In some recent work, I've stumbled across the following identity for $\lambda \vdash n$: $$ n s_\lambda = \sum_{k=1}^n p_k \sum_{\mu \nearrow_k \lambda} (-1)^{\mathrm{ht}(\lambda/\mu)} s_\mu. $$ Here, $\mu \nearrow_k \lambda$ indicates $\lambda/\mu$ is a rim hook of size $k$ and $\mathrm{ht}(\lambda/\mu)$ is one less than the number of rows that $\lambda/\mu$ appears is. The proof is not too tricky. There are $n$ partitions $\mu$ so that $\lambda / \mu$ is a rim hook. Multiplying each $s_\mu$ by the power sum $(-1)^{\mathrm{ht}(\lambda/\mu)}p_k$, we get a contribution of $s_\lambda$ via the Murnaghan-Nakayama rule. We can then show all other terms cancel. Setting $\lambda = (n)$ or $\lambda = (1^n)$, this identity generalizes the Newton identities $$ n h_n = \sum_{k=1}^n h_{n-k} p_k \quad \mbox{and} \quad n e_n = \sum_{k=1}^n (-1)^{k-1} e_{n-k} p_k. $$ Can anyone point me to a reference for this generalization of the Newton identities? I've asked a few people and looked in Macdonald and EC2, but not too carefully.
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4$\begingroup$ I have never seen this nice identity before. $\endgroup$– darij grinbergSep 27, 2018 at 15:36
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2$\begingroup$ I should clarify: this identity was found by my collaborators. $\endgroup$– Zach HSep 27, 2018 at 16:43
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8$\begingroup$ Actually, here is a way to prove your identity quickly (& dirtily): It is a particular case of the more general identity $nf = \sum\limits_{k=1}^n p_k p_k^{\perp} f$, which holds for any homogeneous symmetric function $f$ of degree $n$. And this latter identity falls prey to more-or-less straightforward algebraic approaches, such as observing that both of its sides are derivations of $f$ (recall that the map sending each homogeneous symmetric function $f$ to $nf$ is a derivation), and they are equal when $f$ is one of the power sums $p_1, p_2, p_3, \ldots$. $\endgroup$– darij grinbergSep 27, 2018 at 22:42
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$\begingroup$ Darij, what is $p_k^\perp f$? $\endgroup$– Fedor PetrovSep 28, 2018 at 19:23
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$\begingroup$ @FedorPetrov: $p_k^{\perp}$ is the operation of skewing by $p_k$. See, for example, §2.8 in Darij Grinberg and Victor Reiner, Hopf Algebras in Combinatorics, 11 May 2018. Actually, $p_k^{\perp}$ is also the same as the partial derivative with respect to $p_k$ when the symmetric functions are regarded as polynomials in $p_1, p_2, p_3, \ldots$; this doesn't extend to skewing by other functions. $\endgroup$– darij grinbergSep 29, 2018 at 1:57
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