A248029 Least positive integer m such that m + n divides phi(m)*sigma(n), where phi(.) and sigma(.) are given by A000010 and A000203.

1, 1, 3, 1, 6, 1, 7, 4, 8, 1, 2, 1, 10, 9, 15, 1, 8, 1, 1, 11, 14, 1, 6, 6, 16, 5, 14, 1, 6, 1, 10, 15, 11, 13, 16, 1, 7, 9, 5, 1, 6, 1, 12, 7, 26, 1, 14, 8, 12, 21, 46, 1, 6, 17, 4, 23, 32, 1, 24, 1, 34, 41, 63, 7, 6, 1, 16, 11, 2

Offset: 2

Author

Zhi-Wei Sun, Sep 29 2014

Keywords

• nonn

Comments

Conjecture: For any n > 1, we have a(n) <= n.

The existence of a(n) is easy; in fact, for m = sigma(n) - n, obviously m + n divides phi(m)*sigma(n). - Zhi-Wei Sun, Oct 02 2014

Examples

			a(8) = 7 since 7 + 8 = 15 divides phi(7)*sigma(8) = 6*15 = 90.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 2..10000
• Zhi-Wei Sun, A new theorem on the prime-counting function, arXiv:1409.5685, 2014.

Crossrefs

Cf. A000010, A000203, A248004, A248007, A248008, A248030.

Programs

Mathematica

Do[m=1;Label[aa];If[Mod[EulerPhi[m]*DivisorSigma[1,n],m+n]==0,Print[n," ",m];Goto[bb]];m=m+1;Goto[aa];Label[bb];Continue,{n,2,70}]

PARI

a(n)=m=1;while((eulerphi(m)*sigma(n))%(m+n),m++);m
vector(100,n,a(n)) \\ Derek Orr, Sep 29 2014