Very interesting mathematical solution Mike.
My initial idea to solve this type of sequence formula was using folds.
The idea is to define a fold function g and fold it on a list of _1 and 1,
considering an initial value.
The result was:
g =. ]+2*[
4 ((g~/\.&.|.)@, ]$(_1 1)"_) 5
4 7 15 29 59 117
I found a bit difficult to work this left foldlist in J. Folds from right seem
more natural.
This same idea can be implemented in Mathematica as:
In[1]:= FoldList[2 #1 + #2 &, 4, {-1 , 1 , -1 , 1, -1}]
Out[1]= {4, 7, 15, 29, 59, 117}
________________________________
From: Programming <[email protected]> on behalf of 'Mike
Day' via Programming <[email protected]>
Sent: Wednesday, November 28, 2018 2:08 PM
To: [email protected]
Subject: Re: [Jprogramming] Recursive verbs
Yes, this is one way to get high terms without having to evaluate all
their predecessors.
Bear in mind that it's calculating 2^n twice, and the zero-th term
should surely be the
starting value. Also, it assumes that starting value is 4; just change
"4" to "x" in the
13 : ... expression to make it apply for any starting value.
So here's a lovely compact tacit version which addresses those issues,
at the expense
of returning the wrong sign for a negative starting value:
fa=: (* ([: | -) 3 %~ [: <: ]) _2&^ NB. with a [: to keep Linda
happy
So,
4 fa 200x
5892106162282964343653861005250929542581410977203573729438379
cf
4 (13 :'x p.~(([:(2&^)>:),.~ _1r3 *([:(2&^)>:) + _1&^)y' ) 199x
5892106162282964343653861005250929542581410977203573729438379
If you want a table, the rank needs to be imposed:
4 5 fl"0/ 3 4 5
29 59 117
37 75 149
A possibly easier way to understand why we need ((_2^n) - 1) % 3 is perhaps
to inspect the general term.
We have
a1 = 2a0 - 1 NB. using "Mathematical" notationrather than APL/J
a2 = 2a1 + 1 = 4a0 - 2 + 1
a3 = 2a2 - 1 = 8a0 - 4 + 2 - 1
...
an = (2^n).a0 - (2^[n-1]) + (2^[n-2]) ... + (-1)^n
The absolute sum of all terms except the first is | sum {i=0,n-1} _2^i | ,
which we know from High School maths is | [(_2^n) - 1]%[_2 -1] | ,
ie | [(_2^n) - 1]% 3 |
If you want the correct sign for negative starting values, use
(fa * * [) NB. ... which isn't quite so neat.
Cheers,
Mike
NB - previous correspondence not trimmed
On 28/11/2018 09:49, Linda Alvord wrote:
> Hi Cliff,
>
> It took a while but I got the explicit verb to work.
>
> f=: 13 :'4 p.~(([:(2&^)>:),.~ _1r3 *([:(2&^)>:) + _1&^)y'
>
> f i.10
> 7 15 29 59 117 235 469 939 1877 3755
>
> f
> 4 p.~ ([: 2&^ >:) ,.~ _1r3 * ([: 2&^ >:) + _1&^
>
> Linda
>
> -----Original Message-----
> From: Programming<[email protected]> On Behalf Of
> Cliff Reiter
> Sent: Tuesday, November 27, 2018 12:38 PM
> To:[email protected]
> Subject: Re: [Jprogramming] Recursive verbs
>
> I like that solution. Here is another approach
>
> 4 p.~(2&^@>: ,.~ _1r3 * 2&^@>: + _1&^)i. 10
> 7 15 29 59 117 235 469 939 1877 3755
>
> I found this by composing the linear poly's and noting the constant terms
> satisfy a two step recursion with eigenvalues 2 and _1.
>
> lc=.({.@[+{:@[*{.@]),*&{:
> lc/\.(7$_1 2,:1 2)
> _43 128
> _21 64
> _11 32
> _5 16
> _3 8
> _1 4
> _1 2
>
>
> On 11/27/2018 11:06 AM, Raul Miller wrote:
>> Sure, ... that's a bit bulkier than I prefer, but it works.
>>
>> Here's another alternative:
>>
>> evenodd=: , (_1 2,:_1 4)&p.@{:
>> evenodd evenodd 4
>> 4 7 15 29 59
>>
>> Thanks,
>>
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