287618 A123143

Adjacent sequences: A234197 A234198 A234199 * A234201 A234202 A234203

Keyword

nonn

Author

Zhi-Wei Sun, Dec 21 2013

Status

approved