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(2n-1)*a(2n) = F(4n^2-2)^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)/(x-1)^5. - R. J. Mathar, Jan 04 2011
From G. C. Greubel, Jul 26 2016: (Start)
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
E.g.f.: (1/4)*(4 - (4 - 4*x + 14*x^2 + 8*x^3)*exp(x)). (End)
Sum_{n>=2} 1/a(n) = 49/36 - tanh(sqrt(7)*Pi/2)*Pi/sqrt(7). - Amiram Eldar, Mar 02 2023
MATHEMATICA
s=0; lst={s}; Do[s+=n^3; AppendTo[lst, s], {n, 2, 5!}]; lst (* Vladimir Joseph Stephan Orlovsky, Apr 27 2010 *)
Table[(n-1)*(n+2)*(n^2 + n + 2)/4, {n, 1, 25}] (* G. C. Greubel, Jul 26 2016 *)
LinearRecurrence[{5, -10, 10, -5, 1}, {0, 8, 35, 99, 224}, 40] (* Harvey P. Dale, Jan 21 2019 *)
PROG
(Magma) [(n-1)*(n+2)*(n^2+n+2)/4: n in [1..50]]
(PARI) a(n)=(n-1)*(n+2)*(n^2+n+2)/4 \\ Charles R Greathouse IV, Jul 26 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Nov 30 2009
STATUS
approved