

A248006


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


3



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
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OFFSET

3,1


COMMENTS

Conjecture: For any n > 2, a(n) exists and a(n) <= n.
See also A248007 and A248008 for similar conjectures.  ZhiWei 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

ZhiWei Sun, Table of n, a(n) for n = 3..10000
ZhiWei Sun, A new theorem on the primecounting 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

ZhiWei Sun, Sep 29 2014


STATUS

approved



