A182590 Number of distinct prime factors of 2^n - 1 of the form k*n + 1.

1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 2, 2, 2, 1, 2, 1, 1, 2, 3, 2, 1, 2, 2, 1, 2, 3, 3, 1, 2, 1, 2, 2, 3, 2, 2, 3, 2, 2, 4, 3, 3, 2, 3, 3, 3, 1, 4, 4, 3, 3, 2, 3, 2, 3, 5, 2, 2, 1, 2, 3, 4, 2, 3, 2, 3, 1, 4, 3, 3, 3, 4, 5, 3, 1, 5, 3, 2, 3, 4, 2, 3, 2, 4, 3

Offset: 2

Author

Seppo Mustonen, Nov 22 2010

Keywords

• nonn

Comments

From Thomas Ordowski, Sep 08 2017: (Start)

By Bang's theorem, a(n) > 0 for all n > 1, see A186522.

Primes p such that a(p) = 1 are the Mersenne exponents A000043.

Composite numbers m for which a(m) = 1 are A292079.

a(n) >= A086251(n), where equality is for all prime numbers and for some composite numbers (among others for all odd prime powers p^k with k > 1).

Theorem: if n is prime, then a(n) = A046800(n).

Conjecture: if a(n) = A046800(n), then n is prime.

Problem: is a(n) < A046800(n) for every composite n? (End)

Examples

			For n=10 the prime factors of 2^n - 1 = 1023 are 3, 11 and 31, and 11 = n+1, 31 = 3n+1. Thus a(10)=2.
		

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 2..1200 (terms 2..200 from Seppo Mustonen, terms 201..786 from Michel Marcus)
• S. Mustonen, On prime factors of numbers m^n+-1, 2010.
• Seppo Mustonen, On prime factors of numbers m^n+-1 [Local copy]

Crossrefs

Cf. A046800, A086251, A186522.

Programs

Mathematica

m = 2; n = 2; nmax = 200;
While[n <= nmax, {l = FactorInteger[m^n - 1]; s = 0;
     For[i = 1, i <= Length[l],
      i++, {p = l[[i, 1]];
       If[IntegerQ[(p - 1)/n] == True, s = s + l[[i, 2]]];}];
     a[n] = s;} n++;];
Table[a[n], {n, 2, nmax}]

PARI

a(n) = my(f = factor(2^n-1)); sum(k=1, #f~, ((f[k,1]-1) % n)==0); \\ Michel Marcus, Sep 10 2017

Extensions

Name edited by Thomas Ordowski, Sep 19 2017