A248006 Least positive integer m such that m + n divides phi(m*n), where phi(.) is Euler's totient function.

3, 4, 3, 6, 5, 8, 9, 6, 9, 4, 11, 7, 5, 16, 7, 9, 5, 12, 7, 18, 21, 8, 15, 13, 27, 14, 11, 10, 14, 32, 7, 14, 5, 12, 35, 10, 13, 24, 7, 14, 13, 11, 9, 42, 45, 16, 11, 30, 13, 12, 19, 27, 33, 8, 15, 22, 28, 4, 35, 28, 18, 64, 7, 14, 21, 28, 19, 10

Offset

3,1

Comments

Conjecture: For any n > 2, a(n) exists and a(n) <= n.

See also A248007 and A248008 for similar conjectures. - Zhi-Wei Sun, Sep 29 2014

The conjecture is true: One can show that 2*n divides phi(n^2) for all n > 2. So, a(n) is at most n. - Derek Orr, Sep 29 2014

a(n) >= 3 for all n. - Robert Israel, Sep 29 2014

Links

Zhi-Wei Sun, Table of n, a(n) for n = 3..10000

Zhi-Wei Sun, A new theorem on the prime-counting function, arXiv:1409.5685, 2014.

Example

a(5) = 3 since 3 + 5 divides phi(3*5) = 8.

Maple

f:= proc(n)
local m;
for m from 3 do
  if numtheory:-phi(m*n) mod (m+n) = 0 then return m fi
od
end proc;
seq(f(n), n=3..100); # Robert Israel, Sep 29 2014

Mathematica

Do[m=1; Label[aa]; If[Mod[EulerPhi[m*n], m+n]==0, Print[n, " ", m]; Goto[bb]]; m=m+1; Goto[aa]; Label[bb]; Continue, {n, 3, 70}]

Prog

(PARI)
a(n)=m=1; while(eulerphi(m*n)%(m+n), m++); m
vector(100, n, a(n+2)) \\ Derek Orr, Sep 29 2014

Crossrefs

Cf. A000010, A248004, A248007, A248008.

Sequence in context: A176058 A110738 A175028 * A197699 A005092 A136195

Adjacent sequences: A248003 A248004 A248005 * A248007 A248008 A248009

Keyword

nonn

Author

Zhi-Wei Sun, Sep 29 2014

Status

approved