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 >