linear secod order homogeneous differential equation recursions
- To: mathgroup at smc.vnet.net
- Subject: [mg69852] linear secod order homogeneous differential equation recursions
- From: Roger Bagula <rlbagula at sbcglobal.net>
- Date: Mon, 25 Sep 2006 03:53:29 -0400 (EDT)
Again I have a difficult problem. I have this factorial based recursion: a[n] = (a0*n^2 + b0*n + c0)*a[n - 2]/(n*(n - 1)) a[n]*n!=Integer I want to get a form: b[n]=a[n]*n! I've tried your RSolve , but it doesn't give a form out I can use. M athematica: Clear[a, a0, b0, c0] 0 = 1; b0 = -2; c0 = -1; a[n_] := a[n] = (a0*n^2 + b0*n + c0)*a[n - 2]/(n*(n - 1)); a[0] = 1; a[1] = 1; Table[ExpandAll[a[n]*n!], {n, 0, 30}] {1, 1, -1, 2, -7, 28, -161, 952, -7567, 59024, -597793, 5784352, -71137367, 821377984, -11879940289, 159347328896, -2649226684447, 40474221539584, -760328058436289, 13032699335746048, -272957772978627751, 5187014335626927104, -119828462337617582689, 2500140909772178864128, -63149599651924466077103, 1435080882209230668009472, -39342200583148942366035169, 967244514609021470238384128, -28601779823949281100107567863, 756385210424254789726416388096, -23996893272293446842990249437057} Clear[a] a[n_] := a[n] = -(n^2 - n - 1)*a[n - 2]/(n*(n - 1)); a[0] = 1; a[1] = 1; Table[a[n]*n!, {n, 0, 30}] {1, 1, -1, -5, 11, 95, -319, -3895, 17545, 276545, -1561505, -30143405, 204557155, 4672227775, -37024845055, -976495604975, 8848937968145, 264630308948225, -2698926080284225, -90238935351344725, 1022892984427721275, 37810113912213439775, -471553665821179507775, -19094107525667787086375, 259826069867469908784025, 11437370407875004464738625, -168627119343987970800832225, -8017596655920378129781776125, 127313475104710917954628329875, 6502270887951426663253020437375, -110635409865993787702572018661375} Clear[a] a[n_] := a[n] = (n - 2)*a[n - 2]/(n*(n - 1)); a[0] = 1; a[1] = 1; Table[a[n]*n!, {n, 0, 30}] {1, 1, 0, 1, 0, 3, 0, 15, 0, 105, 0, 945, 0, 10395, 0, 135135, 0, 2027025, 0, 34459425, 0, 654729075, 0, 13749310575, 0, 316234143225, 0, 7905853580625, 0, 213458046676875, 0} I didn't find these on my own: they were recursion examples in an old differential equation text I had. So far none of my results as a[n]*n! haven't been in OEIS. These are solution to differtential equations of the type: y''+p(x)*y'+q(x)*y=0 f(x)=y=Sum[a[n]*(x-x0)^n,{n,0,Infinity}] So in Taylor series terms I'm getting ( I think): b[n]=D[f(x0),{x,n}] a[n]=b[n]/n! I need a gerenal form for the derivatives b[n].
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- Re: linear secod order homogeneous differential equation recursions
- From: Daniel Lichtblau <danl@wolfram.com>
- Re: linear secod order homogeneous differential equation recursions