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User: Tom Copeland

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Tom Copeland has authored 55 sequences. Here are the ten most recent ones:

A356146 Coefficients of the partition polynomials that are binomial convolutions of the partition polynomials of A133314, the refined Euler characteristic polynomials of the permutahedra and coefficient polynomials of reciprocals of Taylor series or e.g.f.s. Irregular triangle read by rows with length given by A000041.

Original entry on oeis.org

1, 1, -3, 1, 12, -9, 1, -60, 72, -9, -12, 1, 360, -600, 180, 120, -30, -15, 1, -2520, 5400, -2700, -1200, 180, 720, 180, -30, -45, -18, 1, 20160, -52920, 37800, 12600, -6300, -12600, -2100, 1260, 840, 1260, 252, -105, -63, -21, 1
Offset: 0

Author

Tom Copeland, Jul 27 2022

Keywords

Examples

			The first few rows of coefficients, with lengths given by A000041, are
0) 1;
1) 1;
2) -3, 1;
3) 12, -9, 1;
4) -60, 72, -9, -12, 1;
5) 360, -600, 180, 120, -30, -15, 1;
6) -2520, 5400, -2700, -1200, 180, 720, 180, -30, -45, -18, 1;
7) 20160, -52920, 37800, 12600, -6300, -12600, -2100, 1260, 840, 1260, 252, -105, -63, -21, 1;
8) -181440, 564480, -529200, -141120, 151200, 201600, 25200, -6300, -50400, -16800, -25200, -3360, 3360, 2520, 3360, 2016, 336, -105, -168, -84, -24, 1;
... .
The first few partition polynomials with monomials in reverse order to those of Abramowitz and Stegun (p. 831-2, see link in A000041) are
P_0 = 1
P_1(a_1)= 1 a_1
P_2(a_1,a_2) = -3 a_1^2 + 1 a_2
P_3(a_1,a_2,a_3) = 12 a_1^3 - 9 a_1 a_2 + 1 a_3
P_4(a_1,a_2,a_3,a_4) = -60 a_1^4 + 72 a_1^2 a_2 - 9 a_2^2 -12 a_1 a_3  + 1 a_4
P_5 = 360 a_1^5 - 600 a_1^3 a_2 + 180 a_1 a_2^2 + 120 a_1^2 a_3 - 30 a_2 a_3 - 15 a_1 a_4  + 1 a_5
P_6 = -2520 a_1^6 + 5400 a_1^4 a_2 - 2700 a_1^2 a_2^2 - 1200 a_1^3 a_3 + 180 a_2^3 + 720 a_1 a_2 a_3  + 180 a_1^2 a_4 - 30 a_3^2 - 45 a_2 a_4  - 18 a_1 a_5 + 1 a_6
P_7 = 20160 a_1^7 - 52920 a_1^5 a_2 + 37800 a_1^3 a_2^2 12600 a_1^4 a_3 + - 6300 a_1 a_2^3 - 12600 a_1^2 a_2 a_3 - 2100 a_1^3 a_4 + 1260 a_2^2 a_3 + 840 a_1 a_3^2 + 1260 a_1 a_2 a_4 + 252 a_1^2 a_5 - 105 a_3 a_4 - 63 a_2 a_5 - 21 a_1 a_6 + 1 a_7
P_8 = -181440 a_1^8 + 564480 a_1^6 a_2 - 529200 a_1^4  a_2^2 - 141120 a_1^5 a_3 + 151200 a_1^2 a_2^3 + 201600 a_1^3 a_2 a_3 + 25200 a_1^4 a_4 - 6300 a_2^4 - 50400 a_1 a_2^2 a_3 - 16800 a_1^2 a_3^2 - 25200 a_1^2 a_2  a_4 - 3360 a_1^3 a_5 + 3360 a_2 a_3^2 + 2520 a_2^2 a_4 + 3360 a_1 a_3 a_4 + 2016 a_1 a_2 a_5 + 336 a_1^2 a_6 - 105 a_4^2 - 168 a_3 a_5 - 84 a_2 a_6 - 24 a_1 a_7  + 1 a_8.
		

Crossrefs

Programs

  • Mathematica
    rows[n_] := {{1}}~Join~With[{s = 1/(1 + Sum[a[k] x^k/k!, {k, n}] + O[x]^(n+1))^2}, -Table[Expand@Coefficient[k! s, x^k Product[a[t], {t, p}]], {k, n}, {p, Reverse@Sort[Sort /@ IntegerPartitions[k]]}]/2];
    rows[7] // Flatten (* Andrey Zabolotskiy, Mar 07 2024 *)

Formula

Rows e.g.f.: (3 - G(x))/2, where G(x) = 1 / (1 + a_1 x + a_2 x^2/2! + a_3 x^3/3! + ...)^2.
d/da_n G(x) = 2 (x^n/n!) (G(x) - 1/2) = 2 (x^n/n!) (-1/2 + P_1 x + P_2 x^2/2! + ...).

Extensions

Ordering in row 7 changed and formula edited by Andrey Zabolotskiy, Mar 07 2024

A356145 Coefficients of the inverse refined Eulerian partition polynomials [E]^{-1}, partitional inverse to A145271. Irregular triangle read by row with lengths A000041.

Original entry on oeis.org

1, 1, -1, 1, 3, -4, 1, -15, 25, -4, -7, 1, 105, -210, 70, 60, -15, -11, 1, -945, 2205, -1120, -630, 70, 350, 126, -15, -26, -16, 1, 10395, -27720, 18900, 7875, -2800, -6930, -1638, 560, 455, 784, 238, -56, -42, -22, 1, -135135, 405405, -346500, -114345, 84700
Offset: 0

Author

Tom Copeland, Jul 27 2022

Keywords

Comments

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).

Examples

			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.
		

Programs

  • 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 (* Andrey Zabolotskiy, Feb 17 2024 *)
  • SageMath
    B. = PolynomialRing(ZZ)
    A. = 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())

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

A356144 Coefficients of the set of partition polynomials [RT] = [P][E]; i.e., coefficients of polynomials resulting from using the set of refined Eulerian polynomials, [E], of A145271 as the indeterminates of the set of permutahedra polynomials, [P], of A133314. Irregular triangle read by rows with lengths given by A000041.

Original entry on oeis.org

1, -1, 1, -1, -1, 2, -1, 1, -3, 2, 1, -1, -1, 4, -4, -2, 5, -1, -1, 1, -5, 8, 2, -4, -2, -4, 5, 4, -4, -1, -1, 6, -12, -3, 8, 18, -6, -14, 13, 2, -16, 14, 0, -8, -1, 1, -7, 18, 3, -20, 0, -15, 8, 18, 57, 6, -54, -15, -12, 84, -30, -48, 14, 14, -8, -13, -1, -1, 8, -24, -4, 32, 51, -27, -16, -6, 171, -42, -177, 50, 90, -18, 456, -276, -246, -15, 30, 154, -42, 124, -166, -113, 42, 6, -21, -19, -1
Offset: 0

Author

Tom Copeland, Jul 27 2022

Keywords

Comments

I stipulate that the row lengths are A000041, but this imposes the insertion of a zero as a coefficient of a monomial for the polynomial RT_7 and for RT_8. The number of nonzero coefficients in each higher order polynomial remains to be determined. The monomials of the partition polynomials are arranged in the order (bottom to top) in Abramowitz and Stegun (starting on p. 831, link in A000041).
The analytic interpretation of these coefficients is related to the e.g.f.s of reciprocals of the derivatives (slopes of tangents) of a pair of compositionally inverse e.g.f.s as explicitly shown in the formulas.
With the notation introduced in the formula section, this set of partition polynomials, [RT], is the e.g.f. counterpart to the special Schur expansion coefficients [b], or [K], of A355201 for o.g.f.s. and is conjugate dual to the Lagrange inversion polynomials [L] of A134685.
For example, as shown in the formulas, [RT] = [P][E] = [P][L][P] where [P] is the set of polynomials of A133314, the refined Euler characteristic polynomials of the permutahedra; [E], the set A145271, the refined Eulerian polynomials; and [L], the set A134685, the classic Lagrange inversion polynomials--all related to transformations of e.g.f.s, or Taylor series, for which [RT], [L], and [E] can each be used to give the compositional inverse and [P], the multiplicative inverse, or reciprocal.
On the other hand, as shown in formulas for A355201, [K] = [R][N] = [R][A][R] where [R] is the set A263633 (mod signs), refined Pascal polynomials; [N], the set A134264, the refined Narayana, or noncrossing partition, polynomials; and [A], the set A133437, the refined Euler characteristic polynomials of the associahedra--all related to transformations of o.g.f.s, or power series, for which [K], [A], and [N] can each be used to give the compositional inverse and [R], the multiplicative inverse, or reciprocal. This is related to three pairs of compositionally inverse series--two pairs of Laurent series and one pair of power series.

