duplicate hunting, pt. 21

[Peter Pein]

Joshua Zucker schrieb:
> On 5/30/07, Andrew Plewe <aplewe at sbcglobal.net> wrote:
>> A099094 and A099465 (sequences are the same as listed, are the terms
>> correct?)
>> http://www.research.att.com/~njas/sequences/?q=id:A099094|id:A099465
> 
> If the first few terms are as given for A099465 (and I see no
> particular reason to start that recursion with 1 0 3 3 ...), then the
> subsequent terms should be
> 1 0 3 3 12 9 45 36 171 135 648 513 2457 1944 9315 7371 35316 27945
> 133893 105948 507627 401679 1924560 1522881 7296561 5773680 27663363
> 21889683 104879772 82990089 397629405 314639316 1507527531 1192888215
> 5715470808 4522582593 21668995017 17146412424 82153397475 65006985051
> 311467177476 246460192425 1180861724853 934401532428 4476986706987
> 3542585174559 16973545295520 13430960120961 64351596007521
> 50920635886560
> 
...
> --Joshua Zucker
> 
> 

Well, if one wants to save A099465, the smallest possible change seems to be:
a(n) = 3*a(n-2) + 3*a(n-3) instead of a(n) = 3*a(n-2) + 3*a(n-4):

I verified this with Mathematica:
an = a[n] /. First[RSolve[{a[n] == 3*a[n - 2] + 3*a[n - 3],
      a[0] == 1, a[1] == 0, a[2] == 3}, a[n], n]];
RootReduce[Table[an, {n, 0, 20}]]
--> {1, 0, 3, 3, 9, 18, 36, 81, 162, 351, 729, 1539, 3240, 6804, 14337,
     30132, 63423, 133407, 280665, 590490, 1242216}

FullSimplify[PowerSum[an, {x, n}]]
--> 1/(1 - 3*x2*(1 + x))

Peter