[seqfan] Re: A003436

[Paul D Hanna]

Yes, the formula in the current version is incorrect: 
    a(n) = 2*n*a(n-1) - 2*(n-1)*a(n-2) - a(n-3) for n>4.  
It was modified from the original in version 13: http://oeis.org/history/view?seq=A003436&v=13 which gives the correct formula:    a(n) = (-2n+4)a(n-2) - a(n-3) + (2n+2)a(n-1). 
This original formula should be qualified by "for n>3" as in: 
   a(n) = (-2*n+4)*a(n-2) - a(n-3) + (2*n+2)*a(n-1) for n>3, starting with [0, 1, 4, 31]. Here is my PARI code that produces all the visible terms in the sequence: 
 
(PARI) {a(n) = if(n<4, [0, 1, 4, 31][n+1], (-2*n+4)*a(n-2) - a(n-3) + (2*n+2)*a(n-1))} 
for(n=1,20,print1(a(n),", "))  
 
---------- Original Message ----------
From: "Harvey P. Dale" <hpd at hpdale.org>
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
Subject: [seqfan] A003436
Date: Sun, 29 Sep 2013 22:07:12 +0000

           I believe the first formula given for the above sequence may be in error.
           Best,
           Harvey

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