%I #29 Sep 24 2021 19:27:49
%S 24,84,924,5544,40194,269874,1864584,12741324,87431844,599001144,
%T 4106310474,28143249834,192901471224,1322153872644,9062210132844,
%U 62113226746824,425730613530834,2918000448971874,20000274149827944,137083914357154044,939587137457703924
%N a(n) = L(n) * L(n+1) * L(n+2) * L(n+3), the product of four consecutive Lucas numbers, A000032.
%C 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.
%C 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
%H Robert Israel, <a href="/A228874/b228874.txt">Table of n, a(n) for n = 0..1193</a>
%H Supriya Mohanty and S. P. Mohanty, <a href="http://www.fq.math.ca/Scanned/28-1/mohanty.pdf">Pythagorean Numbers</a>, Fibonacci Quarterly 28 (1990), 31-42.
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,15,-15,-5,1).
%F 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
%F From _Robert Israel_, Apr 06 2015: (Start)
%F a(n+5) = 5*a(n+4) + 15*a(n+3) - 15*a(n+2) - 5*a(n+1) + a(n).
%F a(n) = -A228873(n+3) + 4*A228873(n+2) + 24*A228873(n+1) - 6*A228873(n) + 4*A228873(n-1) for n >= 2. (End)
%F Sum_{n>=0} 1/a(n) = (10 - 3*sqrt(5))/60. - _Diego Rattaggi_, Aug 16 2021
%p L:= n -> 2*combinat:-fibonacci(n+1)-combinat:-fibonacci(n):
%p seq(mul(L(n+i),i=0..3),n=0..30); # _Robert Israel_, Apr 06 2015
%t Table[LucasL[n] LucasL[n+1] LucasL[n+2] LucasL[n+3], {n, 0, 25}]
%t Times@@@Partition[LucasL[Range[0,30]],4,1] (* _Harvey P. Dale_, Jul 11 2017 *)
%o (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
%Y Cf. A000032 (Lucas numbers), A228873 (similar sequence for Fibonacci numbers).
%Y Cf. A009112 (Pythagorean numbers), A024365, A073120.
%K nonn,easy
%O 0,1
%A _T. D. Noe_, Sep 24 2013