A242753 Number of ordered ways to write n = k + m with 0 < k <= m such that the inverse of k mod prime(k) among 1, ..., prime(k) - 1 is prime and the inverse of m mod prime(m) among 1, ..., prime(m) - 1 is also prime.

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

Offset

1,6

Comments

Conjecture: a(n) > 0 for all n > 3.

This implies that there are infinitely many positive integers k such that k*q == 1 (mod prime(k)) for some prime q < prime(k).

Links

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

Example

a(11) = 1 since 11 = 4 + 7, 4*2 == 1 (mod prime(4)=7) with 2 prime, and 7*5 == 1 (mod Prime(7)=17) with 5 prime.
a(36) = 1 since 36 = 18 + 18, and 18*17 == 1 (mod 61) with 17 prime.
a(46) = 1 since 46 = 6 + 40, 6*11 == 1 (mod prime(6)= 13) with 11 prime, and 40*13 == 1 (mod prime(40)=173) with 13 prime.

Mathematica

p[n_]:=PrimeQ[PowerMod[n, -1, Prime[n]]]
Do[m=0; Do[If[p[k]&&p[n-k], m=m+1], {k, 1, n/2}]; Print[n, " ", m]; Continue, {n, 1, 80}]

Crossrefs

Cf. A000040, A000720, A242425, A242748, A242750, A242752, A242754, A242755.

Sequence in context: A099910 A213325 A043555 * A232443 A118821 A118824

Adjacent sequences: A242750 A242751 A242752 * A242754 A242755 A242756

Keyword

nonn

Author

Zhi-Wei Sun, May 22 2014

Status

approved