OFFSET
1,7
COMMENTS
Sierpiński showed that a(n) = -1 infinitely often. John Selfridge showed that a(78557) = -1 and it is conjectured that a(n) >= 0 for all n < 78557.
Determining a(131072) = a(2^17) is equivalent to finding the next Fermat prime after F_4 = 2^16 + 1. - Jeppe Stig Nielsen, Jul 27 2019
LINKS
T. D. Noe, Table of n, a(n) for n = 1..10000 (with help from the Sierpiński problem website)
Ray Ballinger and Wilfrid Keller, The Sierpiński Problem: Definition and Status
Seventeen or Bust, A Distributed Attack on the Sierpiński problem
EXAMPLE
1*(2^0)+1=2 is prime, so a(1)=0;
3*(2^1)+1=5 is prime, so a(3)=1;
For n=7, 7+1 and 7*2+1 are composite, but 7*2^2+1=29 is prime, so a(7)=2.
MATHEMATICA
Do[m = 0; While[ !PrimeQ[n*2^m + 1], m++ ]; Print[m], {n, 1, 110} ]
sm[n_]:=Module[{k=0}, While[!PrimeQ[n 2^k+1], k++]; k]; Array[sm, 120] (* Harvey P. Dale, Feb 05 2020 *)
CROSSREFS
KEYWORD
easy,nice,sign
AUTHOR
STATUS
approved