Examples

			Arranged by rows, the coefficients are
0)  1;
1) -1;
2)  1, -1;
3) -1, 2, -1;
4)  1, -3, 2, 1, -1;
5) -1, 4, -4, -2, 5, -1, -1;
6)  1, -5, 8, 2, -4, -2, -4, 5, 4, -4, -1;
7) -1, 6, -12, -3, 8, 18, -6, -14, 13, 2, -16, 14, 0, -8, -1;
8)  1, -7, 18, 3, -20, 0, -15, 8, 18, 57, 6, -54, -15, -12, 84, -30, -48, 14, 14, -8, -13, -1;
. . .
The first few partition polynomials are
RT_0 =  1,
RT_1 = -a1,
RT_2 = a1^2  - a2,
RT_3 = -a1^3 + 2 a1 a2 - a3,
Rt_4 = a1^4 - 3 a1^2 a2 + 2 a2^2 + a1 a3 - a4,
RT_5 = -a1^5 + 4 a1^3 a2 - 4 a1 a2^2 - 2 a1^2 a3 + 5 a2 a3 - a1 a4 - a5,
RT_6 = a1^6 - 5 a1^4 a2 + 8 a1^2 a2^2 + 2 a1^3 a3 - 4 a2^3 - 2 a1 a2 a3 - 4 a1^2 a4 + 5 a3^2 + 4 a2 a4 - 4 a1 a5 - a6,
RT_7 = -a1^7 + 6 a1^5 a2 - 12*a1^3 a2^2 - 3 a1^4 a3 + 8 a1 a2^3 + 18 a1^2 a2 a3 - 6 a1^3 a4 - 14 a2^2 a3 + 13 a1 a3^2 + 2 a1 a2 a4 - 16 a1^2 a5 + 14 a3 a4 + 0 a2 a5 - 8 a1 a6 - a7,
RT_8 =  a1^8 - 7 a1^6 a2 + 18 a1^4 a2^2 + 3 a1^5 a3 - 20 a1^2 a2^3 + 0 a1^3 a2 a3 - 15 a1^4 a4 + 8 a2^4 + 18 a1 a2^2 a3 + 57 a1^2 a3^2 + 6 a1^2 a2 a4 - 54 a1^3 a5 - 15 a2 a3^2 - 12 a2^2 a4 + 84 a1 a3 a4 - 30 a1 a2 a5 - 48 a1^2 a6 + 14 a4^2 + 14 a3 a5 - 8 a2 a6 - 13 a1 a7 - a8.
		

Programs

  • Mathematica
    rows[nn_] := {{1}}~Join~With[{s = 1 / D[InverseSeries[Integrate[1/(1 + Sum[c[k] x^k/k!, {k, nn}] + O[x]^(nn+1)), x]], x]}, Table[Coefficient[n! s, x^n Product[c[t], {t, p}]], {n, nn}, {p, Reverse[Sort[Sort /@ IntegerPartitions[n]]]}]];
    rows[7] // Flatten (* Andrey Zabolotskiy, Feb 17 2024 *)
  • SageMath
    B. = PolynomialRing(ZZ)
    A. = PowerSeriesRing(B)
    f = 1/(1 + a1*x + a2*x^2/factorial(2) + a3*x^3/factorial(3) + a4*x^4/factorial(4) + a5*x^5/factorial(5) + a6*x^6/factorial(6) + a7*x^7/factorial(7) + a8*x^8/factorial(8) + a9*x^9/factorial(9) + a10*x^10/factorial(10) )
    g = integrate(f)
    h = g.reverse()
    w = derivative(h,x)
    I = 1 / w
    # Added by # Peter Luschny, Feb 17 2024:
    # The list of coefficients in sparse format (i.e. without the zeros):
    for n, c in enumerate(I.list()[:10]):
        print(f"RT[{n}]", (factorial(n)*c).coefficients())

Formula

Denote this set of partition polynomials by [RT], the permutahedra polynomials of A133314 by [P], the refined Eulerian polynomials of A145271 by [E], and the Lagrange inversion polynomials of A134685 for e.g.f.s by [L]. Let the typically noncommutative product of two sets, e.g., [P][E], represent the substitution of the polynomials of [E] for the indeterminates of [P], i.e., a composition at the level of the indeterminates (see A356145 for examples). Let [I] be the substitutional identity transformation, and mark the substitutional inverse with the superscript -1. Then the following relations hold.
[RT] = [P][E] = [P][L][P] = [P]^{-1}[L][P] = [P][L][P]^{-1} since [P] is an involution, i.e., [P]^2 = [I], or [P] = [P]^{-1}, so [RT] and [L] are conjugate duals.
[RT]^{-1} = ([P][E])^{-1} = [E]^{-1}[P] = ([P][L][P])^{-1} = [P][L][P] = [RT], with [E]^{-1} = A356145, since [L] and [P] are involutions, so is [RT], i.e., [RT]^2 = [I].
RT_n(a_1,a_2,...,a_n) = D_{x=0}^n 1 / [ D_x f^{(-1)}(x)] for which D_x is the derivative w.r.t. x and the indeterminates are defined by 1 / [D_x f(x)] = 1 + a_1 x + a_2 x^2/2! + a_3 x^3/3! + ... with f(x) and f^{(-1)}(x) a compositional inverse pair of formal Taylor series, or e.g.f.s. This is the analytic equivalent of the algebraic relation [RT] = [P][E]. In words, the partition polynomials of row n (initial row is 0) is the n-th coefficient of the formal Taylor series of the reciprocal of the derivative of the compositional inverse of a function in terms of the Taylor series coefficients of the reciprocal of the derivative of that function. Note the correspondence with the analytic interpretation of [E]^{-1} of A356145, consistent with the algebraic identities above.
RT_n(a_1,a_2,...,a_n) = D_{x=0}^n f'(f^{(-1)}(x)) also, by the inverse function theorem, where the prime denotes differentiation with respect to the argument of the function.
With all a_k = (-1)^k, RT_0 = RT_1 = 1, otherwise RT_n = 0. This is determined with f(x) = e^{x}-1 and f^{(-1)}(x) = log(1+x).
With all a_k = 1, RT_0 = 1, RT_1 = -1, otherwise RT_n=0. This is determined with f(x) =1-e^{-x} and f^{(-1)}(x) = -log(1-x).
With all a_k = -1, RT_0 = 1 and RT_n = 2^(n-1) otherwise. This is determined with f(x) = (x - log(2-e^x))/2 and f^{(-1)}(x) = x - log(cosh(x)). (Careful, these are not the row sums of the absolute values of the numerical coefficients, which for the first ten polynomials are 1, 1, 2, 4, 8, 18, 40, 122, 446, and 2428.)
With a_k = k! 2^k, RT_0 = 1 and RT_n = -2*(2(n-1))! / (n-1)! = -2*n!*A000108(n-1) otherwise. This is determined with f(x) = x - x^2 and f^{(-1)}(x) = (1 - sqrt(1-4x))/2. Similar relations hold for the Fuss-Catalan sequences with f(x) = x - x^{m+1} for m > 1.

Extensions

Order of terms in rows 4-6 corrected by Andrey Zabolotskiy, Feb 17 2024

A354622 Irregular triangle read by rows: Refined 3-Narayana triangle. Coefficients of partition polynomials of A134264, a refined Narayana triangle enumerating noncrossing partitions, with all h_k other than h_0, h_3, h_6, ..., h_(3n), ... replaced by zero.

