- #1
Someone2841
- 44
- 6
Let an,m be defined for the non-negative integers n and m such that n ≥ m.
an,0 = 1
am,m = m!
an+1,m+1 = (m+1) * an,m + an,m+1
Is there an explicit formula f such that f(n,m) = an,m?
Here are the first numbers of the sequence:
[itex]
\begin{align}
&m&0&1&2&3&4\\
n\\
0&&1\\
1&&1&1\\
2&&1&3&2\\
3&&1&6&11&6\\
4&&1&10&35&50&24\\
5&&1&5&85&225&274&12\\
\end{align}
[/itex]
an,0 = 1
am,m = m!
an+1,m+1 = (m+1) * an,m + an,m+1
Is there an explicit formula f such that f(n,m) = an,m?
Here are the first numbers of the sequence:
[itex]
\begin{align}
&m&0&1&2&3&4\\
n\\
0&&1\\
1&&1&1\\
2&&1&3&2\\
3&&1&6&11&6\\
4&&1&10&35&50&24\\
5&&1&5&85&225&274&12\\
\end{align}
[/itex]