%I #43 Mar 21 2024 21:21:06
%S 1,1,-1,1,3,-4,1,-15,25,-4,-7,1,105,-210,70,60,-15,-11,1,-945,2205,
%T -1120,-630,70,350,126,-15,-26,-16,1,10395,-27720,18900,7875,-2800,
%U -6930,-1638,560,455,784,238,-56,-42,-22,1,-135135,405405,-346500,-114345,84700
%N Coefficients of the inverse refined Eulerian partition polynomials [E]^{-1}, partitional inverse to A145271. Irregular triangle read by row with lengths A000041.
%C These are the coefficients of the inverse refined Eulerian partitions polynomials, the substitutional inverse to the refined Eulerian partition polynomials [E] of A145271. [E] and [E]^{-1} are a conjugate dual with respect to the permutahedra polynomials [P] of A133314 (see formula section).
%H Tom Copeland, <a href="https://tcjpn.wordpress.com/2022/07/08/iterated-noncrossing-partitions-and-quantum-fields/">Matryoshka Dolls: Iterated noncrossing partitions, the refined Narayana group, and quantum fields</a>, 2022.
%H Tom Copeland, <a href="https://tcjpn.wordpress.com/2022/07/27/one-matrix-to-rule-them-all/#more-9405">One Matrix to Rule Them All</a>, 2022.
%H Tom Copeland, <a href="https://tcjpn.wordpress.com/2022/08/18/the-reduced-inverse-refined-eulerian-polynomials-and-associated-arrays/">The reduced inverse refined Eulerian polynomials and associated arrays</a>, 2022.
%F Given the formal Taylor series or e.g.f. f(x) = x + a_1 x^2/2! + a_2 x^3/3! + ...,
%F 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.
%F 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.
%F [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.
%F 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.
%F 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.
%F 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, ...).
%e 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
%e 0) 1;
%e 1) 1;
%e 2) -1, 1;
%e 3) 3, -4, 1;
%e 4) -15, 25, -4, -7, 1;
%e 5) 105, -210, 70, 60, -15, -11, 1;
%e 6) -945, 2205, -1120, -630, 70, 350, 126, -15, -26, -16, 1;
%e 7) 10395, -27720, 18900, 7875, -2800, -6930, -1638, 560, 455, 784, 238, -56, -42, -22, 1;
%e 8) -135135, 405405, -346500, -114345, 84700, 138600, 24255, -2800, -27300, -11025, -18900, -3780, 1575, 1344, 2142, 1596, 414, -56, -98, -64, -29, 1;
%e ...
%e The first few partition polynomials are
%e E_0^{(-1)} = 1,
%e E_1^{(-1)} = a1,
%e E_2^{(-1)} = -a1^2 + a2,
%e E_3^{(-1)} = 3 a1^3 - 4 a1 a2 + a3,
%e E_4^{(-1)} = -15 a1^4 + 25 a1^2 a2 - 4 a2^2 - 7 a1 a3 + a4,
%e 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 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 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 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,
%e ... .
%e Example substitution identities:
%e With the permutahedra polynomials
%e P_1 = -a_1,
%e P_2 = 2*a_1^2 - a_2,
%e P_3 = -6*a_1^3 + 6*a_2*a_1 - a_3,
%e the refined Eulerian polynomials
%e E_1 = a_1,
%e E_2 = a_1^2 + a_2,
%e E_3 = a_1^3 + 4*a_1*a_2 + a_3,
%e the reciprocal tangent polynomials
%e RT_1 = -a_1,
%e RT_2 = -a_2 + a_1^2,
%e RT_3 = -a_3 + 2*a_1*a_2 - a_1^3,
%e the Lagrange inversion polynomials
%e L_1 = -a_1,
%e L_2 = 3*a_1^2 - a_2,
%e L_3 = -15*a_1^3 + 10*a_1a_2 - a_3,
%e then
%e 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 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 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.
%t 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]]]}]];
%t rows[8] // Flatten (* _Andrey Zabolotskiy_, Feb 17 2024 *)
%o (SageMath)
%o B.<a1,a2, a3, a4, a5, a6, a7, a8, a9, a10> = PolynomialRing(ZZ)
%o A.<x> = PowerSeriesRing(B)
%o 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)
%o g = f.reverse()
%o w = derivative(g,x)
%o I = 1 / w
%o # Added by _Peter Luschny_, Feb 17 2024:
%o for n, c in enumerate(I.list()[:9]):
%o print(f"E[{n}]", (factorial(n)*c).coefficients())
%Y Cf. A000041, A006351, A112493, A124324, A133314, A134685, A134991, A145271, A356144.
%K sign,tabf
%O 0,5
%A _Tom Copeland_, Jul 27 2022