Raul Miller schreef:
> Here's a corrected version of allbut2
>
> allbut2f=:3 :0
> n=.1+i.y
> q=._ q: n
> s=.(*."1 1:=+/)(="1 >./)q
> (*./x:n)%>./(_ q:inv s)*adja n*(1=+/|:q)*.adja+./"1 s
> )
>
> adja=:[: +./ 1 _1 |.!.0"0 _ ]
> last=:</\&.|.
>
Raul I'm in doubt if your verb always produces a correct answer.
So please follow me on this:
Your verb gives
allbut2f 61
9534410355667921512188400
allbut2f 62
9534410355667921512188400
Testing for all divisor pairs consisting of successive
values <= 61
(,>:)2+I. 0<*/"1(]|[:*./(1x+i.61)-."1])"1(,>:)"0[2x+i.59
31 32
-> for 31 & 32 there exists a X for n=61
But for n=62 there are no k,k+1 <= 62 for which
'k and k+1 are not divisors of X' is true.
2+I. 0<*/"1(]|[:*./(1x+i.62)-."1])"1(,<:)"0[2x+i.61
empty
where do I go wrong???
FWIW here's my allbut2 verb with your 'enhanced' lcm
and based on my conjecture.
ab217=: 3 : 0
j=: x:<.&.(2&^.)y
r=. 1x+i.y
s=. 0$0
p=: j-1x
q=: j+1x
v1=: _ q:inv x:>./ _ q: r-.j,p
v2=: _ q:inv x:>./ _ q: r-.j,q
if. (0<j|v1)*.0<p|v1 do. s =. s, v1 end.
if. (0<j|v2)*.0<q|v2 do. s =. s, v2 end.
<./s
)
ab217 61
9534410355667921512188400
ab217 62
_
_10{.":ab217 12197
1212800000
Conjecture:
For a certain n it is possible that there exists at least one
X = lcm {2,3..n} - {p,q} <--- set difference
not divisible by
or 2^m and 2^m + 1 \
> = {p,q}
or 2^m - 1 and 2^m /
in which:
m is <.&.(2&^.)n
and 2^m +/- 1 is a prime or a prime^2 (case: 8,9)
So enough guess work I think and:
assumption is the mother of all disasters, so ...shoot
=@@i
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