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A113253 Corresponds to m = 7 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 A113253 #23 Jun 10 2024 15:33:09
%S A113253 -1,4,131,1681,-8341,68644,369431,923521,-10266601,278289124,
%T A113253 -45142549,385690321,28351798019,545917055044,-2216460177409,
%U A113253 15348835582081,113677067503919,421612384372804,-3999798649362349,75132454060794001
%N A113253 Corresponds to m = 7 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 A113253 Conjecture: a(m, 2*n+1) is a perfect square for all m,n (see A113249).
%H A113253 Colin Barker, <a href="/A113253/b113253.txt">Table of n, a(n) for n = 0..1000</a>
%H A113253 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (-4,0,196,2401).
%F A113253 G.f.: (-1+147*x^2+2401*x^3) / ((7*x+1)*(1-7*x)*(49*x^2+4*x+1)).
%F A113253 a(n) = -4*a(n-1) + 196*a(n-3) + 2401*a(n-4) for n > 3. - _Colin Barker_, May 20 2019
%F A113253 a(n) = 7^(n+1)*(1 - (-1)^n + 2*cos(arccos(-2/7)*(n+1)))/4. - _Eric Simon Jacob_, Jul 30 2023
%t A113253 LinearRecurrence[{-4, 0, 196, 2401}, {-1, 4, 131, 1681}, 25] (* _Paolo Xausa_, Jun 10 2024 *)
%o A113253 (PARI) Vec(-(1 - 147*x^2 - 2401*x^3) / ((1 - 7*x)*(1 + 7*x)*(1 + 4*x + 49*x^2)) + O(x^25)) \\ _Colin Barker_, May 20 2019
%Y A113253 Cf. A000302, A097948, A056450, A113249, A113250, A113251, A113252, A113254, A113255, A113256.
%K A113253 easy,sign
%O A113253 0,2
%A A113253 _Creighton Dement_, Nov 18 2005