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How to solve the following recurrence involving binomial coefficients, where $c$ is a constant non-negative integer, and $n$ and $k$ are non-negative integers ($n \ge k \ge 0$)?

$$ A(n,k) = A(n-1,k) + A(n-1,k-1) + c, \\ A(n,0) = 1, A(n,n) = 1 $$

My Attempt: Without the constant $c$ (or $c=0$), $A(n,k)$ is just $\binom{n}{k}$.

I have tried the generating function $$ g(x,k) = \sum_{n=0}^{\infty} A(n,k) x^n. $$ I need to evaluate $$ \begin{align} \sum_{n=0}^{\infty} A(n+1,k) x^n &= \sum_{n=0}^{\infty}\big(A(n,k)+A(n,k-1) + c\big)x^{n} \\ &= g(x,k) + g(x,k-1) + \frac{c}{1-x} \end{align} $$

On the other hand, $$ \begin{align} \sum_{n=0}^{\infty} A(n+1,k) x^n = \frac{1}{x} \big(\sum_{n=1}^{\infty} A(n,k) x^n\big) = \frac{1}{x}\big(\sum_{n=0}^{\infty} A(n,k) x^n - A(0,k)\big). \end{align} $$ How to handle with $A(0,k)$? How to establish an equation for $g(x,k)$?

hengxin
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1 Answers1

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Let us guess that $A(n,k) = \alpha \binom{n}{k} + \beta$. The recurrence reads $$ \alpha \binom{n}{k} + \beta = \alpha \binom{n-1}{k} + \beta + \alpha \binom{n-1}{k-1} + \beta + c, $$ from which we deduce $\beta = -c$. The base cases are $$ \alpha \binom{n}{n} - c = \alpha \binom{n}{0} - c = 1, $$ From which we deduce $\alpha = c+1$. All in all, the solution is $$ A(n,k) = (c+1) \binom{n}{k} - c. $$

Yuval Filmus
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