A334876 Numbers m with the property that sigma(2^m+1)/(2^m+1) > sigma(2^k+1)/(2^k+1) for all k < m, where sigma is the sum of divisors function, A000203.

1, 3, 5, 9, 15, 45, 135, 315, 945

Offset

1,2

Comments

Set h(m) = sigma(m)/m. Then the sequence lists the numbers m at which record values of h(2^m+1) occur. This sequence is essentially defined similarly to A004394 but restricted to looking just at numbers which are one more than a power of 2.

The sequence is infinite. This can be proved by seeing that we can make h(2^m+1) arbitrarily large. Note that if p is a prime which is 3 (mod 8), then p|2^m+1 for any odd m such that (p-1)/2|m. By the strong version of Dirichlet's theorem the sum of the reciprocals of the primes which are 3 (mod 8) diverges. So we can make h(2^m+1) arbitrarily large by taking m as the largest odd divisor of k! for large k.

It appears that every term in the sequence is odd. This seems likely to be true since if m is even then 2^m+1 is not divisible by 3, which makes it much harder to make h(2^m+1) large.

Links

Table of n, a(n) for n=1..9.

Joshua Zelinsky, On the small prime factors of a non-deficient number, arXiv:2005.12118 [math.NT], 2020-2022.

Mathematica

r[n_] := DivisorSigma[1, 2^n+1]/(2^n + 1); seq = {}; rm = 1; Do[r1 = r[n]; If[r1 > rm, rm = r1; AppendTo[seq, n]], {n, 1, 50}]; seq (* Amiram Eldar, May 15 2020 *)

Prog

(Sage)
def h(n):
    return (sigma(n, 1))/n
def hchecker(k):
    s=0
    for i in range(1, k):
        j=2^i+1
        a=h(j)
        if a> s:
            print(i)
            s=a

Crossrefs

Cf. A000203, A004394, A069061.

Sequence in context: A052007 A117480 A018260 * A337596 A355115 A023656

Adjacent sequences: A334873 A334874 A334875 * A334877 A334878 A334879

Keyword

nonn,more

Author

Joshua Zelinsky, May 13 2020

Extensions

More terms from Amiram Eldar, May 13 2020, using A069061.

Status

approved