$ c(n) = \sum_{i = 1}^{n -1} t(i) c(n - i) $?

Question

It is well known that

$$ a(n) = \sum_{i = 1}^{n-1} a(n - i) $$

$$ b(n) = \sum_{i = 1}^{n-1} i b(n -i) $$

Can be expressed by fibonacci and powers of 2.

So What about

$$ c(n) = \sum_{i = 1}^{n -1} t(i) c(n - i) $$

Where $t(i)$ is the i th triangular number ?

Closed form or asymptotics ?

How about generalizations ?

Answer 1

Use generating functions \begin{eqnarray*} C(x)= \sum_{n=1}^{\infty} c_n x^n. \end{eqnarray*} Now use the recurrence \begin{eqnarray*} \sum_{n=2}^{\infty} c_n x^n =\sum_{n=2}^{\infty} \sum_{i=1}^{n-1} \binom{i+1}{2} c_n x^n \end{eqnarray*} Now invert the order of the sums and add in the first term \begin{eqnarray*} 1+ \sum_{n=2}^{\infty} c_n x^n =1+ \sum_{i=1}^{\infty} \binom{i+1}{2} x^{i}\sum_{n=i+1}^{n-1} c_n x^{n-i} \\ C(x)= 1+\frac{x}{(1-x)^3} C(x) \\ C(x) = \frac{(1-x)^3}{(1-x)^3-x}. \end{eqnarray*} More generally \begin{eqnarray*} m_n=\sum_{i=1}^{n-1} \binom{i+m-1}{m} m_{n-i} \end{eqnarray*} will give \begin{eqnarray*} M(x) = \frac{(1-x)^m}{(1-x)^m-x}. \end{eqnarray*}