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I was exploring some raising and lowering operators related to an infinitesimal generator for fractional integro-derivatives and found an Appell sequence of polynomials, i.e., an infinite sequence of polynomials for which $\frac{d}{dx}p_n(x)=np_{n-1}(x)$, that is defined by the following recursion relation:

$p_{0}(x)=1$, $p_{1}(x)=x+\gamma$, and for $n>0$ $$p_{n+1}(x)=(x+\gamma)p_{n}(x)+\sum_{j=1}^{n}(-1)^j\binom{n}{j}j!\zeta (j+1)p_{n-j}(x)$$

where $\gamma=-\frac{\mathrm{d} }{\mathrm{d} \beta }\beta !\mid_{\beta =0 }$, the Euler-Mascheroni constant, and $\zeta(s)$ is the Riemann zeta function.

They satisfy $$p_{n}(x)=\frac{\mathrm{d^n} }{\mathrm{d} \beta^n }\frac{\exp(\beta x)}{\beta !} \mid_{\beta =0 }.$$

Explicitly,

$$p_2(x)=(x+\gamma)^2-\zeta(2)$$ $$p_3(x)=(x+\gamma)^3-3\zeta(2)(x+\gamma)+2\zeta(3)$$ $$p_4(x)=(x+\gamma)^4-6\zeta(2)(x+\gamma)^2+8\zeta(3)(x+\gamma)+3[\zeta^2(2)-2\zeta(4)]$$ $$p_5=p_1^5-10\zeta(2)p_1^3+20\zeta(3)p_1^2+15[\zeta^2(2)-2\zeta(4)]p_1+4[-5\zeta(2)\zeta(3)+6\zeta(5)]$$

Can anyone provide a reference for these polynomials or point out an interesting combinatorial interpretation?

Background: Rich associations with fractional calculus, digamma function, ladder operators

The fractional integro-derivative can be represented as an exponentiated convolutional infinitesimal generator (cf. MSE-Q125343):

$\displaystyle\frac{d^{-\beta}}{dx^{-\beta}}\frac{x^{\alpha}}{\alpha!}= \displaystyle\frac{x^{\alpha+\beta}}{(\alpha+\beta)!} = exp(-\beta R_x) \frac{x^{\alpha}}{\alpha!}$

where

$$R_xf(x)=\frac{1}{2\pi i}\displaystyle\oint_{|z-x|=|x|}\frac{-ln(z-x)+\lambda}{z-x}f(z)dz$$

$$=(-ln(x)+\lambda)f(x)+\displaystyle\int_{0}^{x}\frac{f\left ( x\right )-f(u)}{x-u}du.$$

with $\lambda=d\beta!/d\beta|_{\beta=0}$. (Note the integrand is related to the q (Jackson) derivative, and the Pincherle derivative / commutator is $[R_x,x]=D_x^{-1}$.)

Then $$exp(-\beta R_x) 1 =\displaystyle\frac{x^\beta}{\beta!} = exp(-\beta\psi_{.}(x)), $$

with $(\psi_{.}(x))^n=\psi_n(x)$, which implies

$$\psi_{n}(x)=(-1)^n \frac{d^n}{d\beta^n}\frac{x^\beta}{\beta!}|_{\beta=0},$$ $$L_x\psi_{n}(x)=n\psi_{n-1}(x)=-x\displaystyle\frac{d}{dx}\psi_{n}(x),$$ $$R_x\psi_{n}(x)=\psi_{n+1}(x).$$

Let $x=e^z$ and $p_n(z)=(-1)^n \psi_{n}(e^z)$. Then

$$exp(-\beta R_z) 1 =\displaystyle\frac{exp(\beta z)}{\beta!} = exp(\beta p_{.}(z)), $$

$$L_z p_{n}(z)=n p_{n-1}(z)=\displaystyle\frac{d}{dz} p_{n}(z),$$ $$R_z p_{n}(z)= p_{n+1}(z)= (z+\gamma)p_n(z)-\displaystyle\int_{-\infty}^{z}\frac{p_n\left ( z\right )-p_n(u)}{e^z-e^u} e^u du$$

with $\gamma=-d\beta!/d\beta|_{\beta=0}$, the Euler-Mascheroni constant.