Original entry on oeis.org

1, 1, 3, 1, 9, 12, 1, 12, 6, 66, 55, 1, 15, 15, 105, 105, 455, 273, 1, 18, 18, 9, 153, 306, 51, 816, 1224, 3060, 1428, 1, 21, 21, 21, 210, 420, 210, 210, 1330, 3990, 1330, 5985, 11970, 20349, 7752, 1, 24, 24, 24, 12, 276, 552, 552, 276, 276, 2024, 6072, 3036, 6072, 506, 10626, 42504, 21252, 42504, 106260, 134596, 43263
Offset: 1

Author

Tom Copeland, Jul 08 2022

Keywords

Comments

A set of partition polynomials with these coefficients and the polynomials of A338135 can be generated by substitution of the refined Narayana, or noncrossing partition, polynomials N_n[h_1,...,h_n] of A134264 (h_0=1) into themselves--once for A338135 and twice for this entry--or by setting the indeterminates h_n of A134264 to zero except for h_0, h_3, h_6, ..., h_(3n), ... with h_0 = 1 and ultimately re-indexing. This is equivalent to recursive use of the Lagrange inversion formula on f(x) = x / h(x) = x / (1 + h_1 x + h_2 x^2 + ...) since its compositional inverse is f^{(-1)}(x) = x + N_1(h_1) x + N_2(h_1,h_2) x^2 + .... The equivalence of the two methods of generation--the substitution and the zeroing out--follows from the general theorems stated by Peter Bala in his presentation of formulas for A108767 in 2008, which stem from a fixed point-iteration formalism of a basic identity for a compositional inverse pair, x* h(f^{(-1)}(x)) = f^{(-1)}(x), where, as above, h(x) = x / f(x).
The sets of refined m-Narayana polynomials are used by Cachazo and Umbert to characterize the scattering amplitudes of a class of quantum fields (see, e.g., section 7.3).
These could also be called the refined 3-Dyck path polynomials. From the interpretation of A134264 as Dyck paths in A125181, or staircases whose steps never rise above the diagonal of a square grid (see illustrations in Weisstein), the monomials of the partition polynomial N_4 = 1 (4') + 4 (1') (3') + 2 (2')^2 + 6 (1')^2 (2') + 1 (1')^4 of A134264 have the following correspondences:
1 (4') --> 1 staircase of one step of height 4,
4 (1') (3') --> 4 staircases of 1 step of height 1 and 1 step of height 3,
2 (2')^2 --> 2 staircases of 2 steps of height 2,
6 (1')^2 (2') --> 6 staircases of 2 steps of height 1 and 1 step of height 2,
1 (1')^4 --> 1 staircase of 4 steps of height 1.
Consequently, the partition polynomials G_{3n} of this entry enumerate staircases of height 3n with steps of heights 3, 6, 9, ..., 3k, ... only.
Diverse combinatorial models of the refined m-Narayana, or m-Dyck, polynomials are inherited from those presented for the refined Narayana, or noncrossing partition, polynomials in A134264 and A125181 and in the references therein.
A127537 gives a combinatorial model (see title and Domb and Barret therein, Table 2, p. 355) that contains the coefficients of the monomials h_1^n and h_1^(n-2) h_2, i.e., A001764 and A003408.

Examples

			Triangle begins:
  1;
  1,  3;
  1,  9, 12;
  1, 12,  6, 66,  55;
  1, 15, 15, 105, 105, 455, 273;
  ...
Row 1: G_3  = g_3
row 2: G_6  = g_6 + 3 g_3^2
row 3: G_9  = g_9 + 9 g_3 g_6 + 12 g_3^3
row 4: G_12 = g_12 + 12 g_3 g_9 + 6 g_6^2 + 66 g_3^2 g_6 + 55 g_3^4
row 5: G_15 = g_15 + 15 g_3 g_12 + 15 g_6 g_9 + 105 g_3^2 g_9 + 105 g_3 g_6^2
              + 455 g_3^3 g_6 + 273 g_3^5.
.
In the notation of Abramowitz and Stegun p. 831 with indices of the partitions above divided by 3 and partition indeterminates h_n denoted (n):
R_1 = (1);
R_2 = (2) + 3 (1)^2;
R_3 = (3) + 9 (1) (2) + 12 (1)^3;
R_4 = (4) + 12 (1) (3) + 6 (2)^2 + 66 (1)^2 (2) + 55 (1)^4;
R_5 = (5) + 15 (1) (4) + 15 (2) (3) + 105 (1)^2 (3) + 105 (1) (2)^2 + 455 (1)^3(2)
          + 273 (1)^5.
		

Crossrefs

The length of row n is equal to A000041(n).
Row sums give A002293, n >= 1.

Programs

  • Mathematica
    Table[Binomial[Total[y], Length[y]-1] (Length[y]-1)! / Product[Count[y, i]!, {i, Max@@y}], {n, 8}, {y, Sort[Sort /@ IntegerPartitions[3n, n, Range[3, 3n, 3]]]}] // Flatten (* Andrey Zabolotskiy, Feb 19 2024, using Gus Wiseman's code for A134264 *)
  • PARI
    \\ Compare with A134264
    C(v)={my(n=vecsum(v), S=Set(v)); n!/((n-#v+1)!*prod(i=1, #S, my(x=S[i]); (#select(y->y==x, v))!))}
    row(n)=[C(3*Vec(p)) | p<-partitions(n)]
    { for(n=1, 7, print(row(n))) } \\ Andrew Howroyd, Feb 19 2024

Formula

Coefficients of the monomials are those of the surviving monomials of the partition polynomials of A134264 after zeroing all indeterminates other than h_0, h_3, h_6, h_9, ..., h_(3n), .... The multinomial coefficients of A125181 also apply for G_n, giving the coefficient of the monomial h_1^e_1 h_2^e_2 ... h_n^n of R_n with se := e_1 + e_2 + ... + e_n as (3n)! / ((3n-se+1)! (e_1)! (e_2)! ... (e_n)!).
1*e_1 + 2*e_2 + ... + n*e_n = n for each monomial of R_n.
The partition polynomials R_n = N_n^3 of this entry can be determined from those of A338135, N_n^2, by substituting the partition polynomials of A134264, N_n, for the indeterminate h_n = (n) of N_n^2 or by doing the same for A134264 twice. E.g., N_1(h_1) = h_1, N_2(h_1,h_2) = h_2 + h_1^2, so N_2^2(h_1,h_2) = N_2(N_1,N_2) = N_2 + N_1 = h_2 + h_1^2 + h_1^2 = h_2 + 2 h_1^2 and N_2^3(h_1,h_2) = N_2^2(N_1,N_2) = N_2 + 2 N_1^2 = h_2 + h_1^2 + 2 h_1^2 = h_2 + 3 h_1^2.
Reduces with all indeterminates h_n = (n) = t to A173020.
The coefficient of the monomial h_1^n is (3*n)! / ((3*n-n+1)! n!) = A001764(n) (see also A179848 and A235534). In general, the coefficients of these monomials of the refined (m+1)-Narayana polynomials are the Fuss-Catalan sequence associated to the row sums of the refined m-Narayana polynomials.
The coefficient of the monomial h_1^(n-2) h_2 is (3n)! / ((3n-n+2)! (n-2)!) = A003408(n-2) for n > 1.
The coefficient of the monomial h_1^(n-3) h_3 is (3n)! / ((3n-n+3)! (n-3)!) = A004321(n) for n > 2.

