A182484 a(n) = Product_{k=1..p-1} k^(2k-p-1), where p = prime(n).

1, 1, 4, 225, 8930250000, 9001015156742400, 377177413291384771899817984000000, 17617791710438789613561393948051882397138944

Offset

1,3

Comments

a(n) is always an integer [Vandendriessche and Lee, Problem A13]. - N. J. A. Sloane, Jan 10 2017

a(9) contains 70 digits; a(10) contains 121 digits; a(11) contains 142 digits; a(12) contains 213 digits; a(13) contains 269 digits.

Conjecture (false!): a(n) = Product_{k=1..p-1} k^(2k-p-1) is a perfect square if, and only if p = prime(n).

The conjecture above is disproved by the counterexample p=63 for which the indicated product is a square, yet 63 is not a prime. [John W. Layman, May 01 2012]

Links

Table of n, a(n) for n=1..8.

Peter Vandendriessche and Hojoo Lee, Problems in elementary number theory, Problem A13

Example

a(3) = 4 because, for prime(3) = 5,
Product_{k=1..4} k^(2k-6) = 1^(-4)*2^(-2)*3^0*4^2 = 4.

Maple

with(numtheory):seq(product(k^(2*k-ithprime(n)-1), k=1.. ithprime(n)-1), n=1..9);

Mathematica

Table[Product[k^(2*k - n - 1), {k, n-1}], {n, Prime[Range[8]]}] (* T. D. Noe, May 01 2012 *)

Crossrefs

Sequence in context: A211610 A364481 A042539 * A159281 A290346 A113255

Adjacent sequences: A182481 A182482 A182483 * A182485 A182486 A182487

Keyword

nonn

Author

Michel Lagneau, May 01 2012

Status

approved