[seqfan] Re: A066411

[Graeme McRae]

A066411(17)=1471123

The way to speed things up is to do some pruning of the recursion tree. If I'm on a branch that can't reach any new leaves, I cut it off. (Can't reach new leaves means of the branch I'm sitting on, all the leaves from the smallest to the largest permutation  have already been seen). 

There would be no additional advantage by taking advantage of the powers of two that arise from the repeated combinations because the second visit to an identical branch would be quickly pruned away.  

Regarding the fact that 2^n is a peak, this is a consequence of the fact that the combinations of 2^n-1 are odd, so only even-numbered sums can be reached. 

Oh, and by the way, I hereby verify the elements 0 through 16 of A066411 that have been reported on the database and in this thread.  

--Graeme McRae,
Palmdale, CA

On Jan 28, 2012, at 8:44 AM, "Heinz, Alois" <alois.heinz at hs-heilbronn.de> wrote:

> 
> a(16) = 683311.
> 
> a(16)/a(15) = 4.32.
> 
> This is compatible with the pattern so far (peaks at n=2^k):
> 
> . . . 
> 
> Am 27.01.2012 19:13, schrieb Ron Hardin:
>> I get 1 1 3 5 23 61 143 215 995 2481 5785 12907 29279 64963 144289 158049 which
>> agrees with the series.
> 
> 
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