

A168566


a(n) = (n1)*(n+2)*(n^2 + n + 2)/4.


1



0, 8, 35, 99, 224, 440, 783, 1295, 2024, 3024, 4355, 6083, 8280, 11024, 14399, 18495, 23408, 29240, 36099, 44099, 53360, 64008, 76175, 89999, 105624, 123200, 142883, 164835, 189224, 216224, 246015, 278783, 314720, 354024, 396899, 443555, 494208
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,2


COMMENTS

The products of two consecutive numbers in this sequence may be evaluated in terms of the Frobenius numbers for 5 consecutive integers, A138985(n) = F(n): for n>0, a(2n1)*a(2n) = F(4n^22)^2  (2n)^2; a(2n)*a(2n+1) = F(4n^2+4n)^2  (2n+1)^2.  Charlie Marion, Jan 23 2012


LINKS

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (5,10,10,5,1).


FORMULA

G.f.: x^2*(8 + 5*x  4*x^2 + x^3)/(x1)^5.  R. J. Mathar, Jan 04 2011
a(n) = A000217(n)^2  1.  Charlie Marion, Sep 27 2011


MATHEMATICA

s=0; lst={s}; Do[s+=n^3; AppendTo[lst, s], {n, 2, 5!}]; lst (* Vladimir Joseph Stephan Orlovsky, Apr 27 2010 *)


PROG

(MAGMA) [(n1)*(n+2)*(n^2+n+2)/4: n in [1..50]]


CROSSREFS

Cf. A000578 (first differences).
Sequence in context: A257454 A100907 A212903 * A058102 A212674 A189592
Adjacent sequences: A168563 A168564 A168565 * A168567 A168568 A168569


KEYWORD

nonn,easy


AUTHOR

Vincenzo Librandi, Nov 30 2009


STATUS

approved



