

A182590


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


4



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
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OFFSET

2,9


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)


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.


EXAMPLE

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.


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}]


PROG

(PARI) a(n) = my(f = factor(2^n1)); sum(k=1, #f~, ((f[k, 1]1) % n)==0); \\ Michel Marcus, Sep 10 2017


CROSSREFS

Cf. A046800, A086251, A186522.
Sequence in context: A220163 A102715 A254687 * A047846 A212632 A025885
Adjacent sequences: A182587 A182588 A182589 * A182591 A182592 A182593


KEYWORD

nonn


AUTHOR

Seppo Mustonen, Nov 22 2010


EXTENSIONS

Name edited by Thomas Ordowski, Sep 19 2017


STATUS

approved



