Here's a little variation. Fibonacci series begins 1 1 ...
Y=: ($:`)(`:6)
N=: 1 :'(u@:<:@:<: + [:u<:)^:(1 < ])'
N=: (5!:1)<'N'
}.N Y("0) i.11
1 1 2 3 5 8 13 21 34 55
Maybe there is a better way to drop the 0.
Linda
-----Original Message-----
From: Programming <[email protected]> On Behalf Of Linda
Alvord
Sent: Wednesday, November 28, 2018 4:49 AM
To: [email protected]
Subject: Re: [Jprogramming] Recursive verbs
Hi Cliff,
It took a while but I got the explicit verb to work.
f=: 13 :'4 p.~(([:(2&^)>:),.~ _1r3 *([:(2&^)>:) + _1&^)y'
f i.10
7 15 29 59 117 235 469 939 1877 3755
f
4 p.~ ([: 2&^ >:) ,.~ _1r3 * ([: 2&^ >:) + _1&^
Linda
-----Original Message-----
From: Programming <[email protected]> On Behalf Of Cliff
Reiter
Sent: Tuesday, November 27, 2018 12:38 PM
To: [email protected]
Subject: Re: [Jprogramming] Recursive verbs
I like that solution. Here is another approach
4 p.~(2&^@>: ,.~ _1r3 * 2&^@>: + _1&^)i. 10
7 15 29 59 117 235 469 939 1877 3755
I found this by composing the linear poly's and noting the constant terms
satisfy a two step recursion with eigenvalues 2 and _1.
lc=.({.@[+{:@[*{.@]),*&{:
lc/\.(7$_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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