Recurrence relation for A043569?
Alonso Del Arte
alonso.delarte at gmail.com
Sun Apr 23 00:37:38 CEST 2006
The differences of the powers of two play an obvious role in this
sequence. What I haven't been able to figure out is a simple way to
exploit this obvious effect.
Thanks for the reference to A023578. From it I was able to put A043569
as a recurrence relation in Mathematica thus
a[0] := 0
a[1] := 1
a[n_] := 2^(Floor[(Sqrt[1 + 8(n - 2)] -
1)/2]) + a[n - Floor[(Sqrt[1 + 8(n - 2)] - 1)/2] - 1]
Table[a[n]*2, {n, 54}]
Is it at all possible to phrase a recurrence relation without the aid
of floors or ceilings?
Al
On 4/21/06, franktaw at netscape.net <franktaw at netscape.net> wrote:
>
>
> This sequence is twice A023758. Take a look at the formulas there.
>
> If you want to generate them in order, take differences of powers of 2, with
> the lower powers in reverse order:
>
> 2^2-2^1, 2^3-2^1, 2^3-2^2, ...
>
> Franklin T. Adams-Watters
>
>
> -----Original Message-----
> From: Alonso Del Arte alonso.delarte at gmail.com
>
> I've just sent through the form a Mathematica command to calculate the
terms
> of A043569. It involves multiplying Mersenne numbers by powers
of 2 and
> sorting. It works but it's not terribly elegant.
Is there a recurrence
> relation for this sequence? I've tried several
different things along the
> lines of
a(1) = 2, a(n) = a(n -1) + log_2 a(n - 1) + log_2 a(n - 2) etc.,
> etc.,
and they work except after a(n - 1) becomes a power of two, or
> before.
Any suggestions?
Al
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