OFFSET
0,5
COMMENTS
LINKS
Tom Copeland, Matryoshka Dolls: Iterated noncrossing partitions, the refined Narayana group, and quantum fields, 2022.
Tom Copeland, One Matrix to Rule Them All, 2022.
Tom Copeland, The reduced inverse refined Eulerian polynomials and associated arrays, 2022.
Tomasz Kania, Pair of identities for coefficients of the exponential transform, answer to question on MathOverflow, 2026.
Tomasz Kania, Closed form for coefficients of (x / A(x))^k, answer to question on MathOverflow, 2026.
Tomasz Kania, Pairs of symmetric recurrences for certain coefficients, answer to question on MathOverflow, 2026.
FORMULA
Given the formal Taylor series or e.g.f. f(x) = x + a_1 x^2/2! + a_2 x^3/3! + ...,
E_n^{-1}(a_1,a_2,...,a_n) = D_{x=0}^n 1 / (D_x f^{(-1)}(x)), where D_x is the derivative w.r.t. x and f^{(-1)}(x) is the (possibly formal) compositional inverse of f(x) about the origin.
E_n^{-1}(a_1,a_2,...,a_n) = D_{x=0}^n 1 f'(f^{(-1)}(x)) by the inverse function theorem, where the prime indicates differentiation w.r.t. the argument of the function f. Note the correspondence to the analytic definitions of the reciprocal tangents [RT] of A356144, consistent with the following algebraic identities.
[E]^{-1} = [P][L] = [P][E][P] = [RT][P], representing, e.g., the substitution of the permutahedra polynomials [P] of A133314 for the indeterminates of the reciprocal tangent polynomials [RT] of A356144. [E] are the refined Eulerian polynomials of A145271, and [L], the classic Lagrange inversion polynomials of A134685.
Since [P]^2 = [L]^2 = [RT]^2 = [I], the substitutional identity, i.e., [P], [L], and [RT] are involutive transformations, many identities follow from the basic ones above, e.g., [L] = [P][E]^{-1} gives an inversion formula for a formal e.g.f. f(x) = x + a_1 x^2/2! + a_2 x^3/3! + ..., and we can identify [E] and [E]^{-1} as a conjugate dual.
With a_n = -x, [E]^{-1} reduces to a signed version of A112493 with an additional initial row, with the row sums of the unsigned coefficients being (1, A006351). A112493 is also given by the diagonals of A124324. See my link above on the reduced polynomials and associated arrays for more detail.
The sequence of row sums of the signed coefficients, i.e., E^{-1}(1,1,...,1), is the sequence (1, 1, 0, 0, 0, 0, ...).
Conjecture: row polynomials are R(n,1) for n > 0 where R(n,k) = R(n-1,k+1) - Sum_{j=1..n-1} binomial(n-1,j-1)*R(j,k)*R(n-j,1) for n > 1, k > 0 with R(1,k) = a_k for k > 0. - Mikhail Kurkov, Mar 22 2025
From Mikhail Kurkov, Apr 22 2026: (Start)
The above conjecture is true, see MathOverflow link.
By Tomasz Kania:
Sum_{j=1..n} S(n,j) * R(j,k) = a_{n+k} where S(n,k) = n! * [x^n] A(x)^k / k! = Sum_{j=1..n-k+1} binomial(n-1,j-1) * a_{j-1} * S(n-j,k-1) with S(n,0) = 0^n, S(n,k) = 0 for k < 0 or k > n and where A(x) = Sum_{n>=1} a_{n-1} * x^n / n! with a_0 = 1.
Sum_{j=k..n} binomial(n,j) * S(j,k) * a_{n-j} = Sum_{j=k..n} binomial(j,k) * S(n,j) * E_{j-k}^{(-1)} = S(n+1,k+1).
E_{n}^{(-1)} = (n-1)! * [x^(n-1)] A(x)'' * (x / A(x))^n for n > 0. (End)
EXAMPLE
The first few rows of coefficients with monomials in reverse order to the partitions of Abramowitz and Stegun (link in A000041, pp. 831-2) are
0) 1;
1) 1;
2) -1, 1;
3) 3, -4, 1;
4) -15, 25, -4, -7, 1;
5) 105, -210, 70, 60, -15, -11, 1;
6) -945, 2205, -1120, -630, 70, 350, 126, -15, -26, -16, 1;
7) 10395, -27720, 18900, 7875, -2800, -6930, -1638, 560, 455, 784, 238, -56, -42, -22, 1;
8) -135135, 405405, -346500, -114345, 84700, 138600, 24255, -2800, -27300, -11025, -18900, -3780, 1575, 1344, 2142, 1596, 414, -56, -98, -64, -29, 1;
...
