# Recurrence relation with summation of previous terms

I am stuck while trying to solve a recurrence relation. I just need the characteristic function to know the dependence of $Y_n$.

My recurrence relation is of the following form where $a$ and $b$ are constant,

$Y_n = a + b \frac{\sum_{j=1}^{n-1}Y_j}{n-1}$

I am unable to figure out the characteristic equation or even the asymptotic nature of the series.

• @Peter $x^n$ seems to work but the $n$ in the denominator doesn't add up. It seems to me that there is something missing. – learner Jan 27 '17 at 17:40
• I suppose, that $n$ is unknown, but fixed. So, we will get a polynomial with coefficients depending on the degree. – Peter Jan 27 '17 at 17:43
• $n$ is known, the second term of the recurrence relation just means taking a mean over the previous terms. – learner Jan 27 '17 at 17:45
• For any concrete $n$, we have concrete coefficients and can determine the characteristic equation as usual. – Peter Jan 27 '17 at 17:46
• $n$ is the sequence index. It is not fixed. This is not a simple linear recurrence, so it does not have a characteristic equation. Characteristic equations occur for recurrences of the form $y_n = f(y_{n-1}, ..., y_{n-k})$ for some fixed $f$. You do not have that here. Your recurrence involves all previous elements of the sequence, not a fixed number of them. The standard linear methods do not apply. I suggest looking that the sequence $X_n = \sum_{j=1}^{n} Y_j$ instead. Once you have it, finding $Y_n$ from it is easy. – Paul Sinclair Jan 27 '17 at 18:35

Using the definition of $X_n$ in the comments by @Paul Sinclair, I ran out the first couple of cases, and then used the Online Encyclopedia of Integer Sequences to formulate the following conjecture: $$X_n = \frac{1}{(n-1)!}\left[\displaystyle\sum_{i=0}^{n-2} \left(ab^i\sum_{j=i+2}^{n} s(n,j)\right)+X_1\prod_{i=1}^{n-1}(b+i)\right]$$ where $s(n,k)$ is the unsigned Stirling number of the first kind.
Using the recurrence relation $X_{n+1}=a+\frac{n+b}{n}X_n$ given by Paul Sinclair, it is an interesting exercise to prove the formula using induction.