This might help see where it goes astray:
(~.,:#/.~)/:~ 10 >./\ dsieve 1000
4 6 8 9 10 12 14 15 16 18 20 21 24 27 28 30 32
2 12 24 12 16 174 16 20 245 158 102 10 160 10 10 10 10
+/ 10 >./\ dsieve 1000
17176
(~.,:#/.~)/:~991{.{. >./ ,:&> 0 -.~ [...] each boxes
4 6 8 9 10 12 14 15 16 18 20 21 24 27 28 30 32
2 12 22 12 18 174 14 20 247 158 102 10 160 10 10 10 10
NB. despite using fixed width font I've had to hand-align these!
I had thought you should be using
<.1000%21 14 10 15 125 27 8 6
for the length of each a.p., ie to get >:i.47, >:i.71 etc, but that
doesn't work either, providing only 990 elements, and sum 17168 .
I can't quite get what you're doing here. (I can talk!) I see that
21* will hoover up all numbers with 3 & 7 as factors, but its list
will include numbers also generated by 14* and 15* etc.
FWIW, 48390900720 is about 5% away from the correct answer.
Got to go out now, so that's it for the time being.
Mike
On 24/10/2014 03:17, 'Pascal Jasmin' via Programming wrote:
got distracted for the day with this approach:
boxes =. (21* >:i.48) ;(14* >:i.77) ;(10* >:i.100) ;(15* >:i.70) ;(125* >:i.8) ; (27*
>:i.36) ; (8* >:i.127) ; 6* >: i.168
+/ 991 {.{. >./ ,:&> 0 -.~ each , each 2 ({:@{. #~ [: -~/ {."1)\ each (< 0 4) , each ( 2
(([: {: {:"1) ,.~ ( 0 10 -~ {."1) {~ [: ( <:/ ) {:"1)\ ]) each|:@:( ,: d) each boxes
17184
its the wrong answer, but close :(
but it can provide a super quick estimate of 1e8, 1e5
+/ 0 -.~ , &> 2 ({:@{. * [: -~/ {."1)\ each (< (, d) 1e8 -1e5) ,~ each (< 0 128) , each ( 10 (
{.@{. , [: ( >./ ) {:"1)\ ]) each |:@:( ,: d) each < 9240* >: i. <: <: 1e8 <.@% 9240
48390900720
timespacex '+/ 0 -.~ , &> 2 ({:@{. * [: -~/ {."1)\ each (< 0 128) , each ( 10 ( {.@{. , [:
( >./ ) {:"1)\ ]) each |:@:( ,: d) each < 9240* >: i. 1e8 >.@% 9240'
0.0413071 1.85434e6
----- Original Message -----
From: 'Pascal Jasmin' via Programming <[email protected]>
To: "[email protected]" <[email protected]>
Cc:
Sent: Tuesday, October 21, 2014 7:15 PM
Subject: Re: [Jprogramming] An easy Euler Project one (485)
a complication with the approach is:
*/ 2 2 2 3 3 5 5 7 11
138600
d */ 2 2 2 3 3 5 5 7 11
144
and so for the ranges starting at 38600 to 63320 there is a greater maximum,
and it also appears that for some multiples of 138600, the number of divisors
exceeds (for part of the range) of multiples of 83160
a number that simplifies the process is:
*/ 2 2 2 3 5 7 11
9240
as it allows needing to examine with d only multiples of it.
----- Original Message -----
From: Raul Miller <[email protected]>
To: Programming forum <[email protected]>
Cc:
Sent: Tuesday, October 21, 2014 6:41 PM
Subject: Re: [Jprogramming] An easy Euler Project one (485)
See also http://oeis.org/A002182
Also Roger Hui's point about a sieve is probably a good one since this
problem only needs you to consider prime factors less than 10000. Still,
those 1229 prime factors multiplied by the 1e8 limit means we would need
something on the order of several terabytes of memory for intermediate
results if we were to store the entire sieve in the obvious way. So some
extra work would be needed, and I'm not sure how that would pan out.
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