11 Sep
2003
11 Sep
'03
2:04 p.m.
[1+sum(i=1,n-1,sigma(i)*part(n-i)}]/part(n-1) = n where part(k) is the k-th partition number. (see http://www.users.globalnet.co.uk/~perry/maths/morepartitionfunction/morepart itionfunction.htm for derivation; { parts=makePartitions(40); partSigma=vector(40); partSigma[1]=1; for (i=2,40,partSigma[i]=i*parts[i]-sum(j=1,i-1,parts[j]*partSigma[i-j])); partSigma } towards the end. Jon Perry perry@globalnet.co.uk http://www.users.globalnet.co.uk/~perry/maths/ http://www.users.globalnet.co.uk/~perry/DIVMenu/ BrainBench MVP for HTML and JavaScript http://www.brainbench.com