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%I #12 Sep 29 2014 11:19:14
%S 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,
%T 14,32,7,14,5,12,35,10,13,24,7,14,13,11,9,42,45,16,11,30,13,12,19,27,
%U 33,8,15,22,28,4,35,28,18,64,7,14,21,28,19,10
%N Least positive integer m such that m + n divides phi(m*n), where phi(.) is Euler's totient function.
%C Conjecture: For any n > 2, a(n) exists and a(n) <= n.
%C See also A248007 and A248008 for similar conjectures. - _Zhi-Wei Sun_, Sep 29 2014
%C 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
%C a(n) >= 3 for all n. - _Robert Israel_, Sep 29 2014
%H Zhi-Wei Sun, <a href="/A248006/b248006.txt">Table of n, a(n) for n = 3..10000</a>
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1409.5685">A new theorem on the prime-counting function</a>, arXiv:1409.5685, 2014.
%e a(5) = 3 since 3 + 5 divides phi(3*5) = 8.
%p f:= proc(n)
%p local m;
%p for m from 3 do
%p if numtheory:-phi(m*n) mod (m+n) = 0 then return m fi
%p od
%p end proc;
%p seq(f(n),n=3..100); # _Robert Israel_, Sep 29 2014
%t 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}]
%o (PARI)
%o a(n)=m=1;while(eulerphi(m*n)%(m+n),m++);m
%o vector(100,n,a(n+2)) \\ _Derek Orr_, Sep 29 2014
%Y Cf. A000010, A248004, A248007, A248008.
%K nonn
%O 3,1
%A _Zhi-Wei Sun_, Sep 29 2014
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