[seqfan] Re: question on a recursively defined sequence
Jim Nastos
nastos at gmail.com
Mon Feb 23 01:04:32 CET 2009
Telescoping once to remove the linear term gives a nicer recurrence
(only makes one recursive call
instead of two.) I'm surprised it's not written in the OEIS page.
t(1) = 1
t(2) = 3
t(n) = 3 + 2^t(n-1)
Can someone get a closed form from this?
JN
On Sun, Feb 22, 2009 at 3:51 PM, T. D. Noe <noe at sspectra.com> wrote:
> At 3:44 PM -0800 2/22/09, Jonathan Post wrote:
>>It's on that page thus:
>>
>>Closed form of a recursive series
>>
>>Is there a closed form with a fixed number of terms for the sequence
>>defined by t0 = 1 and tn = tn - 1 + 2tn - 1? In case anyone's
>>wondering, it originates here. NeonMerlin 08:11, 22 February 2009
>>(UTC)
>>
>> I doubt there's anything useful. There's nothing in the Sloane's
>>entry. Algebraist 09:40, 22 February 2009 (UTC)
>>
>>On Sun, Feb 22, 2009 at 3:34 PM, Brendan McKay <bdm at cs.anu.edu.au> wrote:
>>> At the Mathematics Reference Desk of Wikipedia
>>> http://en.wikipedia.org/wiki/Wikipedia:Reference_desk/Mathematics
>>> someone is asking for a closed form for t[n] where
>>> t[0] = 1, t[n] = t[n-1] + 2^(t[n-1]).
>>> It looks unlikely to me. Does anyone disagree?
>
>
> Sequence A034797: http://www.research.att.com/~njas/sequences/A034797
>
> Tony
>
>
>
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