

A078538


Smallest k > 6 such that sigma_n(k)/phi(k) is an integer.


4



12, 22, 12, 249, 12, 22, 12, 19689, 12, 22, 12, 249, 12, 22, 12
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OFFSET

1,1


COMMENTS

For n = 16, 48, 64, and 80 the solutions are hard to find, exceed 10^6 or even 10^7.
If a(16) exists, it is greater than 2^32. Terms a(17) to a(47) are 12, 22, 12, 249, 12, 22, 12, 9897, 12, 22, 12, 249, 12, 22, 12, 2566, 12, 22, 12, 249, 12, 22, 12, 19689, 12, 22, 12, 249, 12, 22, 12.  T. D. Noe, Dec 08 2013


LINKS



EXAMPLE

These terms appear as 5th entries in A020492, A015759A015774. k = {1, 2, 3, 6} are solutions to Min{k : Mod[sigma[n, k], phi[k]]=0}. First nontrivial solutions are larger: for odd n, k = 12 is solution; for even powers larger numbers arise like 22, 249, 9897, 19689, etc. Certain powersums of divisors proved to be hard to find.


MATHEMATICA

f[k_, x_] := DivisorSigma[k, x]/EulerPhi[x]; Table[fl=1; Do[s=f[k, n]; If[IntegerQ[s]&&Greater[n, 6], Print[{n, k}; fl=0], {n, 100000}, {k, 1, 100}]


PROG

(PARI) ok(n, k)=my(f=factor(n), r=sigma(f, k)/eulerphi(f)); r>=7 && denominator(r)==1
(Python)
from sympy import divisors, totient as phi
def a(n):
k, pk = 7, phi(7)
while sum(pow(d, n, pk) for d in divisors(k, generator=True))%pk != 0:
k += 1
pk = phi(k)
return k


CROSSREFS



KEYWORD

hard,more,nonn


AUTHOR



STATUS

approved



