OFFSET
0,2
COMMENTS
The finite simple continued fraction [1;n-1,n,n+1] has a numerator in the resulting rational number (n^3+2*n+n^2+1)/(n*(n^2+1)) that is the same as a(n). - J. M. Bergot, Sep 29 2011
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
FORMULA
G.f.: ( 1+x+3*x^2+x^3 ) / (x-1)^4 . - R. J. Mathar, Sep 29 2011
a(n) = A053698(n)+n. - Bruno Berselli, Sep 30 2011
MATHEMATICA
Table[n^3 + (n + 1)^2, {n, 0, 45}]
LinearRecurrence[{4, -6, 4, -1}, {1, 5, 17, 43}, 50] (* Harvey P. Dale, Dec 03 2014 *)
PROG
(Magma) [n^3 + (n+1)^2: n in [0..50]]; // Vincenzo Librandi, Sep 30 2011
(PARI) a(n)=n^3+(n+1)^2 \\ Charles R Greathouse IV, Oct 07 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Giovanni Teofilatto, Jan 03 2005
EXTENSIONS
More terms from Mark Hudson and Farideh Firoozbakht, Jan 04 2005
STATUS
approved