Since $p_n(z)$ is an Appell sequence and, consequently, $p_n(x+y)=(p.(x)+y)^n$, umbrally, a change of integration variables $\omega=z-u$ gives

$$R_z p_{n}(z)= p_{n+1}(z)= (z+\gamma)p_n(z)-\displaystyle\int_{0}^{\infty}[p_n(z)-(p_{.}(z)-\omega)^n] \frac{1}{e^{\omega}-1}d\omega$$

from which the recursion formula follows.

In addition, using the operator formalism for Sheffer sequences, of which the Appell is a special case,

$$R_z=z-\frac{\mathrm{d} }{\mathrm{d} \beta}ln[\beta!]\mid _{\beta=\frac{\mathrm{d} }{\mathrm{d} z}=D_z}=z-\Psi(1+D_z)$$

where $\Psi(x)$ is the digamma or Psi function.

UPDATE (Nov. 16, 2012): Just found this exact sequence in the thesis "Regularized Equivariant Euler Classes and Gamma Functions" by R. Lu with a discussion of the relationships to Chern and Pontrjagin classes.

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    $\begingroup$ Curiously, Appell published a "proof" of the irrationality of $\gamma$ in Comptes Rendus in 1926 (he later found an error and retracted). Maybe your sequence is in his paper? $\endgroup$ Commented Nov 1, 2012 at 22:23
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    $\begingroup$ It might be useful for tracking down properties of your polynomials to note that the coefficients are exactly the sizes of conjugacy classes in the Symmetric groups and the $\zeta$ values in each term correspond to the cycle structure of these classes. Have you explored this avenue in trying to determine if they arise elsewhere? $\endgroup$
    – ARupinski
    Commented Nov 2, 2012 at 1:16
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    $\begingroup$ Furthermore the exponent of $(x-\gamma)$ seems to be the number of fixed points in the standard permutation representation of the corresponding conjugacy class, and the sign in front of each term appears to correspond to the value of the sign representation on that conjugacy class. For example, in $p_4$, the term $-6\zeta(2)(x+\gamma)^2$ corresponds to the conjugacy class $1^22^1$ which has class size 6, consists of odd permutations, and has 2 fixed points (the exponent of 1 in the cycle description) in the standard permutation representation. $\endgroup$
    – ARupinski
    Commented Nov 2, 2012 at 1:22
  • $\begingroup$ Of course this description fails for the identity conjugacy class since $\zeta(1) = \infty$ instead of 1 (which is what it would need to be to make the above observations fully hold), but otherwise these relationships seem to indicate your polynomials are strongly related to the symmetric groups in general and so you should probably look in the literature on symmetric groups to see if they arise elsewhere. $\endgroup$
    – ARupinski
    Commented Nov 2, 2012 at 1:25
  • $\begingroup$ @ARupinski, I modified the question, so please feel free to change your comments into an answer. $\endgroup$ Commented Nov 2, 2012 at 2:36

2 Answers 2

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Let $P_i$ be the power sum symmetric function. In your $p_n$, Replace $x+\gamma$ by $P_1$ and $\zeta(i)$ by $P_i$. Then divide the result by $n!$. What you get looks like a well-known symmetric function, which corresponds to the sign representation of the symmetric group $S_n$.

