A228873 a(n) = F(n) * F(n+1) * F(n+2) * F(n+3), the product of four consecutive Fibonacci numbers, A000045.

6, 30, 240, 1560, 10920, 74256, 510510, 3495030, 23965920, 164237040, 1125770256, 7715953440, 52886430870, 362487682830, 2484530961360, 17029219589256, 116720030923320, 800010932051760, 5483356663145790, 37583485265670630, 257601041359736256

Offset: 1

Author

T. D. Noe, Sep 24 2013

Keywords

• nonn

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).

Crossrefs

Cf. A000045 (Fibonacci numbers), A228874 (similar sequence for Lucas numbers).

Cf. A009112 (Pythagorean numbers), A024365, A073120.

Programs

Magma

[Fibonacci(n)*Fibonacci(n+1)*Fibonacci(n+2)*Fibonacci(n+3): n in [1..30]]; // Vincenzo Librandi, Oct 04 2013

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 *)

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

a(n) = 2 * A000217(A059840(n+2)). - Diego Rattaggi, Jan 27 2021

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