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Re: Sum

• To: mathgroup at smc.vnet.net
• Subject: [mg30151] Re: Sum
• From: "marc jeanno" <ts at tsts.com>
• Date: Wed, 1 Aug 2001 02:19:06 -0400 (EDT)
• References: <9k5q80$h7j$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

<BobHanlon at aol.com> ha scritto nel messaggio
news:9k5q80$h7j$1 at smc.vnet.net...
>
>
> There was no question so I am guessing that you want a definition of s[n]
>
> s[n_Integer?Positive] := Sum[P[k]*Log[Sum[P[m], {m, k, n}]], {k, 1, n}];
>
> If not, resend your e-mail.




The method shown above is the one I used. But there is even something
faster:

f=Prime;
n=5000;        (for example)
it=Reverse[Table[k,{k,1,n-1}]];
Log[FoldList[#1+f[#2]&,f[n],it]]


this shows out:

{10.7941, 11.4867,... [cut] }

Now, the problem is to moltiplicate each term of the list whit his
corrispondent prime, and sum the whole thimg...
Any idea?

>
>
> Bob Hanlon
> Chantilly, VA  USA
>