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  • $\begingroup$ And replacing $zeta(i)$ by $x^i$ and $p_{1}(x)$ by 1 seems to give (to low orders at least) oeis.org/A055137 with its interpretations in terms of the characteristic polynomial of the adjacency matrix of the complete n-graph and (Bala's comment in the entry and Rupinski's above) as a sum over permutations in $S_n$ flagged by parity and fixed points. $\endgroup$ Commented Nov 2, 2012 at 9:30
  • $\begingroup$ Thanks. Your point and Rupinski's are clearly illustrated in the Wiki article en.wikipedia.org/wiki/Newton%27s_identities and oeis.org/A036039. $\endgroup$ Commented Nov 2, 2012 at 12:02
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Follow-up on Rupinski's and Chapoton's observations:

To nail down the identification of the $p_n(x)$ with the cycle index polynomials for $S_n$ (or the partition polynomials of the refined Stirling numbers of the first kind A036039), look at the Taylor series rep of the digamma operator for the raising / creation operator for the $p_n(z)$ basis

$$R_z = z-\Psi(1+D_z) = z+\gamma + \sum_{n=1}^{\infty } (-1)^n\zeta (n+1)D_z^n.$$

This is precisely the raising operator for the cycle index polynomials as presented on page 23 of Lagrange à la Lah Part I with $c_1=z+\gamma=p_1(x)$ and $c_{n+1}=(-1)^n\zeta(n+1)$ for $n>0$

$$D^{-1}_{c_1}= :\frac{c_{.}}{1-c_{.}D_{c_1}}: = c_1+\sum_{n=1}^{\infty } c_{n+1}D_{c_1}^n.$$

Alternatively, the Newton identities extrapolated to an entire function as an infinite order polynomial using the Weierstrass factorization maneuver can be applied to see the connections to the power and elementary symmetric polynomial formalism:

$$\exp\left (-\beta p_{.}(z)\right )=\frac{\exp\left (-\beta z \right )}{\left (-\beta \right )!}=\exp\left (-\beta(z+\gamma) \right )\prod_{k=1}^{\infty }\left ( 1-\frac{\beta}{k} \right )\exp\left (\frac{\beta}{k} \right )$$

$$=\exp\left [-(z+\gamma)\beta -\sum_{k=2}^{\infty } \frac{\zeta (k)\beta ^k}{k} \right ]=\exp\left [ :ln(1-a\beta ) :\right ]$$ where $a^1=a_{1}=(z+\gamma)$ and $a^k=a_k=\zeta(k)$ for $k>1$, but this is precisely the umbral form of the e.g.f. for the cycle index polynomials (mod signs).

(Also there are connections to rational zeta series.)

Update (Nov. 16, 2012): The generating series appears on pg. 58 in "Hodge theoretic aspects of mirror symmetry" by L. Katzarkov, M. Kontsevich, and T. Pantev (following Lu's references).

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  • $\begingroup$ And replacing $\zeta(even)$ by -1 and $\zeta(odd)$ by 1 gives the rencontres numbers oeis.org/A008290. $\endgroup$ Commented Nov 5, 2012 at 23:29
  • $\begingroup$ $\gamma$ can be eliminated using $\gamma=\sum_{k=2}^{\infty }(-1)^k\frac{\zeta(k)}{k}=1-\sum_{k=2}^{\infty }\frac{\zeta(k)-1}{k}$. $\endgroup$ Commented Nov 11, 2012 at 22:02
  • $\begingroup$ See also mathoverflow.net/questions/112062/… and mathoverflow.net/questions/111770/… $\endgroup$ Commented Nov 17, 2012 at 6:00
  • $\begingroup$ Also interestingly, $\displaystyle e^{(\omega\:d/dt)}\frac{e^{(t\:z)}}{t!}=\frac{e^{[(t+\omega)\:z]}}{(t+\omega)!}=e^{(\omega\:R_z)}\frac{e^{(t\:z)}}{t!}.$ $\endgroup$ Commented Nov 17, 2012 at 14:00
  • $\begingroup$ See also "An integral lift of the Gamma-genus" and "The motivic Thom isomorphism" by Jack Morava, "Multiple zeta values and Rota-Baxter algebras" by K. Ebrahimi-Fard and L. Guo, "Multiple zeta values" (math.jussieu.fr/~miw/articles/pdf/BoulderVI.pdf) by M. Waldschmidt, and "Double shuffle relations of multiple zeta values" by J. Zhao. $\endgroup$ Commented Nov 20, 2012 at 14:25

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