The first few partition polynomials are
E_0^{(-1)} = 1,
E_1^{(-1)} = a1,
E_2^{(-1)} = -a1^2 + a2,
E_3^{(-1)} = 3 a1^3 - 4 a1 a2 + a3,
E_4^{(-1)} = -15 a1^4 + 25 a1^2 a2 - 4 a2^2 - 7 a1 a3 + a4,
E_5^{(-1)} = 105 a1^5 - 210 a1^3 a2 + 70 a1 a2^2 + 60 a1^2 a3 - 15 a2 a3 - 11 a1 a4 + a5,
E_6^{(-1)} = -945 a1^6 + 2205 a1^4 a2 - 1120 a1^2 a2^2 - 630 a1^3 a3 + 70 a2^3 + 350 a1 a2 a3 + 126 a1^2 a4 - 15 a3^2 - 26 a2 a4 - 16 a1 a5 + a6,
E_7^{(-1)} = 10395 a1^7 - 27720 a1^5 a2 + 18900 a1^3 a2^2 + 7875 a1^4 a3 - 2800 a1 a2^3 - 6930 a1^2 a2 a3 - 1638 a1^3 a4 + 560 a2^2 a3 + 455 a1 a3^2 + 784 a1 a2 a4 + 238 a1^2 a5 - 56 a3 a4 - 42 a2 a5 - 22 a1 a6 + a7,
E_8^{(-1)} = -135135 a1^8 + 405405 a1^6 a2 - 346500 a1^4 a2^2 - 114345 a1^5 a3 + 84700 a1^2 a2^3 + 138600 a1^3 a2 a3 + 24255 a1^4 a4 - 2800 a2^4 - 27300 a1 a2^2 a3 - 11025 a1^2 a3^2 - 18900 a1^2 a2 a4 - 3780 a1^3 a5 + 1575 a2 a3^2 + 1344 a2^2 a4 + 2142 a1 a3 a4 + 1596 a1 a2 a5 + 414 a1^2 a6 - 56 a4^2 - 98 a3 a5 - 64 a2 a6 - 29 a1ma7 + a8,
... .
Example substitution identities:
With the permutahedra polynomials
P_1 = -a_1,
P_2 = 2*a_1^2 - a_2,
P_3 = -6*a_1^3 + 6*a_2*a_1 - a_3,
the refined Eulerian polynomials
E_1 = a_1,
E_2 = a_1^2 + a_2,
E_3 = a_1^3 + 4*a_1*a_2 + a_3,
the reciprocal tangent polynomials
RT_1 = -a_1,
RT_2 = -a_2 + a_1^2,
RT_3 = -a_3 + 2*a_1*a_2 - a_1^3,
the Lagrange inversion polynomials
L_1 = -a_1,
L_2 = 3*a_1^2 - a_2,
L_3 = -15*a_1^3 + 10*a_1a_2 - a_3,
then
E^{-1}_3 = P_3(L_1,L_2,L_3) = -6*(-a_1)^3 + 6*(3*a_1^2 - a_2)*(-a_1) - (-15*a_1^3 + 10*a_1*a_2 - a_3) = 3*a_1^3 - 4*a_2*a_1 + a_3,
E^{-1}_3 = RT_3(P_1,P_2,P_3) = -(-6*a_1^3 + 6*a_2*a_1 - a_3) + 2*(-a_1)*(2*a_1^2 - a_2) - (-a_1)^3 = 3*a_1^3 - 4*a_2*a_1 + a_3,
E{-1}_3(E_1,E_2,E_3) = 3*a_1^3 - 4*a_1*(a_1^2 + a_2) + (a_1^3 + 4*a_1*a_2 + a_3) = a_3.
MATHEMATICA
rows[nn_] := {{1}}~Join~With[{s = 1/D[InverseSeries[x + Sum[c[k - 1] x^k/k!, {k, 2, nn}] + O[x]^(nn + 1)], x]}, Table[Coefficient[n! s, x^n Product[c[t], {t, p}]], {n, nn-1}, {p, Reverse[Sort[Sort /@ IntegerPartitions[n]]]}]];
rows[8] // Flatten (* Andrei Zabolotskii, Feb 17 2024 *)
PROG
(SageMath)
B.<a1, a2, a3, a4, a5, a6, a7, a8, a9, a10> = PolynomialRing(ZZ)
A.<x> = PowerSeriesRing(B)
f = x + a1*x^2/factorial(2) + a2*x^3/factorial(3) + a3*x^4/factorial(4) + a4*x^5/factorial(5) + a5*x^6/factorial(6) + a6*x^7/factorial(7) + a7*x^8/factorial(8) + a8*x^9/factorial(9) + a9*x^10/factorial(10)
g = f.reverse()
w = derivative(g, x)
I = 1 / w
# Added by Peter Luschny, Feb 17 2024:
for n, c in enumerate(I.list()[:9]):
print(f"E[{n}]", (factorial(n)*c).coefficients())
CROSSREFS
KEYWORD
sign,tabf
AUTHOR
Tom Copeland, Jul 27 2022
STATUS
approved