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).
From R. J. Mathar, May 09 2014: (Start)
G.f.: x*(1 -x +3*x^2 -x^3)/((1-x-x^2)*(1-x)^2).
a(n) -2*a(n-1) + a(n-2) = A022120(n-4). (End)
a(n) = 3*Fibonacci(n+1) + 4*Fibonacci(n) - 2*(n+2). - 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(n+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}] (* A192962 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192963 *)
(* Additional programs *)
CoefficientList[Series[(1-x+3x^2-x^3)/((1-x-x^2)(1-x)^2), {x, 0, 40}], x] (* Vincenzo Librandi, May 09 2014 *)
With[{F=Fibonacci}, Table[3*F[n+1]+4*F[n] -2*(n+2), {n, 1, 40}]] (* G. C. Greubel, Jul 12 2019 *)
PROG
(PARI) vector(40, n, f=fibonacci; 3*f(n+1)+4*f(n)-2*(n+2)) \\ G. C. Greubel, Jul 1122019
(Magma) F:=Fibonacci; [3*F(n+1) +4*F(n) -2*(n+2): n in [1..40]]; // G. C. Greubel, Jul 12 2019
(Sage) f=fibonacci; [3*f(n+1) +4*f(n) -2*(n+2) for n in (1..40)] # G. C. Greubel, Jul 12 2019
(GAP) F:=Fibonacci;; List([1..40], n-> 3*F(n+1) +4*F(n) -2*(n+2)); # G. C. Greubel, Jul 12 2019
CROSSREFS
KEYWORD
nonn
AUTHOR
Clark Kimberling, Jul 13 2011
STATUS
approved