|
%I #10 Mar 14 2023 03:39:24
%S 0,0,15,111,3936,28320,999855,7193295,253959360,1827068736,
%T 64504677711,464068265775,16383934179360,117871512438240,
%U 4161454776879855,29938900091047311,1056993129393303936,7604362751613578880
%N Define a(1)=0, a(2)=0, a(3)=15, a(4)=111 then a(n) = 254*a(n-2) - a(n-4) + 126 also sequence such that 7*a(n)*(a(n) + 1) + 1 = a square.
%C This sequence is such that 7*a(n)*(a(n) + 1) + 1 = j^2 = a square integer positive root of a(n)^2 + a(n) - (j^2-1)/7 = 0 is such that 2*a(n) + 1 = sqrt((4*j^2+3)/7), ((4*j^2+3)/7) need to be a square so j^2 = (7*k^2-3)/4, put k=2*a(n)+1 you found that (7*k^2-3)/4 = 7*a(n)^2 + 7*a(n) + 1
%H G. C. Greubel, <a href="/A105051/b105051.txt">Table of n, a(n) for n = 1..825</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,254,-254,-1,1).
%F From _R. J. Mathar_, Aug 28 2008: (Start)
%F a(n) = A105040(n-2).
%F G.f.: 3*x^2*(5+32*x+5*x^2)/((1-x)*(1+16*x+x^2)*(1-16*x+x^2)). (End)
%t LinearRecurrence[{1,254,-254,-1,1}, {0,0,15,111,3936}, 30] (* _G. C. Greubel_, Mar 13 2023 *)
%o (Magma) R<x>:=PowerSeriesRing(Integers(), 30); [0,0] cat Coefficients(R!( 3*x^2*(5+32*x+5*x^2)/((1-x)*(1-254*x^2+x^4)) )); // _G. C. Greubel_, Mar 13 2023
%o (SageMath)
%o @CachedFunction
%o def a(n): # a = A105051
%o if (n<6): return (0,0,0,15,111,3936)[n]
%o else: return a(n-1) +254*a(n-2) -254*a(n-3) -a(n-4) +a(n-5)
%o [a(n) for n in range(41)] # _G. C. Greubel_, Mar 13 2023
%Y Cf. A105040.
%K nonn
%O 1,3
%A _Pierre CAMI_, Apr 04 2005
%E Extended by _R. J. Mathar_, Aug 28 2008
|
