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%I #32 Dec 21 2021 23:49:44
%S 0,0,1,0,1,0,2,1,1,0,1,0,2,1,1,0,3,0,6,1,1,0,1,2,2,1,2,0,1,0,8,3,1,2,
%T 1,0,2,5,1,0,1,0,2,1,2,0,583,1,2,1,1,0,1,1,4,1,2,0,5,0,4,7,1,2,1,0,2,
%U 1,1,0,3,0,2,1,1,4,3,0,2,3,1,0,1,2,4,1,2,0,1,1,8,7,2,582,1,0,2,1,1,0,3,0
%N Smallest m >= 0 such that n*2^m + 1 is prime, or -1 if no such m exists.
%C 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.
%C 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
%H T. D. Noe, <a href="/A040076/b040076.txt">Table of n, a(n) for n = 1..10000</a> (with help from the Sierpiński problem website)
%H Ray Ballinger and Wilfrid Keller, <a href="http://www.prothsearch.com/sierp.html">The Sierpiński Problem: Definition and Status</a>
%H Seventeen or Bust, <a href="http://www.seventeenorbust.com/">A Distributed Attack on the Sierpiński problem</a>
%e 1*(2^0)+1=2 is prime, so a(1)=0;
%e 3*(2^1)+1=5 is prime, so a(3)=1;
%e For n=7, 7+1 and 7*2+1 are composite, but 7*2^2+1=29 is prime, so a(7)=2.
%t Do[m = 0; While[ !PrimeQ[n*2^m + 1], m++ ]; Print[m], {n, 1, 110} ]
%t sm[n_]:=Module[{k=0},While[!PrimeQ[n 2^k+1],k++];k]; Array[sm,120] (* _Harvey P. Dale_, Feb 05 2020 *)
%Y For the corresponding primes see A050921.
%Y Cf. A103964, A040081.
%Y Cf. A033809, A046067 (odd n), A057192 (prime n).
%K easy,nice,sign
%O 1,7
%A _David W. Wilson_
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