A228874 a(n) = L(n) * L(n+1) * L(n+2) * L(n+3), the product of four consecutive Lucas numbers, A000032.

24, 84, 924, 5544, 40194, 269874, 1864584, 12741324, 87431844, 599001144, 4106310474, 28143249834, 192901471224, 1322153872644, 9062210132844, 62113226746824, 425730613530834, 2918000448971874, 20000274149827944, 137083914357154044, 939587137457703924

Offset: 0

Author

T. D. Noe, Sep 24 2013

Keywords

• nonn
• easy

Comments

Mohanty and Mohanty prove in Corollary 2.6 that these numbers are Pythagorean. The number a(n) is primitive Pythagorean if Lucas(n) and Lucas(n+1) have opposite parity. Every third number, starting at a(1) = 84, is not primitive Pythagorean.

Since a(n) = L(n+1)*L(n+2)*(L(n+2)^2-L(n+1)^2), these numbers are in A073120, - Robert Israel, Apr 06 2015

Links

• Robert Israel, Table of n, a(n) for n = 0..1193
• Supriya Mohanty and S. P. Mohanty, Pythagorean Numbers, Fibonacci Quarterly 28 (1990), 31-42.
• Index entries for linear recurrences with constant coefficients, signature (5,15,-15,-5,1).

Crossrefs

Cf. A000032 (Lucas numbers), A228873 (similar sequence for Fibonacci numbers).

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

Programs

Maple

L:= n -> 2*combinat:-fibonacci(n+1)-combinat:-fibonacci(n):
seq(mul(L(n+i),i=0..3),n=0..30); # Robert Israel, Apr 06 2015

Mathematica

Table[LucasL[n] LucasL[n+1] LucasL[n+2] LucasL[n+3], {n, 0, 25}]
Times@@@Partition[LucasL[Range[0,30]],4,1] (* Harvey P. Dale, Jul 11 2017 *)

PARI

Vec(6*(x^4-4*x^3-24*x^2+6*x-4)/((x-1)*(x^2-7*x+1)*(x^2+3*x+1)) + O(x^100)) \\ Colin Barker, Oct 29 2013

Formula

G.f.: 6*(x^4-4*x^3-24*x^2+6*x-4) / ((x-1)*(x^2-7*x+1)*(x^2+3*x+1)). - Colin Barker, Oct 29 2013

From Robert Israel, Apr 06 2015: (Start)

a(n+5) = 5*a(n+4) + 15*a(n+3) - 15*a(n+2) - 5*a(n+1) + a(n).

a(n) = -A228873(n+3) + 4*A228873(n+2) + 24*A228873(n+1) - 6*A228873(n) + 4*A228873(n-1) for n >= 2. (End)

Sum_{n>=0} 1/a(n) = (10 - 3*sqrt(5))/60. - Diego Rattaggi, Aug 16 2021