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Thanks to all that replied. I had forgot you can't do loops, so I guess it
is brute force.
Tom
----- Original Message -----
From: <bellamy_29m@xxxxxxxxx>
To: <Metastockusers@xxxxxxxxxxxxxxx>
Sent: Tuesday, July 20, 2004 7:07 AM
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 :D
>
> <snip>
>
>
>
>
>
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