On Sun, Dec 6, 2009 at 11:39 PM, Aaron S. Meurer <asmeu...@gmail.com> wrote: > p(n, k) (as per MathWorld) gives the number of partitions of n with no > numbers greater than k, but according to the original post, you want the > number of partitions with no numbers less than k, so it probably isn't the > same.
Aaron, my post was in error. At some point I will probably want a function like I described, but currently I am looking for the traditional p(k,n) function. I mistyped. On Mon, Dec 7, 2009 at 7:17 AM, Vinzent Steinberg <vinzent.steinb...@googlemail.com> wrote: > On Dec 7, 5:09 am, smichr <smi...@gmail.com> wrote: >> The web page cited shows a few of the recurrence relationships solved, >> so perhaps that is an option. I don't do much with this so don't >> really know what needs to be done to solve those and whether sympy's >> library might be of use. > > We have rsolve() which does recurrence solving. > > Vinzent Thanks for the tips smichr and Vinzent -- You received this message because you are subscribed to the Google Groups "sympy" group. To post to this group, send email to sy...@googlegroups.com. To unsubscribe from this group, send email to sympy+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/sympy?hl=en.
