Which sum occurs most often?

$a_n^T$ is defined a the number of $n$ -digit numbers, with the sum of digits to be $T$ . For $n < 1$ or $T<1$ , $a_n^T=0$ . The relation below is kinda easy to understand. Having a $(n-1)$ -digit number, you can tack on one digit to the right to make it $n$ -digit, while the added amount to the sum can vary from $0$ to $9$ .

$a_n^T=\sum_{i=T-9}^{T}a_{n-1}^i \ (*)$

Also the following function is defined.

$f_n(x)=\sum_{T=0}^{\infty}a_n^Tx^T$

Both sides of (*) is multiplied by $x^T$ and summed over $T$ .

$\sum_{T} a_n^Tx^T=\sum_{T}\sum_{i=T-9}^{T}a_{n-1}^ix^T \implies$

$f_n(x)=f_{n-1}(x^9+\dots+1)$

Also, one sees that

$f_1(x)=(x^9+\dots+1)$

Therefore,

$f_8(x)=\bigg(\frac{1-x^{10}}{1-x}\bigg)^8$

$f_8(x)$ would have $72$ non-zero terms and, for $T=1$ to $T=72$ and, if you are familiar with multiplication of constant "Random Variables" in probability theory (they approach Gaussian distribution), you would guess that for the value of $T$ in the middle the maximum is achieved. So it is either $T=35$ or $T=36$ .