I was curious if anyone has a reference for a formula giving the values of n and k so that $\binom{n}{k}<\binom{n+j}{k-1}$ for a fixed $j$. 


Clearly this will be true if $k>\frac{n}{2}$ because then one will have that $\binom{n}{k}\le\binom{n}{k-1}<\binom{n+j}{k-1}$. One can improve on this result, and in the case where $j=1$ I have found precise conditions on $k$ in terms of $n$, but my approach is rather blunt and it seems like the general case will be quite tedious using my methods, even though this seems like a question that likely has an elegant combinatorial solution. I was wondering if anyone knows where a solution appears. 



When is (n choose k) < (n+j choose k-1) for fixed j?
It doesn't get that ugly, if you're mainly concerned with large $n$ and $k$. Simple order-of-magnitude stuff indicates that something like $k>\alpha n$ is true (where $\alpha$ depends on $j$). 


The inequality $\binom{n}{k}<\binom{n+j}{k-1}$ is exactly equivalent to $$1 < \frac{k}{n-k+1} \prod\_{i=1}^j \frac{n+i}{n+i-k+1}.$$ Setting $k\approx \alpha n$ and letting $n\to \infty$, this implies the inequality $$\frac{\alpha}{(1-\alpha)^{j+1}} \geq 1.$$ This is easy to solve for $j$. To solve for $\alpha$ is rougher. Since $(1-m/j)^j\to e^{-m}$, one can see that for large $j$ we must have $\alpha>m/j$ (specifically, for each $m$, if $j$ is sufficiently large then $\alpha>m/j$).


It seems to me that $$\frac 1j < \alpha\_j < \frac{\log j}{j}$$ for $j\ge 5$ follows from calculus. So here's the result: Let $\alpha\_j$ be the unique real solution to $\alpha=(1-\alpha)^{j+1}$ with $0<\alpha<1$. If $n,k$ are sufficiently large and $k>\alpha\_j n$, then $\binom{n}{k}<\binom{n+j}{k-1}$. In particular, for each $j\ge 5$, if $n,k$ are sufficiently large with $k>\frac{\log(j)}{j} n $, then $\binom{n}{k}<\binom{n+j}{k-1}$.


