cp's OEIS Frontend

A113252 Corresponds to m = 6 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 A113252 #10 Jun 10 2024 16:04:56
%S A113252 -1,4,92,784,-3856,33856,96704,73984,-418048,59474944,-101917696,
%T A113252 443355136,6249181184,37406654464,-217868812288,2345945595904,
%U A113252 4101714673664,699056521216,52661959000064,3420344569298944,-8264891921072128,41548867031793664
%N A113252 Corresponds to m = 6 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 A113252 Conjecture: a(m, 2*n+1) is a perfect square for all m,n (see A113249).
%H A113252 Colin Barker, <a href="/A113252/b113252.txt">Table of n, a(n) for n = 0..1000</a>
%H A113252 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (-4,0,144,1296).
%F A113252 G.f.: (-1+108*x^2+1296*x^3)/((6*x+1)*(1-6*x)*(36*x^2+4*x+1)).
%F A113252 a(n) = -4*a(n-1) + 144*a(n-3) + 1296*a(n-4) for n>3. - _Colin Barker_, May 20 2019
%t A113252 LinearRecurrence[{-4, 0, 144, 1296}, {-1, 4, 92, 784}, 25] (* _Paolo Xausa_, Jun 10 2024 *)
%o A113252 (PARI) Vec(-(1 - 108*x^2 - 1296*x^3) / ((1 - 6*x)*(1 + 6*x)*(1 + 4*x + 36*x^2)) + O(x^25)) \\ _Colin Barker_, May 20 2019
%Y A113252 Cf. A000302, A097948, A056450, A113249, A113250, A113251, A113253, A113254, A113255, A113256.
%K A113252 easy,sign
%O A113252 0,2
%A A113252 _Creighton Dement_, Nov 18 2005