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

Author

Labos Elemer, Nov 29 2002

Keywords

• hard
• more
• nonn

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

Examples

			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.
		

Crossrefs

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

Programs

Mathematica

f[k_, x_] := DivisorSigma[k, x]/EulerPhi[x]; Table[k=7; While[!IntegerQ[f[n, k]], k++]; k, {n, 1, 15}] (* corrected by Jason Yuen, Jun 27 2025 *)

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