A019514 a(n) = (n!)^3 + 1.

2, 2, 9, 217, 13825, 1728001, 373248001, 128024064001, 65548320768001, 47784725839872001, 47784725839872000001, 63601470092869632000001, 109903340320478724096000001, 241457638684091756838912000001

Offset: 0

Author

R. Muller

Keywords

• nonn
• easy

Comments

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

References

• M. Le, On the Interesting Smarandache Product Sequences, Smarandache Notions Journal, Vol. 9, No. 1-2, 1998, 133-134.
• M. Le, The Primes in Smarandache Power Product Sequences, Smarandache Notions Journal, Vol. 9, No. 1-2, 1998, 96-97.
• F. Iacobescu, Smarandache Partition Type and Other Sequences, Bull. Pure Appl. Sciences, Vol. 16E, No. 2 (1997), pp. 237-240.

Links

• G. C. Greubel, Table of n, a(n) for n = 0..181
• F. Smarandache, Collected Papers, Vol. II
• F. Smarandache, Sequences of Numbers Involved in Unsolved Problems.
• Eric Weisstein's World of Mathematics, Factorial
• Eric Weisstein's World of Mathematics, Smarandache Sequences

Programs

Mathematica

Table[(n!)^3 + 1, {n,0,25}] (* G. C. Greubel, Nov 30 2016 *)