Extensions

Rows 6-8 added by Andrey Zabolotskiy, Feb 19 2024

A355201 Normalized Schur self-convolution expansion coefficients K_{n+1}^n / n giving the coefficients of the Laurent series (compositionally) inverse to f(z) = c_0 z + c_1 + c_2 / z + c_3 / z^2 + ... . Irregular triangle for partition polynomials, with row lengths A000041(n) - 1 except for the first two, which are both of length 1.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 3, 3, 3, 3, 1, 1, 6, 4, 2, 12, 6, 2, 4, 4, 1, 1, 10, 5, 10, 30, 10, 10, 10, 20, 10, 5, 5, 5, 1, 1, 15, 6, 30, 60, 15, 5, 60, 30, 60, 20, 15, 15, 30, 30, 15, 3, 6, 6, 6, 1, 1, 21, 7, 70, 105, 21, 35, 210, 70, 140, 35, 35, 105, 105, 105, 105, 35, 7, 42, 21, 21, 42, 42, 21, 7, 7, 7, 7, 1
Offset: 0

Author

Tom Copeland, Jun 23 2022

Keywords

Comments

For the formal Laurent series f(z) = a_0 z + a_1 + a_2 / z + a_3 / z^2 + ..., the formal compositional inverse is g(z) = b_0 z + b_1 + b_2 / z + b_3 / z^2 + ..., whose coefficients are partition polynomials whose numerical factors are those of this irregular triangle T(n,k). For the Schur coefficients defined in the formula section, -b_n = K_{n}^{n-1} / (n-1) for n > 1.
Analytic proofs of the relationship between the partition polynomials of the compositional inverse pair of Laurent series and Schur's self-convolution expansion coefficients are given in Schur (beware of a sign error) and in Airault and Ren.
Explicit examples (with a_0=1) of K_{n}^{n-1} up through n=5 are in Airault and Bouali on p. 182.
Various formulas for the b_n in terms of the associahedra (A133437), noncrossing (A134264), reciprocal (A263633), and Faber partition (A263916) polynomials are given in Copeland as well as a derivation of the explicit multi-factorial expression in the formula section and a combinatorial model.

Examples

			Triangle begins:
1) 1
2) 1
3) 1
4) 1,  1
5) 1,  1,  2,  1
6) 1,  3,  3,  3,  3,  1
7) 1,  6,  4,  2, 12,  6,  2,  4,  4,  1
8) 1, 10,  5, 10, 30, 10, 10, 10, 20, 10,  5,  5,  5,  1
  ...
The first few partition polynomials, with the monomials in the order of the partitions on p. 831 of Abramowitz and Stegun, are
b0 =    1 / a0
b1 = - a1 / a0
b2 = - a2
b3 = -(a1 a2 + a0 a3)
b4 = -(a1^2 a2 + a0 a2^2 + 2 a0 a1 a3 + a0^2 a4)
b5 = -(a1^3 a2+ 3 a0 a1 a2^2 + 3 a0 a1^2 a_3 + 3 a0^2 a2 a3 + 3 a0^2 a1 a4
      + a0^3 a_5)
b6 = -(a1^4 a2 + 6 a0 a1^2 a2^2 + 4 a0 a1^3 a3 + 2 a0^2 a2^3 + 12 a0^2 a1 a2 a3
      + 6 a0^2 a1^2 a4  + 2 a0^3 a3^2 + 4a0^3 a2 a4 + 4 a0^3 a1 a5 + a0^4 a6)
b7 = -(a1^5 a2 + 10 a_0 a1^3 a2^2 + 5 a0 a1^4 a3 + 10 a0^2 a1 a2^3
      + 30 a0^2 a1^2 a2 a3 + 10 a0^2 a1^3 a4 + 10 a0^3 a2^2 a3 + 10 a0^3 a1 a3^2
      + 20 a0^3 a1 a2 a4 + 10 a0^3 a1^2 a5 + 5 a0^4 a3 a4 + 5 a0^4 a2 a5
      + 5 a0^4 a1 a6 + a0^5 a7)
_____________________
		

Programs

  • Mathematica
    row[0] = row[1] = {1};
    row[n_] := With[{s = Expand[Coefficient[Sum[c[k] x^k, {k, 0, n}]^(n-1), x, n] / (n-1)]}, Table[Coefficient[s, Product[c[t], {t, p}]], {p, Reverse[Sort[Sort /@ IntegerPartitions[n, {n-1}, Range[0, n]]]]}]];
    Table[row[n], {n, 0, 8}] // Flatten (* Andrey Zabolotskiy, Feb 05 2023 *)

Formula

The index notations b(n), b_n, and bn are used interchangeably in this entry for indeterminates.
For n > 1, b_n(a_0,a_1,...,a_n) is a sum of monomials of the form a0^{e0} a1^{e1} a2^{e2} ... an^{en} with e1 * 1 + e2 * 2 + ... + en * n = n. When a_0 is not set to unity, e0 + e1 + ... + en = n - 1. (a1^n is not present.)
From a combinatorial argument in Copeland, the unsigned numerical coefficient of the monomial is given by (n-2)! / [(n - 1 - (e1 + e2 + ... + en))! e1! e2! ... en!].
The partition polynomials are generated by a subset of the Schur self-convolution expansion coefficients as -b_n = K_{n}^{n-1} / (n-1) =(D_{x=0}^n / n!) (a_0 + a_1 x + a_2 x^2 + ... + a_n x^n)^{n-1} / (n-1).
Row sums are the Catalan numbers A000108, ignoring the overall sign, for b_1 onwards.
Reduces to the Narayana triangle A001263 with a_0 = t and all the other indeterminates unity, ignoring the overall sign, for b_2 onwards.
Reduces to A091869 (reversed A091187) with a_1 = t and all the other indeterminates unity, ignoring the overall sign, for b_2 onwards.
b_n(c_1,...,c_n) = - Sum_{k=0}^{n-1} b_k(c_1,...,c_k) N_{n-k}(c_1,...,c_{n-k}) with c_0 = 1 and N_k(c_1,...,c_k) the noncrossing partition polynomials of A134264.
[b] = [R][N], representing the substitution of the noncrossing partition polynomials of A134264 for the indeterminates of the signed reciprocal polynomials of A263633 defined by R_n = 1 / (1 + c_1 x + c_2 x^2 + ...).
Conversely, [R][b] = [N] since the substitution transformation denoted by [R] is an involution, i.e., [R]^2 = [I], the identity substitution.
[b] = [R][A][R], a substitutional conjugation of the set of associahedra partition polynomials of A133147, or A111785, with re-indexing and (1') = 1, e.g., A_0 = 1, A_1 = -c_1, and A_2 = 2 c_1^2 - c_2.
Conversely, [A] = [R][b][R].

Extensions

Rows 8-9 added by Andrey Zabolotskiy, Feb 05 2023

A350499 Unsigned coefficients of free moment partition polynomials determining the free cumulants from the free moments of free probability theory. Irregular triangle with row lengths given by A000041, n >= 1.

Original entry on oeis.org

1, 1, 1, 1, 3, 2, 1, 4, 2, 10, 5, 1, 5, 5, 15, 15, 35, 14, 1, 6, 6, 3, 21, 42, 7, 56, 84, 126, 42, 1, 7, 7, 7, 28, 56, 28, 28, 84, 252, 84, 210, 420, 462, 132, 1, 8, 8, 8, 4, 36, 72, 72, 36, 36, 120, 360, 180, 360, 30, 330, 1320, 660, 792, 1980, 1716, 429
Offset: 1

Author

Tom Copeland, Jan 01 2022

Keywords

Comments

Coefficients are listed in Abramowitz and Stegun order (A036036).
Irregular triangular matrix of the unsigned coefficients of the free moment partition polynomials of free probability theory, for a single variable, that give the free formal cumulants given the free formal moments. This set of partition polynomials together with those of A134264 are the counterparts to the exp-log relations for the classical formal moments and cumulants associated with A036040 and A127671.
Associations with a compositional inverse pair of Laurent series, Kac-Schwarz operators of 2-D quantum theory, Virasoro / Witt / Heisenberg group actions, and KP and KdV integrable hierarchies are noted in references supplied in the MathOverflow link as well as a geometric combinatorial model based upon noncrossing partitions.

