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A248029
Least positive integer m such that m + n divides phi(m)*sigma(n), where phi(.) and sigma(.) are given by A000010 and A000203.
3
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
2,3
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
LINKS
Zhi-Wei Sun, A new theorem on the prime-counting function, arXiv:1409.5685, 2014.
EXAMPLE
a(8) = 7 since 7 + 8 = 15 divides phi(7)*sigma(8) = 6*15 = 90.
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}]
PROG
(PARI)
a(n)=m=1; while((eulerphi(m)*sigma(n))%(m+n), m++); m
vector(100, n, a(n)) \\ Derek Orr, Sep 29 2014
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Sep 29 2014
STATUS
approved