A242879 Least positive integer k < n such that k*p == 1 (mod prime(k)) for some prime p < prime(k) and (n-k)*q == 1 (mod prime(n-k)) for some prime q < prime(n-k), or 0 if such a number k does not exist.

0, 0, 0, 2, 2, 2, 3, 2, 2, 3, 4, 2, 2, 3, 2, 3, 4, 7, 2, 2, 3, 4, 2, 3, 4, 13, 6, 7, 11, 13, 10, 11, 2, 3, 4, 18, 6, 7, 2, 3, 4, 2, 2, 3, 4, 6, 6, 2, 3, 2, 2, 3, 4, 2, 2, 3, 4, 6, 6, 2, 3, 2, 3, 4, 7, 2, 3, 2, 3, 4, 7, 2, 2, 2, 2, 3, 2, 3, 4, 7

Offset: 1

Author

Zhi-Wei Sun, May 25 2014

Keywords

• nonn

Comments

According to the conjecture in A242753, a(n) should be positive for all n > 3.

We have verified that a(n) > 0 for all n = 4, ..., 10^8.

Examples

			a(4) = 2 since 4 = 2 + 2 and 2*2 == 1 (mod prime(2)=3).
a(7) = 3 since 7 = 3 + 4, 3*2 == 1 (mod prime(3)=5) with 2 prime, and also 4*2 == 1 (mod prime(4)=7) with 2 prime, but 5*9 == 1 (mod prime(5)=11) with 9 not prime.
		

Links

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

Crossrefs

Cf. A000040, A242425, A242748, A242753, A242754, A242755.

Programs

Mathematica

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