A132920 a(n) = n*Fibonacci(n) + binomial(n, 2).

1, 3, 9, 18, 35, 63, 112, 196, 342, 595, 1034, 1794, 3107, 5369, 9255, 15912, 27285, 46665, 79610, 135490, 230076, 389873, 659364, 1113108, 1875925, 3156543, 5303637, 8899086, 14913047, 24961635, 41734804, 69706384, 116311602, 193898719, 322961870, 537493302

Offset

1,2

Links

Andrew Howroyd, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (5,-8,2,6,-4,-1,1).

Formula

From Andrew Howroyd, Aug 10 2018: (Start)

a(n) = 5*a(n-1) - 8*a(n-2) + 2*a(n-3) + 6*a(n-4) - 4*a(n-5) - a(n-6) + a(n-7).

G.f.: x*(1 - 2*x + 2*x^2 - 5*x^3 + 5*x^4)/((1 - x)^3*(1 - x - x^2)^2). (End)

Example

a(4) = 18 = 3 + 4 + 5 + 6.

Mathematica

Table[n*Fibonacci[n]+Binomial[n, 2], {n, 36}] (* James C. McMahon, Mar 09 2025 *)
LinearRecurrence[{5, -8, 2, 6, -4, -1, 1}, {1, 3, 9, 18, 35, 63, 112}, 40] (* Harvey P. Dale, Jan 24 2026 *)

Prog

(PARI) a(n) = n*fibonacci(n) + binomial(n, 2); \\ Andrew Howroyd, Aug 10 2018
(PARI) Vec((1 - 2*x + 2*x^2 - 5*x^3 + 5*x^4)/((1 - x)^3*(1 - x - x^2)^2) + O(x^40)) \\ Andrew Howroyd, Aug 10 2018

Crossrefs

Row sums of A132919.

Sequence in context: A293406 A295862 A246695 * A127645 A000241 A028723

Adjacent sequences: A132917 A132918 A132919 * A132921 A132922 A132923

Keyword

nonn,easy

Author

Gary W. Adamson, Sep 05 2007

Extensions

Name changed and terms a(11) and beyond from Andrew Howroyd, Aug 10 2018

Status

approved