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).
From R. J. Mathar, May 09 2014: (Start)
G.f.: (1 -4*x +8*x^2 -3*x^3)/((1-x-x^2)*(1-x)^2).
a(n) - 2*a(n-1) +a(n-2) = A022089(n-3). (End)
a(n) = 6*Fibonacci(n+1) - (2*n+5). - G. C. Greubel, Jul 12 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^2 - 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}] (* A192958 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192959 *)
(* Second program *)
With[{F=Fibonacci}, Table[6*F[n+1]-(2*n+5), {n, 0, 40}]] (* G. C. Greubel, Jul 12 2019 *)
PROG
(PARI) vector(40, n, n--; f=fibonacci; 6*f(n+1)-(2*n+5)) \\ G. C. Greubel, Jul 12 2019
(Magma) F:=Fibonacci; [6*F(n+1)-(2*n+5): n in [0..40]]; // G. C. Greubel, Jul 12 2019
(Sage) f=fibonacci; [6*f(n+1)-(2*n+5) for n in (0..40)] # G. C. Greubel, Jul 12 2019
(GAP) F:=Fibonacci;; List([0..40], n-> 6*F(n+1)-(2*n+5)); # G. C. Greubel, Jul 12 2019
CROSSREFS
KEYWORD
sign
AUTHOR
Clark Kimberling, Jul 13 2011
STATUS
approved