OFFSET
0,3
COMMENTS
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-2,-1,1).
FORMULA
a(n) = 3*a(n-1) - 2*a(n-2) - a(n-3) + a(n-4).
G.f.: (1 - 3*x + 4*x^2)/((1 - x)^2*(1 - x - x^2)). - Colin Barker, May 11 2014
a(n) = Lucas(n+2) + Fibonacci(n+1) - (2*n+3). - G. C. Greubel, Jul 25 2019
MATHEMATICA
(* First program *)
q = x^2; s = x + 1; z = 40;
p[0, x]:= 1;
p[n_, x_]:= x*p[n-1, x] + (n-1)^2;
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}] (* A192981 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192982 *)
(* Additional programs *)
Table[LucasL[n+2]_Fibonacci[n+1]-(2*n+3), {n, 0, 40}] (* G. C. Greubel, Jul 25 2019 *)
PROG
(PARI) vector(40, n, n--; f=fibonacci; f(n+3)+2*f(n+1) -(2*n+3)) \\ G. C. Greubel, Jul 25 2019
(Magma) F:=Fibonacci; [F(n+3)+2*F(n+1) -(2*n+3): n in [0..40]]; // G. C. Greubel, Jul 25 2019
(Sage) f=fibonacci; [f(n+3)+2*f(n+1) -(2*n+3) for n in (0..40)] # G. C. Greubel, Jul 25 2019
(GAP) F:=Fibonacci;; List([0..40], n-> F(n+3)+2*F(n+1) -(2*n+3)); # G. C. Greubel, Jul 25 2019
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Jul 14 2011
STATUS
approved