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A078680
Smallest m > 0 such that n*2^m + 1 is prime, or 0 if no such m exists.
5
1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 3, 2, 1, 1, 4, 3, 1, 6, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 8, 3, 1, 2, 1, 1, 2, 5, 1, 4, 1, 3, 2, 1, 2, 8, 583, 1, 2, 1, 1, 6, 1, 1, 4, 1, 2, 2, 5, 2, 4, 7, 1, 2, 1, 5, 2, 1, 1, 2, 3, 3, 2, 1, 1, 4, 3, 1, 2, 3, 1, 10, 1, 2, 4, 1, 2, 2, 1, 1, 8, 7, 2, 582, 1, 1, 2, 1, 1, 2, 3, 2
OFFSET
1,4
COMMENTS
Sierpiński proved that a(n)=0 for an infinite number of n. The first proven zero is n=78557. There is a conjecture that the first zero is n=65536 (which is equivalent to the statement that 2^(2^k)+1 is composite for k>4). - T. D. Noe, Feb 25 2011 [Edited by Jeppe Stig Nielsen, Jul 01 2020]
LINKS
N. J. A. Sloane, A Nasty Surprise in a Sequence and Other OEIS Stories, Experimental Mathematics Seminar, Rutgers University, Oct 10 2024, Youtube video; Slides [Mentions this sequence]
Eric Weisstein's World of Mathematics, Sierpiński Number of the Second Kind
FORMULA
If a(n) = 0, then a(2n) is also 0. If a(n) = m with m > 1, then a(2n) = m-1. - Jeppe Stig Nielsen, Feb 12 2023
MAPLE
A078680 := proc(n) for m from 1 do if isprime(n*2^m+1) then return m; end if; end do: end proc:
seq(A078680(n), n=1..30) ; # R. J. Mathar, Feb 25 2011
MATHEMATICA
Table[m=1; While[! PrimeQ[n*2^m+1], m++]; m, {n, 100}] (* T. D. Noe, Feb 25 2011 *)
PROG
(PARI) a(n)=if(n<0, 0, m=1; while(isprime(n*2^m+1)==0, m++); m)
CROSSREFS
Cf. A050412, A040076, A078683 (primes n*2^m+1).
Sequence in context: A204901 A016014 A067760 * A296072 A326700 A050412
KEYWORD
nonn,changed
AUTHOR
Benoit Cloitre, Dec 17 2002
EXTENSIONS
Offset corrected by Jaroslav Krizek, Feb 13 2011
STATUS
approved