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Corresponds to m = 5 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.
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%I #20 Aug 18 2023 23:51:27

%S -1,4,59,289,-1381,13924,10079,2209,520439,7628644,-23994301,

%T 149401729,490531859,406344964,-1681645081,149155846849,-249406479121,

%U 1083427010884,9530848465739,30158362505569,-168169798384501,2302905921914404,-239007146013841,2988025311585889

%N Corresponds to m = 5 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 Conjecture: a(m, 2*n+1) is a perfect square for all m,n (see A113249).

%H Colin Barker, <a href="/A113251/b113251.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (-4,0,100,625).

%F G.f.: (-1+75*x^2+625*x^3) / ((5*x+1)*(1-5*x)*(25*x^2+4*x+1)).

%F a(n) = -4*a(n-1) + 100*a(n-3) + 625*a(n-4) for n>3. - _Colin Barker_, May 20 2019

%F a(n) = 5^(n+1)*(1 - (-1)^n + 2*cos(arccos(-2/5)*(n+1)))/4. - _Eric Simon Jacob_, Jul 29 2023

%p with(gfun): seriestolist(series((-1+75*x^2+625*x^3)/((5*x+1)*(1-5*x)*(25*x^2+4*x+1)), x=0,25));

%t LinearRecurrence[{-4,0,100,625},{-1,4,59,289},40] (* _Harvey P. Dale_, Jul 05 2021 *)

%o (PARI) Vec(-(1 - 75*x^2 - 625*x^3) / ((1 - 5*x)*(1 + 5*x)*(1 + 4*x + 25*x^2)) + O(x^30)) \\ _Colin Barker_, May 20 2019

%Y Cf. A000302, A097948, A056450, A113249, A113250, A113252, A113253, A113254, A113255, A113256.

%K easy,sign

%O 0,2

%A _Creighton Dement_, Nov 18 2005