cp's OEIS Frontend

A113251 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.

%I A113251 #20 Aug 18 2023 23:51:27
%S A113251 -1,4,59,289,-1381,13924,10079,2209,520439,7628644,-23994301,
%T A113251 149401729,490531859,406344964,-1681645081,149155846849,-249406479121,
%U A113251 1083427010884,9530848465739,30158362505569,-168169798384501,2302905921914404,-239007146013841,2988025311585889
%N A113251 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 A113251 Conjecture: a(m, 2*n+1) is a perfect square for all m,n (see A113249).
%H A113251 Colin Barker, <a href="/A113251/b113251.txt">Table of n, a(n) for n = 0..1000</a>
%H A113251 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (-4,0,100,625).
%F A113251 G.f.: (-1+75*x^2+625*x^3) / ((5*x+1)*(1-5*x)*(25*x^2+4*x+1)).
%F A113251 a(n) = -4*a(n-1) + 100*a(n-3) + 625*a(n-4) for n>3. - _Colin Barker_, May 20 2019
%F A113251 a(n) = 5^(n+1)*(1 - (-1)^n + 2*cos(arccos(-2/5)*(n+1)))/4. - _Eric Simon Jacob_, Jul 29 2023
%p A113251 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 A113251 LinearRecurrence[{-4,0,100,625},{-1,4,59,289},40] (* _Harvey P. Dale_, Jul 05 2021 *)
%o A113251 (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 A113251 Cf. A000302, A097948, A056450, A113249, A113250, A113252, A113253, A113254, A113255, A113256.
%K A113251 easy,sign
%O A113251 0,2
%A A113251 _Creighton Dement_, Nov 18 2005