Examples

			Triangle begins:
  1;
  1, 1;
  1, 3, 2;
  1, 4, 2, 10,  5;
  1, 5, 5, 15, 15, 35, 14;
  ...
___________
The first few free cumulants in terms of the free moments are
  c_1 = m_1
  c_2 = m_2 - m_1^2
  c_3 = m_3 - 3 m_2 m_1 + 2 m_1^3
  c_4 = m_4 - 2 m_2^2 - 4m_3 m_1 + 10 m_2 m_1^2 - 5 m_1^4
  c_5 = m_5 - 5 m_2  m_3 - 5  m_4 m_1 + 15  m_2^2 m_1 + 15 m_3 m_1^2 - 35 m_2 m_1^3 + 14 m_1^5
___________
Conversely, from A134264, these free moments in terms of the free cumulants are
  m_1 = c_1
  m_2 = c_2 + c_1^2
  m_3 = c_3 + 3 c_2 c_1 + c_1^3
  m_4 = c_4 + + 2 c_2^2 + 4 c_3 c_1 + 6 c_2 c_1^2 + c_1^4
  m_5 = c_5 + 5 c_2 c_3 + 5 c_4 c_1 + 10 c_2^2 c_1 + 10 c_3 c_1^2  + 10 c_2 c_1^3 + c_1^5
___________
		

Programs

  • PARI
    mv(n)={eval(Str("'m",n))}
    Trm(m,v)={my(S=Set(v)); for(i=1, #S, my(x=S[i]); m=polcoef(m, #select(y->y==x, v), mv(x))); m}
    Q(n)={polcoef(-x/serreverse(x*(1 + sum(k=1, n, -x^k*mv(k), O(x*x^n)))), n)}
    row(n)={my(q=Q(n)); [Trm(q,Vec(v)) | v<-partitions(n)]}
    { for(n=1, 7, print(row(n))) } \\ Andrew Howroyd, Feb 01 2022
    
  • PARI
    C(v)={my(n=vecsum(v), S=Set(v)); (n+#v-2)!/(n-1)!/prod(i=1, #S, my(x=S[i]); (#select(y->y==x, v))!)}
    row(n)=[C(Vec(p)) | p<-partitions(n)]
    { for(n=1, 7, print(row(n))) } \\ Andrew Howroyd, Feb 01 2022

Formula

O.g.f.: C(x) = 1 + c_1 x + c_2 x^2 + ... = x / (x + m_1 x^2 + m_2 x^3 + m_3 x^4 + ...)^(-1) = x / M^(-1)(x), the shifted reciprocal of the compositional inverse of a power series M(x) = x + m_1 x^2 + m_2 x^3 + ..., the o.g.f. of the free moments m_n in free probability theory.
Row sums: big Schroeder numbers A006318.
Refinement of A060693 and A088617, i.e., by letting m_n = -t and removing all resulting signs, the elements of these two lower triangular matrices are generated.
The coefficients of the highest order terms in m_1^n of the free moment partition polynomials are the signed Catalan numbers A000108.
Taking the derivative with respect to the indeterminate m_1 generates the Lagrange inversion partition polynomials, with shifted indices, of A133437 and A111785 with an overall scale factor. These Lagrange inversion polynomials are the refined Euler characteristic polynomials of the associahedra. E.g.,
D_{m_1} c_5 = 5 (- m_4 + 3 m_2^2 + 6 m_3 m_1 - 21 m_2 m_1^2 + 14 m_1^4). An analogous differential formula that applies to the classical formal cumulants in relation to the permutahedra is stated in my 2012 comment in A127671.
The o.g.f. satisfies the partial differential equations D_{m_1} (x / C(x)) = -(1/3) D_x (x / C(x))^3 and D_{m_1} (C(x) / x) = D_x (x / C(x)), where D_{m_1} and D_x are the partial derivatives with respect to m_1 and x.
More generally, D_{m_n} (x / C(x)) = -(1/(n+2)) D_x (x / C(x))^{n+2), equivalent to D_{m_n} M^(-1)(x) = -(1/(n+2)) D_x (M^(-1)(x))^{n+2). Equations of this type are found in Zhou (see eqn. 44 on p. 11), characterizing the KdV hierarchy. These differential equations can be transformed into the inviscid Burgers-Hopf partial differential equation (see, e.g., A133437, A086810, A001764, A002293, A133932, A134685, and A276850).
The formal free cumulants when identified as the indeterminates of the noncrossing Lagrange inversion partition polynomials NCP_n(c_1,c_2,...,c_n) = m_n of A134264 (as in the example section) satisfy the partial differential equations D_{m_k} NCP_n(c_1, ..., c_n) = d(m_n)/dm_k = delta_{n-k}, where delta_{n} is the Kronecker delta which is zero for all integers n other than n = 0, for which it evaluates as unity. This provides a recursion method for determining the partial derivatives dc_n/dm_k from the partial derivatives dc_p/dm_k and cumulants c_p with k <= p < n. For example, dc_k/dm_j = 0 for j > k and dc_k/dm_k = 1, so dm_3/dm_2 = 0 = D_{m_2} (c_3 + 3 c_2 c_1 + c_1^3) = dc_3/dm_2 + 3 c_1 dc_2/dm_2 = dc_3/dm_2 + 3 c_1 , implying dc_3/dm_2 = -3 c_1 = -3 m_1.
T(n,k) = (n+j-2)!/((n-1)!*Product_{i>=1} s_i!), where (1*s_1 + 2*s_2 + ... = n) is the k-th partition of n and j = s_1 + s_2 + ... is the number of parts. - Andrew Howroyd, Feb 01 2022
Conjecture: free cumulants in terms of the free moments are R(n,1) for n > 0 where R(n,k) = R(n-1,k+1) - Sum_{j=1..n-1} R(j,k)*R(n-j,1) for n > 1, k > 0 with R(1,k) = m_k for k > 0. - Mikhail Kurkov, Mar 30 2025

Extensions

Terms a(19) and beyond from Andrew Howroyd, Feb 01 2022

A344678 Coefficients for normal ordering of (x + D)^n and for the unsigned, probabilist's (or Chebyshev) Hermite polynomials H_n(x+y).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 1, 3, 3, 3, 3, 1, 1, 4, 6, 6, 12, 4, 3, 6, 1, 1, 5, 10, 10, 30, 10, 15, 30, 5, 15, 10, 1, 1, 6, 15, 15, 60, 20, 45, 90, 15, 90, 60, 6, 15, 45, 15, 1, 1, 7, 21, 21, 105, 35, 105, 210, 35, 315, 210, 21, 105, 315, 105, 7, 105, 105, 21, 1
Offset: 0

