Solving inequality equation involving sum of binomial coefficients

Question

I have a function $f(k,\,i)$ involving binomial coefficients: $$f(k,\,i)\,=\left(\begin{matrix}k+i \\ k\end{matrix}\right)=\frac{(k+i)!}{k!\,i!}$$ And the following sum over this function (expansion taken from answer by @user17762): $$S(k,x)=\sum_{i=0}^xf(k,\,i)=\left(\begin{matrix}k+x+1 \\ k+1\end{matrix}\right)$$ Now, given some number $y\ge 0$, I need to find $x$ such that: $$S(k,x-1)\le y<S(k,x)$$ Is it possible to solve this problem by coming up with some formula $g(k,y)$ such that $$S(k,g(k,y)-1)\le y<S(k,g(k,y))$$

Let me add an example:

Given for $k=2$: $$f(2,\,0\,\dots)=1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, \,\dots$$ $$S(2, 0\,\dots)=1,4,10,20,35,56,84,120,165,220,286, \,\dots$$ I need the following: $$g(2,y)=\left\{\begin{matrix}0 & \text{if} & y <1 \\ 1 & \text{if} & 1\le y <4 \\ 2 & \text{if} & 4\le y <10 \\ 3 & \text{if} & 10\le y <20 \\ 4 & \text{if} & 20\le y <35 \\ 5 & \text{if} & 35\le y <56 \\ 6 & \text{if} & 56\le y <84 \\ 7 & \text{if} & 84\le y <120 \\ 8 & \text{if} & 120\le y <165 \\ 9 & \text{if} & 165\le y <220 \\ 10 & \text{if} & 220\le y <286 \\ & \dots\end{matrix}\right.$$

Answer 1

If $x \in \mathbb{Z}$, we have $$\sum_{i=0}^x \dbinom{k+i}{k} = \dbinom{k+x+1}{k+1}$$ I trust this should help you.

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For smaller values of $k$, say for instance, $k=2$, we have $$S(2,x)=\dbinom{k+x+1}{k+1} = \dbinom{x+3}3 = \dfrac{(x+3)(x+2)(x+1)}6$$ a polynomial of degree $3$. In general, $S(k,x)$ is a polynomial in degree $k+1$.

Answer 2

This is not an answer but it seems to be too long for a comment.

In the same spirit as user17762's answer, we can "see" that we have polynomials and, as a result, $$S(k,x)=\frac 1{(k+1)!} \,\prod_{i=1}^{k+1}(x+i)$$ $$S(k,x-1)=\frac 1{(k+1)!} \,\prod_{i=1}^{k+1}(x+i-1)$$ So, in other words, for given values of $k$ and $y$, you search for an integer $x$ such that $$\frac 1{(k+1)!} \,\prod_{i=1}^{k+1}(x+i-1)\leq y \lt \frac 1{(k+1)!} \,\prod_{i=1}^{k+1}(x+i)$$ Onr thing which may be could help is that $$\frac{S(k,x)}{S(k,x-1)}=1+\frac{k+1}x$$