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A105051 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.

%I A105051 #10 Mar 14 2023 03:39:24
%S A105051 0,0,15,111,3936,28320,999855,7193295,253959360,1827068736,
%T A105051 64504677711,464068265775,16383934179360,117871512438240,
%U A105051 4161454776879855,29938900091047311,1056993129393303936,7604362751613578880
%N A105051 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 A105051 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 A105051 G. C. Greubel, <a href="/A105051/b105051.txt">Table of n, a(n) for n = 1..825</a>
%H A105051 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,254,-254,-1,1).
%F A105051 From _R. J. Mathar_, Aug 28 2008: (Start)
%F A105051 a(n) = A105040(n-2).
%F A105051 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 A105051 LinearRecurrence[{1,254,-254,-1,1}, {0,0,15,111,3936}, 30] (* _G. C. Greubel_, Mar 13 2023 *)
%o A105051 (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 A105051 (SageMath)
%o A105051 @CachedFunction
%o A105051 def a(n): # a = A105051
%o A105051     if (n<6): return (0,0,0,15,111,3936)[n]
%o A105051     else: return a(n-1) +254*a(n-2) -254*a(n-3) -a(n-4) +a(n-5)
%o A105051 [a(n) for n in range(41)] # _G. C. Greubel_, Mar 13 2023
%Y A105051 Cf. A105040.
%K A105051 nonn
%O A105051 1,3
%A A105051 _Pierre CAMI_, Apr 04 2005
%E A105051 Extended by _R. J. Mathar_, Aug 28 2008