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%I #72 Dec 17 2022 06:05:38
%S 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,
%T 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,
%U 4,3,3,3,4,5,3,1,5,3,2,3,4,2,3,2,4,3
%N Number of distinct prime factors of 2^n - 1 of the form k*n + 1.
%C From _Thomas Ordowski_, Sep 08 2017: (Start)
%C By Bang's theorem, a(n) > 0 for all n > 1, see A186522.
%C Primes p such that a(p) = 1 are the Mersenne exponents A000043.
%C Composite numbers m for which a(m) = 1 are A292079.
%C 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).
%C Theorem: if n is prime, then a(n) = A046800(n).
%C Conjecture: if a(n) = A046800(n), then n is prime.
%C Problem: is a(n) < A046800(n) for every composite n? (End)
%H Charles R Greathouse IV, <a href="/A182590/b182590.txt">Table of n, a(n) for n = 2..1200</a> (terms 2..200 from Seppo Mustonen, terms 201..786 from Michel Marcus)
%H S. Mustonen, <a href="http://www.survo.fi/papers/PrimeFactors2010.pdf">On prime factors of numbers m^n+-1</a>, 2010.
%H Seppo Mustonen, <a href="/A182590/a182590.pdf">On prime factors of numbers m^n+-1</a> [Local copy]
%e 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.
%t m = 2; n = 2; nmax = 200;
%t While[n <= nmax, {l = FactorInteger[m^n - 1]; s = 0;
%t For[i = 1, i <= Length[l],
%t i++, {p = l[[i, 1]];
%t If[IntegerQ[(p - 1)/n] == True, s = s + l[[i, 2]]];}];
%t a[n] = s;} n++;];
%t Table[a[n], {n, 2, nmax}]
%o (PARI) a(n) = my(f = factor(2^n-1)); sum(k=1, #f~, ((f[k,1]-1) % n)==0); \\ _Michel Marcus_, Sep 10 2017
%Y Cf. A046800, A086251, A186522.
%K nonn
%O 2,9
%A _Seppo Mustonen_, Nov 22 2010
%E Name edited by _Thomas Ordowski_, Sep 19 2017
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