cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A145271 Coefficients for expansion of (g(x)d/dx)^n g(x); refined Eulerian numbers for calculating compositional inverse of h(x) = (d/dx)^(-1) 1/g(x); iterated derivatives as infinitesimal generators of flows.

Original entry on oeis.org

1, 1, 1, 1, 1, 4, 1, 1, 11, 4, 7, 1, 1, 26, 34, 32, 15, 11, 1, 1, 57, 180, 122, 34, 192, 76, 15, 26, 16, 1, 1, 120, 768, 423, 496, 1494, 426, 294, 267, 474, 156, 56, 42, 22, 1, 1, 247, 2904, 1389, 4288, 9204, 2127, 496, 5946, 2829, 5142, 1206, 855, 768, 1344, 1038, 288, 56, 98, 64, 29, 1
Offset: 0

Views

Author

Tom Copeland, Oct 06 2008

Keywords

Comments

For more detail, including connections to Legendre transformations, rooted trees, A139605, A139002 and A074060, see Mathemagical Forests p. 9.
For connections to the h-polynomials associated to the refined f-polynomials of permutohedra see my comments in A008292 and A049019.
From Tom Copeland, Oct 14 2011: (Start)
Given analytic functions F(x) and FI(x) such that F(FI(x))=FI(F(x))=x about 0, i.e., they are compositional inverses of each other, then, with g(x) = 1/dFI(x)/dx, a flow function W(s,x) can be defined with the following relations:
W(s,x) = exp(s g(x)d/dx)x = F(s+FI(x)) ,
W(s,0) = F(s) ,
W(0,x) = x ,
dW(0,x)/ds = g(x) = F'[FI(x)] , implying
dW(0,F(x))/ds = g(F(x)) = F'(x) , and
W(s,W(r,x)) = F(s+FI(F(r+FI(x)))) = F(s+r+FI(x)) = W(s+r,x) . (See MF link below.) (End)
dW(s,x)/ds - g(x)dW(s,x)/dx = 0, so (1,-g(x)) are the components of a vector orthogonal to the gradient of W and, therefore, tangent to the contour of W, at (s,x) . - Tom Copeland, Oct 26 2011
Though A139605 contains A145271, the op. of A145271 contains that of A139605 in the sense that exp(s g(x)d/dx) w(x) = w(F(s+FI(x))) = exp((exp(s g(x)d/dx)x)d/du)w(u) evaluated at u=0. This is reflected in the fact that the forest of rooted trees assoc. to (g(x)d/dx)^n, FOR_n, can be generated by removing the single trunk of the planted rooted trees of FOR_(n+1). - Tom Copeland, Nov 29 2011
Related to formal group laws for elliptic curves (see Hoffman). - Tom Copeland, Feb 24 2012
The functional equation W(s,x) = F(s+FI(x)), or a restriction of it, is sometimes called the Abel equation or Abel's functional equation (see Houzel and Wikipedia) and is related to Schröder's functional equation and Koenigs functions for compositional iterates (Alexander, Goryainov and Kudryavtseva). - Tom Copeland, Apr 04 2012
g(W(s,x)) = F'(s + FI(x)) = dW(s,x)/ds = g(x) dW(s,x)/dx, connecting the operators here to presentations of the Koenigs / Königs function and Loewner / Löwner evolution equations of the Contreras et al. papers. - Tom Copeland, Jun 03 2018
The autonomous differential equation above also appears with a change in variable of the form x = log(u) in the renormalization group equation, or Beta function. See Wikipedia, Zinn-Justin equations 2.10 and 3.11, and Krajewski and Martinetti equation 21. - Tom Copeland, Jul 23 2020
A variant of these partition polynomials appears on p. 83 of Petreolle et al. with the indeterminates e_n there related to those given in the examples below by e_n = n!*(n'). The coefficients are interpreted as enumerating certain types of trees. See also A190015. - Tom Copeland, Oct 03 2022

Examples

			From _Tom Copeland_, Sep 19 2014: (Start)
Let h(x) = log((1+a*x)/(1+b*x))/(a-b); then, g(x) = 1/(dh(x)/dx) = (1+ax)(1+bx), so (0')=1, (1')=a+b, (2')=2ab, evaluated at x=0, and higher order derivatives of g(x) vanish. Therefore, evaluated at x=0,
R^0 g(x) =  1
R^1 g(x) =  a+b
R^2 g(x) = (a+b)^2 + 2ab = a^2 + 4 ab + b^2
R^3 g(x) = (a+b)^3 + 4*(a+b)*2ab = a^3 + 11 a^2*b + 11 ab^2 + b^3
R^4 g(x) = (a+b)^4 + 11*(a+b)^2*2ab + 4*(2ab)^2
         =  a^4 + 26 a^3*b + 66 a^2*b^2 + 26 ab^3 + b^4,
etc., and these bivariate Eulerian polynomials (A008292) are the first few coefficients of h^(-1)(x) = (e^(ax) - e^(bx))/(a*e^(bx) - b*e^(ax)), the inverse of h(x). (End)
Triangle starts:
  1;
  1;
  1,   1;
  1,   4,    1;
  1,  11,    4,    7,    1;
  1,  26,   34,   32,   15,   11,    1;
  1,  57,  180,  122,   34,  192,   76,  15,   26,   16,    1;
  1, 120,  768,  423,  496, 1494,  426, 294,  267,  474,  156,   56,  42,  22,    1;
  1, 247, 2904, 1389, 4288, 9204, 2127, 496, 5946, 2829, 5142, 1206, 855, 768, 1344, 1038, 288, 56, 98, 64, 29, 1;
		

