

A096502


a(n) = k is the smallest exponent k such that 2^k  (2n+1) is a prime number, or 0 if no such k exists.


14



2, 3, 3, 39, 4, 4, 4, 5, 6, 5, 5, 6, 5, 5, 5, 7, 6, 6, 11, 7, 6, 29, 6, 6, 7, 6, 6, 7, 6, 6, 6, 8, 8, 7, 7, 10, 9, 7, 8, 9, 7, 8, 7, 7, 8, 7, 8, 10, 7, 7, 26, 9, 7, 8, 7, 7, 10, 7, 7, 8, 7, 7, 7, 47, 8, 14, 9, 11, 10, 9, 10, 8, 9, 8, 8, 31, 8, 8, 15, 8, 10, 9
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

0,1


COMMENTS

As D. W. Wilson observes, this is similar to the Riesel/Sierpinski problem and there is e.g. no prime of the form 2^k  777149, which is divisible by 3,5,7,13,19,37 or 73 if k is in 1+2Z, 2+4Z, 4+12Z, 8+12Z, 12+36Z, 0+36Z resp. 24+36Z. Already for n=935 it is difficult to find a solution. Is this linked to the fact that 2n+1=1871 is member of a prime quadruple (A007530) and quintuple (A022007)?  M. F. Hasler, Apr 07 2008


LINKS

T. D. Noe, Table of n, a(n) for n = 0..934
F. Firoozbakht, M. F. Hasler, Variations on Euclid's formula for Perfect Numbers, JIS 13 (2010) #10.3.1.


EXAMPLE

a(0)=A000043(1)=2, a(1)=A050414(1)=3, a(2)=A059608(1)=3, a(3)=A059609(1)=39.
For n=110 and n=111 even these smallest exponents are rather large: a(110)=714, a(111)=261 which mean that 2^714221 and 2^261223 are the least corresponding prime numbers.


MATHEMATICA

Table[k = 1; While[2^k < n  ! PrimeQ[2^k  n], k++]; k, {n, 1, 1869, 2}] (* T. D. Noe, Mar 18 2013 *)


PROG

(PARI) A096502(n, k)={ k  k=log(n)\log(2)+1; n=2*n+1; while( !ispseudoprime(2^k++n), ); k } /* will take a long time for n=935... */  M. F. Hasler, Apr 07 2008


CROSSREFS

Cf. A000043, A007530, A022007, A050414, A059608, A059609.
Sequence in context: A281786 A319354 A100650 * A101462 A345751 A242786
Adjacent sequences: A096499 A096500 A096501 * A096503 A096504 A096505


KEYWORD

nonn,hard


AUTHOR

Labos Elemer, Jul 09 2004


STATUS

approved



