Tangent numbers are divisible by $2^{n}$ Let us consider a $$\tan(z) = \sum_{n=1}^{\infty}{T_{2n-1} \cdot \frac{z^{2n-1}}{(2n-1)!}}$$.
So, it can be shown that $$T_{2n+1}=\frac{(-1)^{n} 4^{n+1}(4^{n+1}-1) B_{2n+2}}{2n+2} $$ where $B_{2n+2}$ is the $2n+2$ th Bernoulli number.
How to prove that $T_{2n+1}$ is divisible by $2^{n}$?
The first idea is to derive it using induction, but the induction step seem to be too much complicated. Are there another possible approaches?
Any help would be much appreciated.
 A: Take a look at the first several derivatives of $\tan(z)$:
$$
\begin{align}
\tan(z)&=\tan(z)\\
\sec^2(z)&=\tan(z)^2+1\\
2\sec^2(z)\tan(z)&=2\tan^3(z)+2\tan(z)\\
6\tan^2(z)\sec^2(z)+2\sec^2(z)&=6\tan^4(z)+8\tan^2(z)+2\\
24\tan^3(z)\sec^2(z)+16\tan(z)\sec^2(z)&=24\tan^5(z)+40\tan^3(z)+16\tan(z)\\
\end{align}
$$ 
This suggests that the $n$th derivative of tangent is a certain polynomial in $\tan(z)$. That is, $$\frac{d^n}{dz^n}\tan(z)=P_n(\tan(z))$$ These polynomials $P$ satisfy a recursion that is easy to prove: $$P_n(X)=\frac{d}{dX}P_{n-1}(X)\cdot\left(X^2+1\right)$$ and in fact proving this recursion proves the existence of the polynomials.
We can give names to the coefficients of these polynomials: $$P_n(X)=\sum_{k=0}^{n+1}c_{n,k}X^k$$ and be sure to define $c_{n,k}$ to be $0$ for negative $k$ or $k$ larger than $n+1$. The recursion tells us $$c_{n,j}=(j-1)c_{n-1,j-1}+(j+1)c_{n-1,j+1}$$ 
Inductively, assume that $2^n$ divides $c_{2n+1,k}$ for all $k$. (This is true when $n=1$, as is apparent from the fourth row [third derivative] above.) Then $$\begin{align}
c_{2n+3,k} 
&=(k-1)c_{2n+2,k-1}+(k+1)c_{2n+2,k+1}\\
&= (k-1)(k-2)c_{2n+1,k-2}+(k-1)(k)c_{2n+1,k}+(k+1)(k)c_{2n+1,k}+(k+1)(k+2)c_{2n+1,k+2}
\end{align}
$$
Since each of the $c_{2n+1,*}$ are divisible by $2^n$ by the induction assumption, and since each of $(k-1)(k-2)$, $(k-1)(k)$, $k(k+1)$, and $(k+1)(k+2)$ are even, it follows that $c_{2n+3,k}$ is divisible by the next higher power of $2$, namely $2^{n+1}$. So by induction, each $c_{2n+1,k}$ is divisible by $2^n$. 
You are specifically asking about  $c_{2n+1,0}$, the values of $\left.P_{2n+1}(X)\right|_{X=0}=\left.\frac{d^{2n+1}}{dz^{2n+1}}\tan(z)\right|_{z=0}$, which are of course the Maclaurin series coefficients for $\tan(z)$ (prior to division by a factorial).
A: I don't have enough points to comment.
There are (at least) three pertinent OEIS sequences, namely:
(1) http://oeis.org/A000182 the tangent numbers
(2) http://oeis.org/A101921 the exponent of 2 in the sequence of tangent numbers, including a recursive definition for that
(3) and A093049 which essentially gives, with the appropriate offset, the number of 2's that remain after dividing out n of them.
Whereas this is not a proof, it should, I hope still be helpful.
