OFFSET
0,2
COMMENTS
Inverse binomial transform is 1, 13, 27, 14, 0, 0, 0, ... (0 continued).
LINKS
Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
FORMULA
O.g.f.: (1 + 10*x + 4*x^2 - x^3)/(1 - x)^4.
E.g.f.: (6 + 78*x + 81*x^2 + 14*x^3)*exp(x)/6.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
a(n) = (n + 2)*(14*n^2 + 11*n + 3)/6. Therefore:
a(6*k + r) = 504*k^3 + 18*(14*r + 13)*k^2 + (42*r^2 + 78*r + 25)*k + a(r), with 0 <= r <= 5. Example: for r=5, a(6*k + 5) = (6*k + 7)*(84*k^2 + 151*k + 68).
MAPLE
seq((n + 2)*(14*n^2 + 11*n + 3)/6, n=0..50); # Peter Luschny, Feb 21 2018
MATHEMATICA
Table[(n + 2) (14 n^2 + 11 n + 3)/6, {n, 0, 50}]
(* Second program: *)
LinearRecurrence[{4, -6, 4, -1}, {1, 14, 54, 135}, 41] (* Jean-François Alcover, Feb 21 2018 *)
PROG
(Sage) [(n+2)*(14*n^2+11*n+3)/6 for n in (0..50)]
(Maxima) makelist((n+2)*(14*n^2+11*n+3)/6, n, 0, 50);
(Magma) [(n+2)*(14*n^2+11*n+3)/6: n in [0..50]];
(GAP) List([0..50], n -> (n+2)*(14*n^2+11*n+3)/6);
(PARI) a(n)=(n+2)*(14*n^2+11*n+3)/6 \\ Charles R Greathouse IV, Feb 21 2018
(PARI) Vec((1 + 10*x + 4*x^2 - x^3)/(1 - x)^4 + O(x^60)) \\ Colin Barker, Feb 22 2018
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Bruno Berselli, Feb 20 2018
STATUS
approved