Author

Tom Copeland, May 26 2021

Keywords

Comments

This irregular triangular array contains the integer coefficients of the normal ordering of the expansions of the differential operator R = (x + D)^n with D = d/dx. This operator is the raising/creation operator for the unsigned, modified Hermite polynomials H_n(x) of A099174; i.e., R H_n(x) = H_{n+1}(x). The lowering/annihilation operator L is D; i.e, L H_n(x) = D H_n(x) = n H_{n-1}(x).
Generalizing to the ladder operators for S_n(x) a general Sheffer polynomial sequence, (L + R)^n 1 = H_n(S.(x)), where the umbral composition is defined by H_n(S.(x)) = (h. + S.(x))^n = Sum_{k = 0..n} binomial(n,k) h_{n-k} S_k(x), where h_n are the Taylor series coefficients of e^{t^2/2}; i.e., e^{h.t} = e^{t^2/2}.
Another interpretation is that the n-th row contains the coefficients of the terms in the normal ordering of the 2^n permutations of two binary symbols given by (L + R)^n under the Leibniz condition LR = RL + 1 equivalent to the commutator relation [L,R] = LR - RL = 1. For example, (x + D)^2 = xx + xD + Dx + DD = x^2 + 2 xD + 1 + D^2.
As in the examples, the coefficients are ordered as those for the terms x^k D^m, ordered first from higher k to lower k and subordinately from lower m to higher m.
The number sequences associated to these polynomials are intimately related to the complete graphs K_n, which have n vertices and (n-1) edges. H_n(x) are the independence polynomials for K_n. The moments, h_n, of the H_n(x) Appell polynomials are the aerated double factorials A001147, the number of perfect matchings in the complete graph K_{n}, zero for odd n. The row lengths, A002620, give the number of maximal strokes on the complete graph K_n. The triangular numbers--the sum of two consecutive row lengths--give the number of edges of K_n. The row sums, A005425, are the number of matchings of the corona K'_n of the complete graph K_n and the complete graph K_1. K_n can be viewed as the projection onto a plane of the edges of the regular (n-1)-dimensional simplex, whose face polynomials are (x+1)^n - 1 (cf. A135278 and A074909).
The coefficients represent the combinatorics of a switchboard problem in which among n subscribers, a subscriber may talk to one of the others, someone on an outside line, or not at all--no conference calls allowed. E.g., the coefficient 12 for H_4(x+y) is the number of ways among 4 subscribers that a pair of subscribers can talk to each other while another is on an outside line and the remaining subscriber is disconnected. The coefficient 3 for the polynomial corresponds to the number of ways two pairs of subscribers can talk among themselves. This can be regarded as a dinner table problem also where a diner may exchange positions with another diner; remain seated; or get up, change plans, and sit back down in the same chair--no more than one exchange per diner.
From Peter Luschny, May 28 2021: (Start)
If we write the monomials of the row polynomials in degree-lexicographic order, we observe: The coefficient triangle appears as a series of concatenated subtriangles. The first one is Pascal's triangle A007318. Appending the rows of triangle A094305 begins in row 2. In row 4, the next triangle starts, which is A344565. This scheme seems to go on indefinitely. [This is now set out in A344911.] (End)
From Tom Copeland, May 29 2021: (Start)
Simplex model: H_n(x+y) = (h. + x + y)^n with h_n the aerated odd double factorials (h_0 = 1, h_1 = 0, h_2 = 1, h_3 = 0, h_4 = 3, h_5 = 0, h_6 = 15, ...), the number of perfect matchings for the n vertices of the (n-1)-Dimensional simplices, i.e., the hypertriangles, or hypertetrahedrons. This multinomial enumerates permutations of three families of objects--vertices labeled with either x or y, or perfect (pair) matchings of the unlabeled vertices for the n vertices of the (n-1)-D simplex. For example, (x+D)^2 = xx + xD + Dx + D^2 = x^2 + 2xD + D^2 + 1 corresponds to H_2(x+y) = (h.+ x + y)^2 = h_0 (x+y)^2 + 2 h_1 (x+y) + h_2 = x^2 + 2xy + y^2 + 1, which, in turn, corresponds to a line segment, the 1-D simplex, with both vertices labeled with x's; or one with an x, the other a y; or both vertices labeled with y's; or one unlabeled matched pair. The exponent k of x^k y^m represents the number of these vertices to which an x is assigned, and m, those with a y assigned. The remaining unlabeled vertices (a sub-simplex) form an independent edge set of (single color) colored edges, i.e., no colored edge is touching another colored edge (a perfect matching for the sub-simplex) nor the vertices assigned an x or y. Accordingly, the coefficients for k=m=0 represent the number of ways to assign a perfect matching to the simplex--zero for simplices with an odd number of vertices and the odd double factorials (1, 3, 15, 105, ...) for the simplices with an even number. For the simplices with an odd number of vertices, the coefficients for k+m=1 are the odd double factorials. (Note the polynomials are invariant upon interchange of the variables x and y.) (End)

Examples

			(x + D)^0 = 1,
(x + D)^1 = x + D,
(x + D)^2 = x^2 + 2 x D + 1 + D^2,
(x + D)^3 = x^3 + 3 x^2 D + 3 x + 3 x D^2 + 3 D + D^3,
(x + D)^4 = x^4 + 4 x^3 D + 6 x^2 + 6 x^2 D^2 + 12 x D + 4 x D^3 + 3 + 6 D^2 + D^4.
(x + D)^5 = x^5 + 5 x^4 D + 10 x^3 + 10 x^3 D^2 + 30 x^2 D + 10 x^2 D^3 + 15 x + 30 x D^2 + 5 x D^4 + 15 D + 10 D^3 + D^5
H_6(x + y) = x^6 + 6 x^5 y + 15 x^4 + 15 x^4 y^2 + 60 x^3 y + 20 x^3 y^3 + 45 x^2 + 90 x^2 y^2 + 15 x^2 y^4 + 90 x y + 60 x y^3 + 6 x y^5 + 15 + 45 y^2 + 15 y^4 + y^6
H_7(x + y) = x^7 + 7 x^6 y + 21 x^5 + 21 x^5 y^2 + 105 x^4 y + 35 x^4 y^3 + 105 x^3 + 210 x^3 y^2 + 35 x^3 y^4 + 315 x^2 y + 210 x^2 y^3 + 21 x^2 y^5 + 105 x + 315 x y^2 + 105 x y^4 + 7 x y^6 + 105 y + 105 y^3 + 21 y^5 + y^7
		

Programs

  • Mathematica
    Last /@ CoefficientRules[#, {x, y}] & /@ Table[Simplify[(-y)^n (-2)^(-n/2) HermiteH[n, (x + 1/y)/Sqrt[-2]]], {n, 0, 7}] // Flatten (* Andrey Zabolotskiy, Mar 08 2024 *)

Formula

The bivariate e.g.f. is e^{t^2/2} e^{t(x + y)} = Sum_{n >= 0} H_n(x+y) t^n/n! = e^{t H.(x+y)} = e^{t (x + H.(y))}, as described below.
The coefficient of x^k D^m is n! h_{n-k-m} / [(n-k-m)! k! m!] with 0 <= k,m <= n and (k+m) <= n with h_n, as defined in the comments, aerated A001147.
Row lengths, r(n): 1, 2, 4, 6, 9, 12, 16, 20, 25, ... A002620(n).
Row sums: A005425 = 1, 2, 5, 14, 43, 142, ... .
The recursion H_{n+1}(x+y) = (x+y) H_n(x+y) + n H_{n-1}(x+y) follows from the differential raising and lowering operations of the Hermite polynomials.
The Baker-Campbell-Hausdorff-Dynkin expansion leads to the disentangling relation e^{t (x + D)} = e^{t^2/2} e^{tx} e^{tD} from which the formula above for the coefficients may be derived via differentiation with respect to t.
The row bivariate polynomials P_n(x,y) with y a commutative analog of D, or L, have the e.g.f. e^{-xy} e^{t(x + D)} e^{xy} = e^{-xy} e^{t^2/2} e^{tx} e{tD} e^{xy} = e^{t^2/2} e^{t(x + y)} = e^{t(h. + x + y)} = e^{t (x + H.(y))} = e^{t H.(x +y)}, so P_n(x,y) = H_n(x + y) = (x + H.(y))^n, the Hermite polynomials mentioned in the comments along with the umbral composition. The row sums are H_n(2), listed in A005425. For example, P_3(x,y) = (x + H.(y))^3 = x^3 H_0(y) + 3 x^2 H_1(y) + 3 x H_2(y) + H_3(y) = H_3(x+y) = (x+y)^3 + 3(x+y) = x^3 + 3 x^2 y + 3 x + 3 x y^2 + 3 y + y^2.
Alternatively, P_n(x,y) = H_n(x+y) = (z + d/dz)^n 1 with z replaced by (x+y) after the repeated differentiations since (x + D)^n 1 = H_n(x).
With initial index 1, the lengths r(n) of the rows of nonzero coefficients are the same as those for the polynomials given by 1 + (x+y)^2 + (x+y)^4 + ... + (x+y)^n for n even and for those for (x+y)^1 + (x+y)^3 + ... + (x+y)^n for n odd since the Hermite polynomials are even or odd polynomials. Consequently, r(n)= O(n) = 1 + 2 + 4 + ... n for n odd and r(n) = E(n) = 2 + 4 + ... + n for n even, so O(n) = ((n+1)/2)^2 and E(n) = (n/2)((n/2)+1) = n(n+2)/4 = 2 T(n/2) where T(k) are the triangular numbers defined by T(k) = 0 + 1 + 2 + 3 + ... + k = A000217(k). E(n) corresponds to A002378. Additionally, r(n) + r(n+1) = 1 + 2 + 3 + ... + n+1 = T(n+1).
From Tom Copeland, May 31 2021: (Start)
e^{D^2/2} = e^{h.D}, so e^{D^2/2} x^n = e^{h. D} x^n = (h. + x)^n = H_n(x) and e^{D^2/2} (x+y)^n = e^{h. D} (x+y)^n = (h. + x + y)^n = H_n(x+y).
From the Appell Sheffer polynomial calculus, the umbral compositional inverse of the sequence H_n(x+y), i.e., the sequence HI_n(x+y) such that H_n(HI.(x+y)) = (h. + HI.(x+y))^n = (h. + hi. + x + y)^n = (x+y)^n, is determined by e^{-t^2/2} = e^{hi. t}, so hi_n = -h_n and HI_n(x+y) = (-h. + x + y)^n = (-1)^n (h. - x - y)^n = (-1)^n H_n(-(x+y)). Then H_n(-H.(-(x+y))) = (x+y)^n.
In addition, HI_n(x) = (x - D) HI_{n-1}(x) = (x - D)^n 1 = e^{-D^2/2} x^n = e^{hi. D} x^n = e^{-h. D} x^n with the e.g.f. e^{-t^2/2} e^{xt}.
The umbral compositional inversion property follows from x^n = e^{-D^2/2} e^{D^2/2} x^n = e^{-D^2/2} H_n(x) = H_n(HI.(x)) = e^{D^2/2} e^{-D^2/2} x^n = HI_n(H.(x)). (End)
The umbral relations above reveal that H_n(x+y) = (h. + x + y)^n = Sum_{k = 0..n} binomial(n,k) h_k (x+y)^{n-k}, which gives, e.g., for n = 3, H_3(x+y) = h_0 * (x+y)^3 + 3 h_1 * (x+y)^2 + 3 h_2 * (x+y) + h_3 = (x+y)^3 + 3 (x+y), the n-th through 0th rows of the Pascal matrix embedded within the n-th row of the Pascal matrix modulated by h_k. - Tom Copeland, Jun 01 2021
Varvak gives the coefficients of x^(n-m-k) D^{m-k} as n! / ( 2^k k! (n-k-m)! (m-k)! ), referring to them as the Weyl binomial coefficients, and derives them from rook numbers on Ferrers boards. (No mention of Hermite polynomials nor matchings on simplices are made.) Another combinatorial model and equivalent formula are presented in Blasiak and Flajolet (p. 16). References to much earlier work are given in both papers. - Tom Copeland, Jun 03 2021

