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"Maybe the original enquirer could
be so kind to elaborate on what this code is supposed to
do."
Dusant, if you read the thread, I posted an
explanation of what the code is supposed to do.
Follow the thread.
Tom
<BLOCKQUOTE
>
----- Original Message -----
<DIV
>From:
<A title=dusant@xxxxxxxxxxxxxxxxxx
href="">Dusant
To: <A
title=Metastockusers@xxxxxxxxxxxxxxx
href="">Metastockusers@xxxxxxxxxxxxxxx
Sent: Wednesday, July 21, 2004 3:19
AM
Subject: Re: [Metastockusers] Re: formula
question
Brute force is fine. :) except there seems
something missing.
There should be a seeding number to begin with,
to bring it down to starting point or else the calculated values would shoot
though the roof, as we keep on adding the 1/3 values over 20
periods.
Maybe the original enquirer could be so kind to
elaborate on what this code is supposed to do.
DusantChief Architect<A
href="">http://www.candlestrength.com/
<BLOCKQUOTE
>
----- Original Message -----
<DIV
>From:
<A title=bellamy_29m@xxxxxxxxx
href="">bellamy_29m@xxxxxxxxx
To: <A
title=Metastockusers@xxxxxxxxxxxxxxx
href="">Metastockusers@xxxxxxxxxxxxxxx
Sent: Tuesday, July 20, 2004 5:37
PM
Subject: [Metastockusers] Re: formula
question
When neat coding fails, try brute
force:n:=20;((n)-(n+1)/3)*C+((n-1)-(n+1)/3)*Ref(C,n-(n+1))+((n-2)-(n+1)/3)*Ref(C,n-(n+2))+((n-3)-(n+1)/3)*Ref(C,n-(n+3))+((n-4)-(n+1)/3)*Ref(C,n-(n+4))+((n-5)-(n+1)/3)*Ref(C,n-(n+5))+((n-6)-(n+1)/3)*Ref(C,n-(n+6))+((n-7)-(n+1)/3)*Ref(C,n-(n+7))+((n-8)-(n+1)/3)*Ref(C,n-(n+8))+((n-9)-(n+1)/3)*Ref(C,n-(n+9))+((n-10)-(n+1)/3)*Ref(C,n-(n+10))+((n-11)-(n+1)/3)*Ref(C,n-(n+11))+((n-12)-(n+1)/3)*Ref(C,n-(n+12))+((n-13)-(n+1)/3)*Ref(C,n-(n+13))+((n-14)-(n+1)/3)*Ref(C,n-(n+14))+((n-15)-(n+1)/3)*Ref(C,n-(n+15))+((n-16)-(n+1)/3)*Ref(C,n-(n+16))+((n-17)-(n+1)/3)*Ref(C,n-(n+17))+((n-18)-(n+1)/3)*Ref(C,n-(n+18))+((n-19)-(n+1)/3)*Ref(C,n-(n+19));wabbit
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