A244855 a(n) = Fibonacci(n)^4-1.

0, 0, 15, 80, 624, 4095, 28560, 194480, 1336335, 9150624, 62742240, 429981695, 2947295520, 20200652640, 138458409999, 949005240560, 6504586067280, 44583076827135, 305577005139120, 2094455819300624, 14355614096087055, 98394841894789440, 674408281676875200

Offset: 1

Author

Michel Lagneau, Jul 25 2014

Keywords

• nonn
• easy

Comments

For n > 1, a(n) = F(n-2)*F(n-1)*F(n+1)*F(n+2) with the property F(n-2)+F(n-1)+ F(n+1)+ F(n+2) = F(n) + F(n+3) = 2*F(n+2).

F(n)^2 - 1 = F(n-1)*F(n+1) if n odd, and F(n)^2 - 1 = F(n-2)*F(n+2)if n even ;

F(n)^2 + 1 = F(n-2)*F(n+2) if n odd, and F(n)^2 + 1 = F(n-1)*F(n+1) if n even, hence the product (F(n)^2 - 1)*(F(n)^2 + 1)= F(n-2)*F(n-1)*F(n+1)*F(n+2).

Examples

			a(5) = Fibonacci(5)^4-1 = 624 = 3*8*2*13 because Fibonacci(5)^2-1=3*8 and Fibonacci(5)^2+1 = 2*13.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..200
• Hideyuki Ohtsuka, Problem B-1174, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 53, No. 3 (2015), p. 273; A Summation of Recipricals, Solution to Problem B-1174 by Steve Edwards, ibid., Vol. 54, No. 3 (2016), pp. 277-278.
• Index entries for linear recurrences with constant coefficients, signature (5,15,-15,-5,1).

Crossrefs

Cf. A000045, A001622, A056571, .

Programs

Magma

[Fibonacci(n)^4-1: n in [1..30]]; // Vincenzo Librandi, Jul 26 2014

Maple

with(numtheory):with(combinat, fibonacci):nn:=30:for i from 1 to nn do:x:=fibonacci(i)^4-1: printf(`%d, `, x):od:

Mathematica

Table[(Fibonacci[n]^4 - 1), {n, 40}] (* Vincenzo Librandi, Jul 26 2014 *)
LinearRecurrence[{5,15,-15,-5,1},{0,0,15,80,624},30] (* Harvey P. Dale, Dec 01 2019 *)

PARI

a(n) = fibonacci(n)^4-1; \\ Michel Marcus, Oct 20 2020

Formula

From R. J. Mathar, Nov 02 2014: (Start)

G.f.: x^3*(-15-5*x+x^2) / ( (x-1)*(x^2-7*x+1)*(x^2+3*x+1) ).

a(n) = A056571(n)-1. (End)

Sum_{n>=3} 1/a(n) = 35/18 - 5*sqrt(5)/6 = 25/9 - 5*phi/3, where phi is the golden ratio (A001622). - Amiram Eldar, Oct 20 2020

Sum_{n>=3} (-1)^(n+1)/a(n) = 1/18 (Ohtsuka, 2015). - Amiram Eldar, Dec 09 2021