OFFSET
0,3
COMMENTS
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-5,1,2,-1).
FORMULA
a(n) = 4*a(n-1) - 5*a(n-2) + a(n-3) + 2*a(n-4) - a(n-5).
G.f.: x*(1-x+2*x^2)/((1-x)^3*(1-x-x^2)). - Colin Barker, May 12 2014
a(n) = 3*Fibonacci(n+4) - n*(n+4) - 9. - Ehren Metcalfe, Jul 13 2019
MATHEMATICA
(* First program *)
p[0, x]:= 1; p[n_, x_]:= x*p[n-1, x] +n^2 +1;
Table[Expand[p[n, x]], {n, 0, 7}]
reduce[{p1_, q_, s_, x_}]:= FixedPoint[(s PolynomialQuotient @@ #1 + PolynomialRemainder @@ #1 &)[{#1, q, x}] &, p1]
t = Table[reduce[{p[n, x], q, s, x}], {n, 0, z}];
u1 = Table[Coefficient[Part[t, n], x, 0], {n, 1, z}] (* A192953 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192389 *)
(* Additional programs *)
CoefficientList[Series[x*(1-x+2*x^2)/((1-x)^3*(1-x-x^2)), {x, 0, 40}], x] (* Vincenzo Librandi, May 13 2014 *)
Table[3*Fibonacci[n+4] -n*(n+4)-9, {n, 0, 40}] (* G. C. Greubel, Jul 24 2019 *)
PROG
(PARI) Vec(x*(1-x+2*x^2)/((1-x)^3*(1-x-x^2)) + O(x^40)) \\ Colin Barker, May 12 2014
(PARI) vector(40, n, n--; 3*fibonacci(n+2)-n*(n+4)-9) \\ G. C. Greubel, Jul 24 2019
(Magma) [3*Fibonacci(n+2)-n*(n+4)-9: n in [0..40]]; // G. C. Greubel, Jul 24 2019
(Sage) [3*fibonacci(n+2)-n*(n+4)-9 for n in (0..40)] # G. C. Greubel, Jul 24 2019
(GAP) List([0..40], n-> 3*Fibonacci(n+2)-n*(n+4)-9); # G. C. Greubel, Jul 24 2019
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Jul 13 2011
STATUS
approved