A229912 a(n) = Fibonacci(n) * (2*Fibonacci(n) + 1).

0, 3, 3, 10, 21, 55, 136, 351, 903, 2346, 6105, 15931, 41616, 108811, 284635, 744810, 1949325, 5102415, 13356696, 34965703, 91537215, 239640778, 627376753, 1642475955, 4300029216, 11257576275, 29472642291, 77160257866, 202007981253, 528863443111, 1384581955240

Offset

0,2

Comments

a(n) = A014105(A000045(n)), so as in A014105, the sum of squares of n+1 consecutive integers equals the sum of squares of consecutive n integers. [In this present sequence, the sum of the Fibonacci(n)+1 consecutive squares starting with a(n)^2 equals the sum of the next Fibonacci(n) consecutive squares. - Jon E. Schoenfield, Feb 08 2015]

Links

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

Index entries for linear recurrences with constant coefficients, signature (3,1,-5,-1,1).

Formula

G.f.: -x*(3 - 6*x - 2*x^2 + 3*x^3) / ( (1+x)*(x^2 - 3*x + 1)*(x^2 + x - 1) ). - R. J. Mathar, Oct 06 2013

a(n) = A000045(n) + 2 * A007598(n). - R. J. Mathar, Oct 06 2013

Example

a(5) = 55 since 55^2 + 56^2 + 57^2 + 58^2 + 59^2 + 60^2 = 61^2 + 62^2 + 63^2 + 64^2 + 65^2.

Mathematica

Table[Fibonacci[n]*(2*Fibonacci[n] + 1), {n, 0, 25}] (* T. D. Noe, Oct 06 2013 *)
CoefficientList[Series[-x (3 - 6 x - 2 x^2 + 3 x^3)/((1 + x) (x^2 - 3 x + 1) (x^2 + x - 1)), {x, 0, 50}], x] (* Vincenzo Librandi , Oct 07 2013 *)
#(2#+1)&/@Fibonacci[Range[0, 30]] (* or *) LinearRecurrence[{3, 1, -5, -1, 1}, {0, 3, 3, 10, 21}, 40] (* Harvey P. Dale, Dec 18 2022 *)

Prog

(Magma) [Fibonacci(n)*(2*Fibonacci(n)+1): n in [0..30]]; // Vincenzo Librandi, Oct 07 2013
(PARI) vector(50, n, fibonacci(n-1)*(2*fibonacci(n-1)+1)) \\ Derek Orr, Feb 07 2015

Crossrefs

Cf. A000045, A014105.

Sequence in context: A025519 A360651 A076987 * A321354 A019153 A049975

Adjacent sequences: A229909 A229910 A229911 * A229913 A229914 A229915

Keyword

nonn,easy

Author

Carmine Suriano, Oct 03 2013

Status

approved