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 08 2014: (Start)
G.f.: x*(1 -x +3*x^2)/((1-x-x^2)*(1-x)^2).
a(n) - a(n-1) = A192746(n-2). (End)
a(n) = 4*Fibonacci(n+2) - (3*n+4). - 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] + 3n - 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}] (* A192746 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192952 *)
(* Second program *)
With[{F=Fibonacci}, Table[4*F[n+2]-(3*n+4), {n, 0, 40}]] (* G. C. Greubel, Jul 12 2019 *)
PROG
(PARI) vector(40, n, n--; f=fibonacci; 4*f(n+2)-(3*n+4)) \\ G. C. Greubel, Jul 12 2019
(Magma) F:=Fibonacci; [4*F(n+2)-(3*n+4): n in [0..40]]; // G. C. Greubel, Jul 12 2019
(Sage) f=fibonacci; [4*f(n+2)-(3*n+4) for n in (0..40)] # G. C. Greubel, Jul 12 2019
(GAP) F:=Fibonacci;; List([0..40], n-> 4*F(n+2)-(3*n+4)); # G. C. Greubel, Jul 12 2019
CROSSREFS
KEYWORD
nonn
AUTHOR
Clark Kimberling, Jul 13 2011
STATUS
approved