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A040076
Smallest m >= 0 such that n*2^m + 1 is prime, or -1 if no such m exists.
22
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
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
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. A033809, A046067 (odd n), A057192 (prime n).
Sequence in context: A257510 A305445 A225721 * A019269 A204459 A035155
KEYWORD
easy,nice,sign
STATUS
approved