Let $T\_k(x\_1,\ldots,x\_n)$ be the Todd polynomials, $e\_k(x\_1,\ldots,x\_n)$ the elementary symmetric polynomials and $p\_k(x\_1,\ldots, x\_n)$ the power sums of degree $k$.


We have the following generating formulas
\begin{align\*}
\sum\_{k\geq 0}T\_k(x\_1,\ldots,x\_n)t^k = \prod\_{i=1}^n\frac{tx\_i}{1-e^{-tx^i}}\,,\\
\sum\_{k\geq 0}e\_k(x\_1,\ldots,x\_n)t^k = \prod\_{i=1}^n(1+tx\_i)\,,\\
\sum\_{k\geq 0}\frac{1}{k!}p\_k(x\_1,\ldots,x\_n)t^k = \sum\_{i=1}^ne^{tx\_i}\,.
\end{align\*}
There is an explicit relation between $(1/k!)p\_k$ and $e\_k$ in terms of Newton's identities which can be expressed using generating series (see e.g. [wikipedia](https://en.wikipedia.org/wiki/Chebyshev_polynomials)). Is there some similar expression for $T\_k$ in terms of $e\_k$ or $p\_k$?


For example if $X$ is a hyperkähler complex manifold. Replacing $x\_i$ with $\alpha\_i$ the roots of $c(TX)$, one can show that (see [(3.13)](https://arxiv.org/pdf/math/0101062.pdf)):
$$
td(X) = \text{exp}\Big[-2\sum\_{n\geq0} b\_{2n}\text{ch}\_{2n}(TX)\Big]\,,
$$
where $b\_{2n}$ are the modified Bernoulli numbers. Is there an explicit formula without any assumptions on the geometry?



Todd polynomials
We have


$$\log \sum\_{k \ge 0} T\_k t^k = \sum\_{i=1}^n \log \frac{x\_i t}{1 - e^{-x\_i t}}$$


so if we write


$$\log \frac{x\_i t}{1 - e^{-x\_i t}} = \log \sum\_{k \ge 0} B\_k^{+} x\_i^k \frac{t^k}{k!} = \sum\_{k \ge 1} b\_k x\_i^k \frac{t^k}{k!}$$


(using the sign conventions explained on [Wikipedia](https://en.wikipedia.org/wiki/Bernoulli_number#Notation)) then we straightforwardly have


$$\sum\_{k \ge 0} T\_k t^k = \exp \left( \sum\_{k \ge 1} b\_k p\_k \frac{t^k}{k!} \right).$$


[WolframAlpha](https://www.wolframalpha.com/input/?i=taylor+series+log%28x%2F%281+-+e%5E%28-x%29%29%29) gives that the generating function of the series $b\_k$ begins


$$b(t) = \log \frac{t}{1 - e^{-t}} = \frac{t}{2} - \frac{t^2}{24} + \frac{t^4}{2880} - \frac{t^6}{181440} \pm $$


which gives $b\_2 = - \frac{1}{12}, b\_4 = \frac{1}{120}, b\_6 = - \frac{1}{252}$. Plugging the denominators into the OEIS gives [A006953](http://oeis.org/A006953), the sequence of denominators of $\frac{B\_{2k}}{2k}$, which suggests the following. We have


$$b'(t) = \frac{d}{dt} \left( \log t - \log (1 - e^{-t}) \right) = \frac{1}{t} - \frac{e^{-t}}{1 - e^{-t}} = \frac{1}{t} \left( 1 - \frac{t}{e^t - 1} \right) = - \sum\_{k \ge 1} B\_k^{-} \frac{t^{k-1}}{k!}$$


which gives $b\_k = - \frac{B\_k^{-}}{k}$ for $k \ge 1$. Altogether we have


$$\boxed{ \sum\_{k \ge 0} T\_k t^k = \exp \left( \sum\_{k \ge 1} - \frac{B\_k^{-}}{k} p\_k \frac{t^k}{k!} \right) }.$$


As a sanity check, expanding the terms in WolframAlpha [up to $t^4$](https://www.wolframalpha.com/input/?i=exp%28c_1+*+t%2F2+-+%28c_1%5E2+-+2c_2%29+*+t%5E2%2F24+%2B+%28c_1%5E4+-+4+c_2*c_1%5E2+%2B+4+c_3*c_1+%2B+2+c_2%5E2+-+4c_4%29+*+t%5E4%2F2880%29+expand+in+t) in terms of elementary symmetric polynomials / Chern classes gives


$$T\_1 = \frac{c\_1}{2}$$
$$T\_2 = \frac{c\_1^2 + c\_2}{12}$$
$$T\_3 = \frac{c\_1 c\_2}{24}$$
$$T\_4 = \frac{-c\_1^4 + 4c\_2 c\_1^2 + c\_3 c\_1 + 3c\_2^2 - c\_4}{720}$$


which agrees with the formulas for the first few terms of the Todd class on [Wikipedia](https://en.wikipedia.org/wiki/Todd_class).


**Edit:** I believe this is the same as the result you quote except that the linear term in your result is zero (equivalently, $c\_1$ vanishes for a hyperkahler manifold).


