A078538 Smallest k > 6 such that sigma_n(k)/phi(k) is an integer.

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

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

Table of n, a(n) for n=1..15.

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

Cf. A000203, A001157, A001158, A000010, A015759-A015774, A020492.

Sequence in context: A212958 A340688 A031186 * A278030 A286094 A348187

Adjacent sequences: A078535 A078536 A078537 * A078539 A078540 A078541

Keyword

hard,more,nonn

Author

Labos Elemer, Nov 29 2002

Status

approved