A338135 Irregular triangle read by rows: Row p gives number of non-overlapping clusters of 2q-plets joining 2p points on a circle, i.e., number of noncrossing partitions from A134264 with h_k for k odd replaced by zero.

Original entry on oeis.org

1, 1, 2, 1, 6, 5, 1, 8, 4, 28, 14, 1, 10, 10, 45, 45, 120, 42, 1, 12, 12, 6, 66, 132, 22, 220, 330, 495, 132, 1, 14, 14, 14, 91, 182, 91, 91, 364, 1092, 364, 1001, 2002, 2002, 429
Offset: 1

Author

Tom Copeland, Oct 11 2020

Keywords

Comments

This combinatorial problem arises in relating connected and disconnected Green functions associated to a "zero-dimensional" quantum field theory presented by Brezin et al. in "Planar Diagrams" via Eqn. 31 on p. 42.
Appears to be a refinement of A120986 and A108767 in that summing the coefficients of partitions with the same sum of exponents gives the rows or reverse rows of the two entries; for example, row 4 here becomes x + 8 xx + 4 x^2 + 28 x^2x + 14 x^4 = x + 12 x^2 + 28 x^3 + 14 x^4, which is row 4 of A108767. In short, replace each g_k or (k) by x in the formula section here to obtain the coarser entry or its reverse from this refined entry, apparently.
This also gives the relationship between moments and free cumulants in free probability theory restricted to an even number of noncrossing partitions as given by restricting the similar enumeration formuia on p. 34 of Novak and LaCroix to b_{2n} = G_{2n} and K_{2q} = g_{2q}. This is consistent with setting h_k to zero for odd k in A134264, e.g., doing so for the coefficients of t^7 for g(t) there gives G_6 here.
A125181 is another version of A134264, providing interpretations in terms of Dyck paths and trees.

Examples

			row 1: G_2  = g_2
row 2: G_4  = g_4  +  2 g_2^2
row 3: G_6  = g_6  +  6 g_2 g_4 +  5 g_2^3
row 4: G_8  = g_8  +  8 g_2 g_6 +  4 g_4^2   +  28 g_2^2 g_4 + 14 g_2^4
row 5: G_10 = g_10 + 10 g_2 g_8 + 10 g_4 g_6 +  45 g_2^2 g_6 + 45 g_2 g_4^2
              + 120 g_2^3 g_4  + 42 g_2^5
_____________
In the notation of Abramowitz and Stegun p. 831 with indices of the partitions above divided by 2;
R_1  = (1)
R_2  = (2)  +  2 (1)^2
R_3  = (3)  +  6 (1) (2) +  5 (1)^3
R_4  = (4)  +  8 (1) (3) +  4 (2)^2   +  28 (1)^2 (2) + 14 (1)^4
R_5  = (5)  + 10 (1) (4) + 10 (2) (3) +  45 (1)^2 (3) + 45 (1) (2)^2
        + 120 (1)^3 (2) + 42 (1)^5
______________
		

Crossrefs

Programs

  • Mathematica
    Table[(2 n)!/((2 n + 1 - Length@p)! Product[r!, {r, Last /@ Tally[p]}]), {n, 5}, {p, Sort[Sort /@ IntegerPartitions[n]]}] // Flatten (* Andrey Zabolotskiy, Mar 07 2024 *)

Formula

Under the constraint 2p = Sum_{q} 2q r_q, then G_{2p} = Sum_{r_q >= 0} [(2p)! / (2p + 1 - Sum_{q} r_q)! ] (g_2^r_1 /(r_1)!) (g_4^r_2 / (r_2)!) ... (g_{2q}^r_q / (r_q)!) where g_{2k} are the connected Green functions.
With R_p = G_{2p} and N_q = g_{2q}, then R_p = Sum_{r_q >= 0} [(2p)! / (2p + 1 - Sum_{q} r_q)! ] (N_1^r_1 /(r_1)!) (N_2^r_2 / (r_2)!) ... (N_{q}^r_q / (r_q)!) where N_q are the partitions in Abramowitz and Stegun on p. 831.
Coefficients of the final terms g_{2}^p = (1)^p are the Catalan numbers A000108.

Extensions

Rows 6-7 from Andrey Zabolotskiy, Mar 07 2024

A298673 Inverse matrix of A135494.

Original entry on oeis.org

1, 1, 1, 4, 3, 1, 26, 19, 6, 1, 236, 170, 55, 10, 1, 2752, 1966, 645, 125, 15, 1, 39208, 27860, 9226, 1855, 245, 21, 1, 660032, 467244, 155764, 32081, 4480, 434, 28, 1, 12818912, 9049584, 3031876, 635124, 92001, 9576, 714, 36, 1, 282137824, 198754016, 66845340, 14180440, 2108085, 230097, 18690, 1110, 45, 1
Offset: 1

Author

Tom Copeland, Jan 24 2018

Keywords

Comments

Since this is the inverse matrix of A135494 with row polynomials q_n(t), first introduced in that entry by R. J. Mathar, and the row polynomials p_n(t) of this entry are a binomial Sheffer polynomial sequence, the row polynomials of the inverse pair are umbral compositional inverses, i.e., p_n(q.(t)) = q_n(p.(t)) = t^n. For example, p_3(q.(t)) = 4q_1(t) + 3q_2(t) + q_3(t) = 4t + 3(-t + t^2) + (-t -3t^2 +t^3) = t^3. In addition, both sequences possess the umbral convolution property (p.x) + p.(y))^n = p_n(x+y) with p_0(t) = 1.
This is the inverse of the Bell matrix generated by A153881; for the definition of the Bell matrix see the link. - Peter Luschny, Jan 26 2018

Examples

			Matrix begins as
     1;
     1;    1;
     4,    3,    1;
    26,   19,    6,    1;
   236,  170,   55,   10,    1;
  2752, 1966,  645,  125,   15,    1;
		

Crossrefs

Programs

  • Maple
    # The function BellMatrix is defined in A264428. Adds (1,0,0,0, ..) as column 0.
    BellMatrix(n -> `if`(n=0, 1, -1), 9): MatrixInverse(%); # Peter Luschny, Jan 26 2018
  • Mathematica
    BellMatrix[f_, len_] := With[{t = Array[f, len, 0]}, Table[BellY[n, k, t], {n, 0, len - 1}, {k, 0, len - 1}]];
    B = BellMatrix[Function[n, If[n == 0, 1, -1]], rows = 12] // Inverse;
    Table[B[[n, k]], {n, 2, rows}, {k, 2, n}] // Flatten (* Jean-François Alcover, Jun 28 2018, after Peter Luschny *)

