My understanding of Mathematica is very limited since I just started using it, I would appreciate any feed back or help.

I am trying to find the closed form of a recurrent function with 2 variables:

A[n_,k_] := 1/n*Sum[A[n-i, k-1], {i,1,n}];
A[0,1] :=0;
A[n_, 1] :=1/n;

I tried doing it using RSolve:

a[n,k] == (1/n)*Sum[a[n-i, k-1], {i,1,n}], 
a[0,k] ==0, 
a[n,k], {n, k}]

But i just get as a response the RSolve input: https://i.stack.imgur.com/GVRX4.jpg

Am I doing the RSolve equation correctly or am I missing something out here?

  • 1
    $\begingroup$ Try this again in a fresh kernel; it seems you had old definitions hanging around. $\endgroup$ Dec 22, 2016 at 1:33
  • $\begingroup$ Thanks for you reply! I tried that and now I don't get the recurrsion error. Although when i enter the RSolve, the output I get is the same RSolve formula again. Not the closed form. $\endgroup$
    – t0m9er
    Dec 22, 2016 at 9:39
  • $\begingroup$ It means RSolve[] has no idea what to do; sadly, for partial difference equations, this seems to be par for the course. $\endgroup$ Dec 22, 2016 at 12:38
  • $\begingroup$ There goes my hope! Thanks for the help though :) $\endgroup$
    – t0m9er
    Dec 22, 2016 at 13:33

2 Answers 2


Here is the recursion with seems to work

A[n_][k_] := 
  A[n][k] = If[k > 1, 1/n*Sum[A[n - i][k - 1], {i, 1, n}], 1/n];
A[0][1] := 0;
A[0][k_ /; k > 1] := 0;

I have tried with


which gives



360968703235711654233892612988250163157207/\ 6972037522971247716453380893531230355680000



it's also working --- it gives 0.110791.

You can test or play with your recursion like this :

p = Table[N[A[500][i]], {i, 1, 20}]
q = Table[N[A[i][3]], {i, 1, 20}]

enter image description here

  • $\begingroup$ Thanks for the help! I tried using the if clause inside the RSolve: imgur.com/a/T1TMH But got the same result. Is that were you were referring to? or am I doing it wrong? $\endgroup$
    – t0m9er
    Dec 22, 2016 at 17:03
  • $\begingroup$ Sorry I have seen an error but to be able to correct It I must close my kernel and I cannot do it right now I will come back to this point later. $\endgroup$ Dec 22, 2016 at 17:54
  • $\begingroup$ Thanks a lot! I would really appreciate it! $\endgroup$
    – t0m9er
    Dec 22, 2016 at 19:50
  • $\begingroup$ I have corrected the code. It seems to work properly. But it will be advised to test manually for small values. $\endgroup$ Dec 22, 2016 at 21:09
  • $\begingroup$ Hey! Thanks for the corrected code. I can now play with the recursion, but I am looking for a closed form of the equation. You have any idea how to get to the closed from using the formula? I think usually RSolve does it in Mathematica, but it doesn't seem to work for me. $\endgroup$
    – t0m9er
    Dec 22, 2016 at 21:25

It turns out this recurrent function have been researched before. It can be solved by: $$\frac{S(n,k)}{n!}$$ Where $S(n,k)$ is Sterling number of the first kind: https://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind The Stirling numbers of the first kind have known generating functions and are much more researched.

@Sil deserves the credit for finding it.


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