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 A040076 Smallest m >= 0 such that n*2^m+1 is prime, or -1 if no such m exists. 20
 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, 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, 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 (list; graph; refs; listen; history; text; internal format)
 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 For the corresponding primes see A050921. Cf. A103964, A040081. Cf. A033809, A046067 (odd n), A057192 (prime n). Sequence in context: A257510 A305445 A225721 * A019269 A204459 A035155 Adjacent sequences:  A040073 A040074 A040075 * A040077 A040078 A040079 KEYWORD easy,nice,sign AUTHOR STATUS approved

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Last modified July 29 09:41 EDT 2021. Contains 346344 sequences. (Running on oeis4.)