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a(n) = (n!)^3 + 1.
3

%I #13 Nov 30 2016 12:57:04

%S 2,2,9,217,13825,1728001,373248001,128024064001,65548320768001,

%T 47784725839872001,47784725839872000001,63601470092869632000001,

%U 109903340320478724096000001,241457638684091756838912000001

%N a(n) = (n!)^3 + 1.

%C Since this is a sum of two cubes, it can be factorized. So all terms are divisible by n!+1. Thus only two primes occur in this sequence: a(0) and a(1). - _Dmitry Kamenetsky_, Sep 30 2008

%D M. Le, On the Interesting Smarandache Product Sequences, Smarandache Notions Journal, Vol. 9, No. 1-2, 1998, 133-134.

%D M. Le, The Primes in Smarandache Power Product Sequences, Smarandache Notions Journal, Vol. 9, No. 1-2, 1998, 96-97.

%D F. Iacobescu, Smarandache Partition Type and Other Sequences, Bull. Pure Appl. Sciences, Vol. 16E, No. 2 (1997), pp. 237-240.

%H G. C. Greubel, <a href="/A019514/b019514.txt">Table of n, a(n) for n = 0..181</a>

%H F. Smarandache, <a href="http://www.gallup.unm.edu/~smarandache/CP2.pdf">Collected Papers, Vol. II</a>

%H F. Smarandache, <a href="http://www.gallup.unm.edu/~smarandache/Sequences-book.pdf">Sequences of Numbers Involved in Unsolved Problems</a>.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Factorial.html">Factorial</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/SmarandacheSequences.html">Smarandache Sequences</a>

%t Table[(n!)^3 + 1, {n,0,25}] (* _G. C. Greubel_, Nov 30 2016 *)

%K nonn,easy

%O 0,1

%A R. Muller