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A035050 a(n) is the smallest k such that k*2^n + 1 is prime. 21
1, 1, 1, 2, 1, 3, 3, 2, 1, 15, 12, 6, 3, 5, 4, 2, 1, 6, 3, 11, 7, 11, 25, 20, 10, 5, 7, 15, 12, 6, 3, 35, 18, 9, 12, 6, 3, 15, 10, 5, 6, 3, 9, 9, 15, 35, 19, 27, 15, 14, 7, 14, 7, 20, 10, 5, 27, 29, 54, 27, 31, 36, 18, 9, 12, 6, 3, 9, 31, 23, 39, 39, 40, 20, 10, 5, 58, 29, 15, 36, 18, 9, 13 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,4
COMMENTS
From Ulrich Krug (leuchtfeuer37(AT)gmx.de), Jun 05 2010: (Start)
If a(i) = 2 * m then a(i+1) = m.
Proof: (I) a(i) = 2*m, 2*m * 2^i + 1 = m*2^(i+1) + 1 prime, so a(i+1) <= m;
(II) if a(i+1) = m-d for an integer d > 0, (m-d) * 2^(i+1) + 1 = (2*m-2*d) * 2^i + 1 prime;
(2m-2d) < 2m contradiction to a(i) = 2 * m, d = 0.
(End)
Conjecture: for n > 0, a(n) = k < 2^n, so k*2^n + 1 is a Proth prime A080076. - Thomas Ordowski, Apr 13 2019
LINKS
Gareth A. Jones and Alexander K. Zvonkin, Groups of prime degree and the Bateman-Horn conjecture, 2021.
FORMULA
a(n) << 19^n by Xylouris' improvement to Linnik's theorem. - Charles R Greathouse IV, Dec 10 2013
EXAMPLE
a(3)=2 because 1*2^3 + 1 = 9 is composite, 2*2^3 + 1 = 17 is prime.
a(99)=219 because 2^99k + 1 is not prime for k=1,2,...,218. The first term which is not a composite number of this arithmetic progression is 2^99*219 + 1.
MATHEMATICA
a = {}; Do[k = 0; While[ ! PrimeQ[k 2^n + 1], k++ ]; AppendTo[a, k], {n, 0, 100}]; a (* Artur Jasinski *)
PROG
(PARI) a(n) = {my(k = 1); while (! isprime(2^n*k+1), k++); k; }
(Magma) sol:=[]; m:=1; for n in [0..82] do k:=0; while not IsPrime(k*2^n+1) do k:=k+1; end while; sol[m]:=k; m:=m+1; end for; sol; // Marius A. Burtea, Jun 05 2019
CROSSREFS
Analogous case is A034693. Special subscripts (n's for a(n)=1) are the exponents of known Fermat primes: A000215. See also Fermat numbers A000051.
Sequence in context: A245326 A241534 A337137 * A198790 A306995 A212907
KEYWORD
nonn
AUTHOR
STATUS
approved

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Last modified April 23 14:32 EDT 2024. Contains 371914 sequences. (Running on oeis4.)