

A128164


Least k>2 such that (n^k1)/(n1) is prime, or 0 if no such prime exists.


7



3, 3, 0, 3, 3, 5, 3, 0, 19, 17, 3, 5, 3, 3, 0, 3, 25667, 19, 3, 3, 5, 5, 3, 0, 7, 3, 5, 5, 5, 7, 0, 3, 13, 313, 0, 13, 3, 349, 5, 3, 1319, 5, 5, 19, 7, 127, 19, 0, 3, 4229, 103, 11, 3, 17, 7, 3, 41, 3, 7, 7, 3, 5, 0, 19, 3, 19, 5, 3, 29, 3, 7, 5, 5, 3, 41, 3, 3, 5, 3, 0, 23, 5, 17, 5, 11, 7, 61, 3, 3
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OFFSET

2,1


COMMENTS

a(n) = A084740(n) for all n except n = p1, where p is an odd prime, for which A084740(n) = 2.
All nonzero terms are odd primes.
a(n) = 0 for n = {4,9,16,25,32,36,49,64,81,100,121,125,144,...}, which are the perfect powers with exceptions of the form n^(p^m) where p>2 and (n^(p^(m+1))1)/(n^(p^m)1) are prime and m>=1 (in which case a(n^(p^m))=p).  Max Alekseyev, Jan 24 2009
a(n) = 3 for n in A002384, i.e., for n such that n^2 + n + 1 is prime.
a(152) > 20000.  Eric Chen, Jun 01 2015


LINKS

Max Alekseyev, Table of n, a(n) for n = 2..151
Eric Chen, Table of n, a(n) for n = 2..1001 status
H. Dubner, Generalized repunit primes, Math. Comp., 61 (1993), 927930.
Eric Weisstein's World of Mathematics, Repunit


PROG

(PARI) a052409(n) = if(ispower(n), ispower(n), n>1)
a052410(n) = n^(1/a052409(n))
is(n) = issquare(n)  (ispower(n) && !ispseudoprime((n^a052410(n)1)/(n1)))
a(n) = if(is(n), 0, forprime(p=3, 6000, if(ispseudoprime((n^p1)/(n1)), return(p)))) \\ Eric Chen, Jun 01 2015


CROSSREFS

Cf. A084738, A065854, A084740, A084741, A065507, A084742.
Sequence in context: A084055 A084103 A036477 * A260636 A245256 A140686
Adjacent sequences: A128161 A128162 A128163 * A128165 A128166 A128167


KEYWORD

nonn


AUTHOR

Alexander Adamchuk, Feb 20 2007


EXTENSIONS

a(18) = 25667 found by Henri Lifchitz, Sep 26 2007


STATUS

approved



