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The partition polynomials of OEIS A263633 give the coefficients of the power series/o.g.f of the multiplicative inverse (reciprocal) of a power series/o.g.f. and so give the Newton identities for transforming between complete homogeneous symmetric polynomials/functions and elementary symmetric polynomials/functions. Certain Koszul duals are related to this.

After reading Stanley's answer, I guess I need to rephrase the question to lean away from algebraic combinatorics towards geometric combinatorics. For example, the square of the lower triangular Pascal matrix is a convolution of binomial coefficients and also the face vectors of the hypersquares, or hypercubes. Another example, the refined associahedra partition face polynomials are combinations of averaged factorials that give a prescription for compositional inversion of o.g.f.s--it's the connection to the combinatorics of the faces--the geometry--that I would like emphasized.

On my blog "Shadows of Simplicity" I posted "Squaring Triangles" to explain to motivated high school students the relation between the algebraic and geometric combinatorics. Ideally, that is what I'm looking for here.

The algebraic combinatorics of the complementary reciprocal of a Taylor series/e.g.f. is governed by the antipode/refined Euler characteristic classes of the permutahedra or, equivalently, by surjective mappings, so I have an indirect geometric combinatorial interpretation of 'scaled' versions of the Newton identities, but I'm looking for more direct interpretations.

What geometric structures are enumerated by the integer coefficients of these partition polynomials for conversion of an o.g.f. into a reciprocal o.g.f.?

One answer from A133314 is colored surjections, where the arrows mapping into an element of the image induce a linear order by color that are then permuted. The resulting partition polynomials of A263633 are then the coefficients of the e.g.f. enumerating these mappings, i.e. the scaled, signed face partition polynomials of the permutahedra divided by n!.

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  • $\begingroup$ in the links, I show that the Pascal lower triangular matrix itself, ignoring the initial column of ones, contains the face vectors of the hypertriangles. E.g., (4,6,4,1) is the f-vector for the tetrahedron, encompassing 4 vertices, 6 edges, 4 triangles as facets, and 1 tetrahedron. $\endgroup$
    – Tom Copeland
    Commented Jun 17, 2022 at 19:18
  • $\begingroup$ Related: mathoverflow.net/questions/243185/… $\endgroup$
    – Tom Copeland
    Commented Jun 21, 2022 at 4:51
  • $\begingroup$ This is the first I've heard of that particular blog, and I notice that you've got lines that say things like $ n!\sum_{k=0}^n \binom{n-n}{n-k} \frac{(-x)^k}{k!} $ instead of $ \displaystyle n!\sum_{k=0}^n \binom{n-n}{n-k} \frac{(-x)^k}{k!}. $ Note not only the difference in sizes of some things, but the difference in formatting of the subscripts and superscripts on the summation sign, thus $\sum_{k=0}^n,$ with the subscript and superscript to the right of $\sum,$ versus $\displaystyle \sum_{k=0}^n.$ Why put them in inline style rather than display style when they're on a line by themselves? $\endgroup$
    – Michael Hardy
    Commented Dec 11, 2023 at 18:10
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    $\begingroup$ @MichaelHardy, limited personal time and availability of free, nice, efficient apps for publishing large pdf files in LaTex. I've got tons of written material that I haven't published yet. Any recommendations for stand-alone apps or for add-ons for Google docs? Of course, I use cut and paste for related entries / use (MO, MSE, OEIS, pdf, WordPress, Desmos, Wolfram Alpha) and don't regard it a necessity to perfect the formatting for each venue for aesthetic tastes. $\endgroup$
    – Tom Copeland
    Commented Dec 11, 2023 at 18:50
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    $\begingroup$ (cont.) In fact, some users have edited some of my posts beautifully but making the lines then extremely difficult to cut and paste for use in other venues or even for calculations via substitutions in specific applications. // Some twenty years ago I used a free app that made it very easy to produce reasonably-sized pdfs with nicely formatted multi-colored equations, but it disappeared a couple of years later and I haven't found a comparable one since. $\endgroup$
    – Tom Copeland
    Commented Dec 11, 2023 at 19:09

1 Answer 1

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Since $$ \frac{1}{1+a_1x+a_2x^2+\cdots} = \sum_{k\geq 0}(-1)^k(a_1x+a_2x^2+\cdots)^k, $$ the coefficient of $a_1^{c_1}a_2^{c_2}\cdots x^n$, where $n=\sum ic_i$, is just $(-1)^k$ times the multinomial coefficient ${k\choose c_1,c_2,\dots}$, where $k=\sum c_i$.

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  • $\begingroup$ Thanks. Yes, that is the algebraic combinatorics. I refined my question to be clearer about what I'm looking for. $\endgroup$
    – Tom Copeland
    Commented Sep 8, 2020 at 1:55

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