A123194 a(n) = (n+1)*Fibonacci(n+2) + 3.

4, 7, 12, 23, 43, 81, 150, 275, 498, 893, 1587, 2799, 4904, 8543, 14808, 25555, 43931, 75261, 128538, 218923, 371934, 630457, 1066467, 1800603, 3034828, 5106871, 8580900, 14398415, 24129163, 40388073, 67527582, 112786499, 188195274, 313733813, 522562323

Offset

0,1

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Philip Matchett Wood, A bijective proof of f_{n+4}+f_1+2f_2+...+nf_n=(n+1)f_{n+2}+3, Integers 6 (2006), A2, 4 pp.

Index entries for linear recurrences with constant coefficients, signature (3,-1,-3,1,1).

Formula

G.f.: (4 - 5*x - 5*x^2 + 6*x^3 + 3*x^4)/((1 - x)*(1 - x - x^2)^2). - Ilya Gutkovskiy, Feb 24 2017

From Colin Barker, Feb 25 2017: (Start)

a(n) = 3 - 2^(-1-n)*((1-sqrt(5))^n*(-5+3*sqrt(5)) - (1+sqrt(5))^n*(5+3*sqrt(5)))/5*(1+n).

a(n) = 3*a(n-1) - a(n-2) - 3*a(n-3) + a(n-4) + a(n-5) for n>4.

(End)

Mathematica

Table[(n + 1) Fibonacci[n+2] + 3, {n, 0, 40}] (* Vincenzo Librandi, Feb 25 2017 *)
LinearRecurrence[{3, -1, -3, 1, 1}, {4, 7, 12, 23, 43}, 40] (* Harvey P. Dale, Jan 12 2018 *)

Prog

(PARI) a(n) = (n+1)*fibonacci(n+2) + 3; \\ Michel Marcus, Feb 25 2017
(PARI) Vec((4 - 5*x - 5*x^2 + 6*x^3 + 3*x^4)/((1 - x)*(1 - x - x^2)^2) + O(x^50)) \\ Colin Barker, Feb 25 2017
(Magma) [(n+1)*Fibonacci(n+2) + 3: n in [0..40]]; // Vincenzo Librandi, Feb 25 2017

Crossrefs

Cf. A023607.

Sequence in context: A010901 A187211 A023624 * A372449 A208668 A243860

Adjacent sequences: A123191 A123192 A123193 * A123195 A123196 A123197

Keyword

nonn

Author

N. J. A. Sloane, Oct 04 2006

Status

approved