OFFSET
1,2
COMMENTS
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..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.: x*(1+3*x^2)/((1-x-x^2)*(1-x)^2). - R. J. Mathar, May 11 2014
a(n) = Lucas(n+4) - Fibonacci(n-1) - 2*(2*n+3). - Ehren Metcalfe, Jul 13 2019
MATHEMATICA
(* First program *)
q = x^2; s = x + 1; z = 40;
p[0, x]:= 1;
p[n_, x_]:= x*p[n-1, x] + 2*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}] (* A192973 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192974 *)
(* Additional programs *)
LinearRecurrence[{3, -2, -1, 1}, {1, 3, 10, 23}, 50] (* Vincenzo Librandi, Jul 14 2019 *)
With[{F = Fibonacci}, Table[F[n+4]+3*F[n+2] -2*(2*n+3), {n, 40}]] (* G. C. Greubel, Jul 24 2019 *)
PROG
(Magma) [Lucas(n+4)-Fibonacci(n-1)-2*(2*n+3): n in [1..40]]; // Vincenzo Librandi, Jul 14 2019
(PARI) vector(40, n, f=fibonacci; f(n+4)+3*f(n+2) -2*(2*n+3)) \\ G. C. Greubel, Jul 24 2019
(Sage) f=fibonacci; [f(n+4)+3*f(n+2) -2*(2*n+3) for n in (1..40)] # G. C. Greubel, Jul 24 2019
(GAP) F:=Fibonacci;; List([1..40], n-> F(n+4)+3*F(n+2) -2*(2*n+3)); # G. C. Greubel, Jul 24 2019
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Jul 13 2011
STATUS
approved