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The tangent numbers $(T_{2n+1})=(1,2,16,272,7936,...)$ (cf. OEIS: A000182) satisfy many recurrences. I would be interested to find references for the following which I think must be very old: $T_3 -2T_1=0$, $T_5 -8T_3 =0,$ $T_7 -18T_5 +8T_3 =0,...$ or more generally

$${T_{2n + 1}} = \sum\limits_{j \ge 1} {}{(-1)}^{j-1}{2^{2j}} {\binom{n+1}{2j}} {\frac{n+1-j}{n+1}}T_{2n - 2j + 1}.$$

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The following is too long for a comment, so let me type it as an answer though it does not literally answer your question.

Using the standard formula $$ T\_{2k-1}=(-1)^{k-1}2^{2k}(2^{2k}-1)\frac{B_{2k}}{2k}, $$ your formula can be rewritten as $$ (2^{2n+2}-1)B_{2n+2}=\sum_{j\ge1}(-1)^{n-j+1}\binom{n+1}{2j}(2^{2n-2j+2}-1)B_{2n-2j+2}, $$ which looks "simpler", and also might be more recognisable by specialists, since identities for Bernoulli numbers are usually more "popular".

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    $\begingroup$ @Vladimir: Thank you for the idea to reformulate the identity with other known numbers. I now tried it with the Genocchi numbers $G_{2n}$. They satisfy $nT_{2n-1}=2^{2n-2}G_{2n}$. Then the identity reduces to $\sum{(-1)^j}{\binom{n}{2j}}G_{2n-2j}$. This is Seidel’s identity for the Genocchi numbers. So I accept your comments as answer to my question. $\endgroup$
    – Johann Cigler
    Commented Jul 25, 2012 at 20:17
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    $\begingroup$ Ah, that's nice. I noticed that the identity for Bernoulli numbers I ended up with becomes especially simple if one considers the sequence of numbers $\{(2^{2n}-1)B_{2n}\}$ (which is very close to the sequence of Genocchi numbers as I now see), but I did not explore it further. Glad I could help you to figure it all out! $\endgroup$
    – Vladimir Dotsenko
    Commented Jul 25, 2012 at 20:28

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