

A252168


Smallest k>0 such that (2n1)2^k is prime, or 1 if no such k exists.


2



2, 3, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 4, 1, 1, 2, 4, 1, 2, 1, 1, 2, 1, 2, 4, 1, 2, 4, 1, 1, 2, 3, 1, 2, 1, 1, 2, 3, 1, 2, 1, 2, 4, 1, 2, 4, 3, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 4, 3, 4, 4, 47, 1, 2, 1, 2, 6, 1, 1, 2, 3, 3, 8, 1, 1, 2, 3, 1, 2, 5, 1, 2, 1, 2, 4, 1, 2, 4, 1, 1, 2, 3, 3, 6, 1, 1, 2, 1, 1, 2, 3, 3, 4
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OFFSET

1,1


COMMENTS

It is known that a(254602) = 1, because 5092032^k is always divisible by 3, 5, 7, 13, 17, or 241. a(1147) is the first unknown term.
a((A101036(n)+1)/2) = 1, so there are infinitely many n such that a(n) = 1.
a((A133122(n)+1)/2) = A096502((A133122(n)1)/2).


LINKS

Table of n, a(n) for n=1..105.


EXAMPLE

a(12) = 2 because 2*121 = 23 and that 232^1 = 21 is not prime but 232^2 = 19 is.
a(69) = 6 because 2*691 = 137, 1372^k is composite for k = 1, 2, 3, 4, 5 and prime for k = 6.
Even the smallest k can be also very large. For example, a(169) = 791.
a(1147) > 65536.


MATHEMATICA

Table[k = 1; While[!PrimeQ[Abs[(2*n1)  2^k]], k++]; k, {n, 1, 1000}]


PROG

(PARI) A252168(n, k)={ k  k=1; n=2*n1; while(!ispseudoprime(abs((n2^k++)), ); k }


CROSSREFS

Cf. A096502, A006285, A133122, A188903.
Cf. A046067, A046069, A067760.
Sequence in context: A086195 A086197 A139336 * A100619 A211984 A275471
Adjacent sequences: A252165 A252166 A252167 * A252169 A252170 A252171


KEYWORD

nonn,hard


AUTHOR

Eric Chen, Dec 14 2014


STATUS

approved



