A186522 -id:A186522 - cp's OEIS frontend

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

Seppo Mustonen, Nov 22 2010

nonn

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)

			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.

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]

Cf. A046800, A086251, A186522.

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}]

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

Name edited by Thomas Ordowski, Sep 19 2017

A186283 Least number k such that k*n+1 is a prime dividing 2^n-1.

Original entry on oeis.org

1, 2, 1, 6, 1, 18, 2, 8, 1, 2, 1, 630, 3, 2, 1, 7710, 1, 27594, 2, 6, 1, 2, 10, 24, 105, 9728, 1, 8, 1, 69273666, 8, 18166, 1285, 2, 1, 6, 4599, 2, 1, 326, 1, 10, 2, 14, 1, 50, 2, 90462791808, 5, 2, 1, 120, 1615, 16, 2, 568, 1, 3050, 1, 37800705069076950, 11545611, 2, 4, 126, 1, 2891160, 2, 145690999102, 1

Offset: 2

Bill McEachen, Feb 16 2011

nonn

The smallest prime factor of 2^n-1 of the form k*n+1 is A186522(n).

By Fermat's little theorem, a(n) = 1 if and only if n+1 is an odd prime. Further, for prime p, a(p) = 2 if and only if p is in A002515. - Thomas Ordowski, Sep 03 2017

			For n=8, 2^n-1 = 255 = 3 * 5 * 17.  The smallest prime factor of the form k*n+1 is 17 = 2*8+1. Hence, a(8) = 2.

Kenneth H. Rosen, Elementary Number Theory and Its Applications, 3rd Ed, Theorem 6.12, p. 225

Max Alekseyev, Table of n, a(n) for n = 2..1236 (terms to a(300) from David A. Corneth)
Will Edgington, Mersenne Page [from Internet Archive Wayback Machine]
Eric W. Weisstein, MathWorld: Mersenne Number
Eric W. Weisstein, MathWorld: Mersenne Prime

Cf. A000225, A049479, A079324.

Table[p=First/@FactorInteger[2^n-1]; (Select[p, Mod[#1,n] == 1 &, 1][[1]] - 1)/n, {n, 2, 70}]

a(n) = {if(isprime(n+1),return(1)); my(f = factor(2^n - 1)[,1]); for(i=1,#f, if(f[i]%n == 1, return((f[i]-1) / n)))} \\ David A. Corneth, Sep 03 2017

a(n) = (A186522(n)-1)/n.

cp's OEIS Frontend

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

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