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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 (list; graph; refs; listen; history; text; internal format)
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, A015759-A015774. 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 power-sums 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
a(n)=my(k=7); while(!ok(k, n), k++); k \\ Charles R Greathouse IV, Nov 27 2013
(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
print([a(n) for n in range(1, 16)]) # Michael S. Branicky, Dec 22 2021
CROSSREFS
Sequence in context: A212958 A340688 A031186 * A278030 A286094 A348187
KEYWORD
hard,more,nonn
AUTHOR
Labos Elemer, Nov 29 2002
STATUS
approved

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Last modified July 17 21:57 EDT 2024. Contains 374377 sequences. (Running on oeis4.)