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

Author

Zhi-Wei Sun, May 22 2014

Keywords

• nonn

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).

Examples

			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.
		

Links

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

Crossrefs

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

Programs

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}]