Exists a straight evidence of this lcm identification?

The identification

$\displaystyle (n+1) \text{lcm} \left( {n \choose 0}, {n \choose 1}, ... {n \choose n} \right) = \text{lcm}(1, 2, ... n+1)$

is possibly not popular. The only means I recognize just how to confirm it is by utilizing Kummer's theorem that the power of $p$ separating ${a+b \choose a}$ is the variety of lugs required to add $a$ and also $b$ in base $p$. Exists an extra straight evidence, e.g., by revealing that each side separates the various other?

2019-05-05 04:17:46
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Answers: 3

Consider Leibniz harmonic triangle -- a table that resembles "Pascal triangular turned around" : on it's sides exist numbers $\frac{1}{n}$ and also each number is the amount of 2 under it (see the picture).

One can conveniently proove by induction that m - th number in n - th row of Leibniz triangular is $\frac{1}{(n+1)\binom{n}{m}}$. So LHS of our identification is simply lcd of portions in n - th row of the triangular.

Yet it's not tough to see that any kind of such number is an integer straight mix of portions on triangular's sides (i.e. $1/1,1/2,\dots,1/n$)-- and also the other way around. So LHS amounts to $lcd(1/1,\dots,1/n)$-- which is specifically RHS.

2019-05-09 03:49:39

It will certainly require something brilliant. For $n + 1 = 6$, you require $6$ times the lcm of $1, 5$ and also $10$ so as to get adequate powers of 2 (and also 3 ). Exists a characterization of those $n + 1$ where the whole variable of $n + 1$ is required, and also not the lcm, for the left hand side?

2019-05-08 04:57:14

More usually, for $0 \leq k \leq n$, there is an identification

$(n+1) {\rm lcm} ({n \choose 0}, {n \choose 1}, \dots {n \choose k}) = {\rm lcm} (n+1,n,n-1, \dots n+1-k)$.

This is merely the reality that any kind of integer straight mix of $f(x), \Delta f(x), \Delta^2 f(x), \dots \Delta^k f(x)$ is an integer straight mix of $f(x), f(x-1), f(x-2), \dots f(x-k)$ where $\Delta$ is the distinction driver, $f(x) = 1/x$, and also $x = (n+1)$.

2019-05-08 02:37:55