OFFSET
1,1
COMMENTS
Mohanty and Mohanty prove in Corollary 2.5 that these numbers are Pythagorean. The number a(n) is primitive Pythagorean if F(n) and F(n+1) have opposite parity. Every third number, starting at a(1) = 6, is not primitive Pythagorean.
Since a(n) = F(n+1)*F(n+2)*(F(n+2)^2 - F(n+1)^2), a(n) is in A073120. - Robert Israel, Apr 06 2015
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Supriya Mohanty and S. P. Mohanty, Pythagorean Numbers, The Fibonacci Quarterly, Vol. 28, No. 1 (1990), pp. 31-42.
H.-J. Seiffert, Problem B-1020, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 44, No. 4 (2006), p. 278; Two Fibonacci Identities, Solution to Problem B-1020 by Harris Kwong, ibid., Vol. 45, No. 2 (2007), pp. 182-184.
Index entries for linear recurrences with constant coefficients, signature (5,15,-15,-5,1).
FORMULA
G.f.: -6*x/((x-1)*(x^2+3*x+1)*(x^2-7*x+1)). - Alois P. Heinz, Oct 02 2013
a(n+5) = 5*a(n+4)+15*a(n+3)-15*a(n+2)-5*a(n+1)+a(n). - Robert Israel, Apr 06 2015
Sum_{n>=1} 1/a(n) = (12-5*sqrt(5))/4. - Diego Rattaggi, Aug 16 2021
a(n) = 3 * Sum_{k=1..n} 2^(n-k)*F(k)^2*F(k+1)*F(k+2) (Seiffert, 2006). - Amiram Eldar, Jan 11 2022
MAPLE
seq(mul(combinat:-fibonacci(i), i=n..n+3), n=1..30); # Robert Israel, Apr 06 2015
MATHEMATICA
Table[Fibonacci[n] Fibonacci[n+1] Fibonacci[n+2] Fibonacci[n+3], {n, 25}]
CoefficientList[Series[-6/((x - 1) (x^2 + 3 x + 1) (x^2 - 7 x + 1)), {x, 0, 30}], x] (* Vincenzo Librandi, Oct 04 2013 *)
Times@@@Partition[Fibonacci[Range[30]], 4, 1] (* Harvey P. Dale, Dec 23 2013 *)
LinearRecurrence[{5, 15, -15, -5, 1}, {6, 30, 240, 1560, 10920}, 30] (* Harvey P. Dale, Jul 24 2021 *)
PROG
(Magma) [Fibonacci(n)*Fibonacci(n+1)*Fibonacci(n+2)*Fibonacci(n+3): n in [1..30]]; // Vincenzo Librandi, Oct 04 2013
CROSSREFS
KEYWORD
nonn
AUTHOR
T. D. Noe, Sep 24 2013
STATUS
approved