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A113256 Corresponds to m = 10 in a family of 4th-order linear recurrence sequences given by a(m,n) = m^4*a(n-4) + (2*m)^2*a(n-3) - 4*a(m-1), a(m,0) = -1, a(m,1) = 4, a(m,2) = -13 + 6*(m-1) + 3*(m-1)^2, a(m,3) = (-8+m^2)^2.

%I A113256 #13 Jun 10 2024 16:39:00
%S A113256 -1,4,284,8464,-42256,322624,4935104,47997184,-485499136,7142278144,
%T A113256 39980801024,125848981504,-2501476028416,97421005963264,
%U A113256 60463578988544,16045087719424,13889461750267904,942837644226985984,-3160296751934734336,18357422585040338944
%N A113256 Corresponds to m = 10 in a family of 4th-order linear recurrence sequences given by a(m,n) = m^4*a(n-4) + (2*m)^2*a(n-3) - 4*a(m-1), a(m,0) = -1, a(m,1) = 4, a(m,2) = -13 + 6*(m-1) + 3*(m-1)^2, a(m,3) = (-8+m^2)^2.
%C A113256 Conjecture: a(m, 2*n+1) is a perfect square for all m,n (see A113249).
%H A113256 Colin Barker, <a href="/A113256/b113256.txt">Table of n, a(n) for n = 0..900</a>
%H A113256 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (-4,0,400,10000).
%F A113256 G.f.: (-1+300*x^2+10000*x^3) / ((10*x+1)*(1-10*x)*(100*x^2+4*x+1)).
%F A113256 a(n) = -4*a(n-1) + 400*a(n-3) + 10000*a(n-4) for n > 3. - _Colin Barker_, May 20 2019
%t A113256 LinearRecurrence[{-4, 0, 400, 10000}, {-1, 4, 284, 8464}, 25] (* _Paolo Xausa_, Jun 10 2024 *)
%o A113256 (PARI) Vec(-(1 - 300*x^2 - 10000*x^3) / ((1 - 10*x)*(1 + 10*x)*(1 + 4*x + 100*x^2)) + O(x^20)) \\ _Colin Barker_, May 20 2019
%Y A113256 Cf. A000302, A097948, A056450, A113249, A113250, A113251, A113252, A113253, A113254, A113255.
%K A113256 easy,sign
%O A113256 0,2
%A A113256 _Creighton Dement_, Nov 18 2005