Formula

E.g.f.: e^[p.(t)x] = e^[t*h(x)] = exp[t*[(x-1)/2 + T{ (1/2) * exp[(x-1)/2] }], where T is the tree function of A000169 related to the Lambert function. h(x) = sum(j=1,...) A000311(j) * x^j / j! = exp[xp.'(0)], so the first column of this entry's matrix is A000311(n) for n > 0 and the second column of the full matrix for p_n(t) to n >= 0. The compositional inverse of h(x) is h^(-1)(x) = 1 + 2x - e^x.
The lowering operator is L = h^(-1)(D) = 1 + 2D - e^D with D = d/dt, i.e., L p_n(t) = n * p_(n-1)(t). For example, L p_3(t) = (D - D^2! - D^3/3! - ...) (4t + 6t^ + t^3) = 3 (t + t^2) = 3 p_2(t).
The raising operator is R = t * 1/[d[h^(-1)(D)]/dD] = t * 1/[2 - e^D)] = t (1 + D + 3D^2/2! + 13D^3/3! + ...). The coefficients of R are A000670. For example, R p_2(t) = t (1 + D + 3D^2/2! + ...) (t + t^2) = 4t + 3t^2 + t^3 = p_3(t).
The row sums are A006351, or essentially 2*A000311.
Conjectures from Mikhail Kurkov, Mar 01 2025: (Start)
T(n,k) = Sum_{j=0..n-k} binomial(n+j-1, k-1)*A269939(n-k, j) for 1 <= k <= n.
T(n,k) = A(n-1,k,0) for n > 0, k > 0 where A(n,k,q) = A(n-1,k,q+1) + 2*(q+1)!*Sum_{j=0..q} A(n-1,k,j)/j! for n >= 0, k > 0, q >= 0 with A(0,k,q) = Stirling1(q+1,k) for k > 0, q >= 0 (see A379458). In other words, T(n,k) = Sum_{j=0}^{n-1} A379460(n-j-1,j)*Stirling1(j+1,k) for n > 0, k > 0.
Recursion for the n-th row (independently of other rows): T(n,k) = 1/(n-k)*Sum_{j=2..n-k+1} b(j-1)*binomial(-k,j)*T(n,k+j-1)*(-1)^j for 1 <= k < n with T(n,n) = 1 where b(n) = 1 + 4*Sum_{i=1..n} A135148(i).
Recursion for the k-th column (independently of other columns): T(n,k) = 1/(n-k)*Sum_{j=2..n-k+1} c(j-1)*binomial(n,j)*T(n-j+1,k) for 1 <= k < n with T(n,n) = 1 where c(n) = A000311(n+1) + (n-1)*A000311(n). (End)

A295380 Number of canonical forms for separation coordinates on hyperspheres S_n, ordered by increasing number of independent continuous parameters.

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 3, 8, 5, 1, 6, 20, 22, 8, 1, 11, 49, 73, 46, 11, 1, 23, 119, 233, 206, 87, 15, 1, 46, 288, 689, 807, 485, 147, 19, 1, 98, 696, 1988, 2891, 2320, 1021, 236, 24, 1, 207, 1681, 5561, 9737, 9800, 5795, 1960, 356, 29, 1, 451, 4062, 15322, 31350, 38216, 28586, 13088, 3525, 520, 35, 1, 983, 9821, 41558, 97552, 139901, 127465, 74280, 27224, 5989, 730, 41, 1
Offset: 1

Author

Tom Copeland, Nov 21 2017

Keywords

Comments

Table 1 of the Schöbel and Veselov paper with initial 1 added. Reverse of Table 2 of the Devadoss and Read paper.
Apparently A032132 contains the row sums.
From Petros Hadjicostas, Jan 28 2018: (Start)
In this triangle, which is read by rows, for 0 <= k <= n-1 and n>=1, let T(n,k) be the number of inequivalent canonical forms for separation coordinates of the hypersphere S^n with k independent continuous parameters. It is the mirror image of sequence A232206, that is, T(n, k) = A232206(n+1, n-k) for 0 <= k <= n-1 and n>=1. (Triangular array A232206(N, K) is defined for N >= 2 and 1 <= K <= N-1.)
If B(x,y) = Sum_{n,k>=0} T(n,k)*x^n*y^k (with T(0,0) = 1, T(0,k) = 0 for k>=1, and T(n,k) = 0 for 1 <= n <= k), then B(x,y) = 1 + (x/2)*(B(x,y)^2/(1-x*y*B(x,y)) + (1 + x*y*B(x,y))*B(x^2,y^2)/(1-x^2*y^2*B(x^2,y^2))). This can be derived from the bivariate g.f. of A232206. See the comments for that sequence.
Let S(n) := Sum_{k>=0} T(n,k). The g.f. of S(n) is B(x, y=1). If we let y=1 in the above functional equation, we get x*B(x,1) = x + (1/2)*((x*B(x,1))^2/(1-x*B(x,1)) + (1 + x*B(x,1))*x^2*B(x^2,1)/(1-x^2*B(x^2,1))). After some algebra, we get 2*x*B(x,1) = x + (1/2)(x*B(x,1)/(1-x*B(x,1)) + (x*B(x,1) + x^2*B(x^2,1))/(1-x^2*B(x,1))), i.e., 2*x*B(x,1) = x + BIK(x*B(x,1)), where we have the "BIK" (reversible, indistinct, unlabeled) transform of C. G. Bower. This proves that S(n) = A032132(n+1) for n>=0, which is Copeland's claim above.
Note that for the second column we have T(n,k=2) = A048739(n-2) for 2 <= n < = 10, but T(11,2) = 4062 <> 4059 = A048739(9). In any case, they have different g.f.s (see the formula section below).
(End)

Examples

			From _Petros Hadjicostas_, Jan 27 2018: (Start)
Triangle T(n,k) begins:
n\k      0     1     2     3     4     5     6    7   8  9
----------------------------------------------------------------
(S^1)    1,
(S^2)    1,    1,
(S^3)    2,    3,    1,
(S^4)    3,    8,    5,    1,
(S^5)    6,   20,   22,    8,    1,
(S^6)   11,   49,   73,   46,   11,    1,
(S^7)   23,  119,  233,  206,   87,   15,    1,
(S^8)   46,  288,  689,  807,  485,  147,   19,   1,
(S^9)   98,  696, 1988, 2891, 2320, 1021,  236,  24,  1,
(S^10) 207, 1681, 5561, 9737, 9800, 5795, 1960, 356, 29, 1,
...
(End)
		

Crossrefs

Formula

From Petros Hadjicostas, Jan 28 2018: (Start)
G.f.: If B(x,y) = Sum_{n,k>=0} T(n,k)*x^n*y^k (with T(0,0) = 1, T(0,k) = 0 for k>=1, and T(n,k) = 0 for 1 <= n <= k), then B(x,y) = 1 + (x/2)*(B(x,y)^2/(1-x*y*B(x,y)) + (1 + x*y*B(x,y))*B(x^2,y^2)/(1-x^2*y^2*B(x^2,y^2))).
If c(N,K) = A232206(N,K) and C(x,y) = Sum_{N,K>=0} c(N,K)*x^N*y^K (with c(1,0) = 1 and c(N,K) = 0 for 0 <= N <= K), then C(x,y) = x*B(x*y, 1/y) and B(x,y) = C(x*y, 1/y)/(x*y).
Setting y=0 in the above functional equation, we get x*B(x,0) = x + (1/2)*((x*B(x,0))^2 + x^2*B(x^2,0)), which is the functional equation for the g.f. of the first column. This proves that T(n,k=0) = A001190(n+1) for n>=0 (assuming T(0,0) = 1).
The g.f. of the second column is B_1(x,0) = Sum_{n>=0} T(n,2)*x^n = lim_{y->0} (B(x,y)-B(x,0))/y, where B(x,0) = 1 + x + x^2 + ... is the g.f. of the first column. We get B_1(x,0) = x*B(x,0)*(B(x,0) - 1)/(1 - x*B(x,0)).
(End)

Extensions

Typo for T(11,3)=15322 corrected by Petros Hadjicostas, Jan 28 2018