Re: Sum

*To*: mathgroup at smc.vnet.net*Subject*: [mg30147] Re: Sum*From*: "Orestis Vantzos" <atelesforos at hotmail.com>*Date*: Tue, 31 Jul 2001 04:27:20 -0400 (EDT)*Organization*: National Technical University of Athens, Greece*References*: <9k2doa$2fn$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

S[n_] := Module[{P = Array[Prime, n], s}, s[0] = 0; s[1] = First@P; s[i_] := (s[i] = s[i - 1] + P[[i]]); Log@(-Table[s[i], {i, 0, n - 1}] + s[n]).P] ..does the trick ;-) In[10]:= Timing[S[5000];] Out[10]= {0.99 Second, Null} Orestis PS. This is an application of "dynamic programming" in Mathematica. "marc jeanno" <ts at tsts.com> wrote in message news:9k2doa$2fn$1 at smc.vnet.net... > Dear Sirs, > > let n be an integer >=1. Let us consider the following sums: > > s(1)= P[1]Log[P[1]] > s(2)= P[2]Log[P[2]]+P[1]log[P[2]+P[1]] > s(3)= P[3]Log[P[3]]+P[2]Log[P[3]+P[2]]+P[1]Log[P[3]+P[2]+P[1]] > . > . > . > s(n)= P[n]Log[P[n]]+...+P[1]Log[P[n]+...+P[1]] . > > > P[k] is the expression Prime[k]. > > I used a method to evaluate the expression in function of n (n must be > assigned a priori). But my PC was frozen during the > operation(zzzzzz...zzzzz...) for large n values. > How could I evaluate s(n) in a very very fast way? > Do I have to try with 2 Fold lines? > I'd like to know Your opinions. > Thanks. > > > > > > >