Tangent numbers are divisible by $2^{n}$

Question

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.

Answer 1

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).

Answer 2

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.