References

  • D. S. Alexander, A History of Complex Dynamics: From Schröder to Fatou to Julia, Friedrich Vieweg & Sohn, 1994.
  • T. Mansour and M. Schork, Commutation Relations, Normal Ordering, and Stirling Numbers, Chapman and Hall/CRC, 2015.

Crossrefs

Cf. (A133437, A086810, A181289) = (LIF, reduced LIF, associated g(x)), where LIF is a Lagrange inversion formula. Similarly for (A134264, A001263, A119900), (A134685, A134991, A019538), (A133932, A111999, A007318).
Second column is A000295, subdiagonal is A000124, row sums are A000142, row lengths are A000041. - Peter Luschny, Jul 21 2016

Programs

  • Maple
    with(LinearAlgebra): with(ListTools):
    A145271_row := proc(n) local b, M, V, U, G, R, T;
    if n < 2 then return 1 fi;
    b := (n,k) -> `if`(k=1 or k>n+1,0,binomial(n-1,k-2)*g[n-k+1]);
    M := n -> Matrix(n, b):
    V := n -> Vector[row]([1, seq(0,i=2..n)]):
    U := n -> VectorMatrixMultiply(V(n), M(n)^(n-1)):
    G := n -> Vector([seq(g[i], i=0..n-1)]);
    R := n -> VectorMatrixMultiply(U(n), G(n)):
    T := Reverse([op(sort(expand(R(n+1))))]);
    seq(subs({seq(g[i]=1, i=0..n)},T[j]),j=1..nops(T)) end:
    for n from 0 to 9 do A145271_row(n) od; # Peter Luschny, Jul 20 2016

Formula

Let R = g(x)d/dx; then
R^0 g(x) = 1 (0')^1
R^1 g(x) = 1 (0')^1 (1')^1
R^2 g(x) = 1 (0')^1 (1')^2 + 1 (0')^2 (2')^1
R^3 g(x) = 1 (0')^1 (1')^3 + 4 (0')^2 (1')^1 (2')^1 + 1 (0')^3 (3')^1
R^4 g(x) = 1 (0')^1 (1')^4 + 11 (0')^2 (1')^2 (2')^1 + 4 (0')^3 (2')^2 + 7 (0')^3 (1')^1 (3')^1 + 1 (0')^4 (4')^1
R^5 g(x) = 1 (0') (1')^5 + 26 (0')^2 (1')^3 (2') + (0')^3 [34 (1') (2')^2 + 32 (1')^2 (3')] + (0')^4 [ 15 (2') (3') + 11 (1') (4')] + (0')^5 (5')
R^6 g(x) = 1 (0') (1')^6 + 57 (0')^2 (1')^4 (2') + (0')^3 [180 (1')^2 (2')^2 + 122 (1')^3 (3')] + (0')^4 [ 34 (2')^3 + 192 (1') (2') (3') + 76 (1')^2 (4')] + (0')^5 [15 (3')^2 + 26 (2') (4') + 16 (1') (5')] + (0')^6 (6')
where (j')^k = ((d/dx)^j g(x))^k. And R^(n-1) g(x) evaluated at x=0 is the n-th Taylor series coefficient of the compositional inverse of h(x) = (d/dx)^(-1) 1/g(x), with the integral from 0 to x.
The partitions are in reverse order to those in Abramowitz and Stegun p. 831. Summing over coefficients with like powers of (0') gives A008292.
Confer A190015 for another way to compute numbers for the array for each partition. - Tom Copeland, Oct 17 2014
Equivalent matrix computation: Multiply the n-th diagonal (with n=0 the main diagonal) of the lower triangular Pascal matrix by g_n = (d/dx)^n g(x) to obtain the matrix VP with VP(n,k) = binomial(n,k) g_(n-k). Then R^n g(x) = (1, 0, 0, 0, ...) [VP * S]^n (g_0, g_1, g_2, ...)^T, where S is the shift matrix A129185, representing differentiation in the divided powers basis x^n/n!. - Tom Copeland, Feb 10 2016 (An evaluation removed by author on Jul 19 2016. Cf. A139605 and A134685.)
Also, R^n g(x) = (1, 0, 0, 0, ...) [VP * S]^(n+1) (0, 1, 0, ...)^T in agreement with A139605. - Tom Copeland, Jul 21 2016
A recursion relation for computing each partition polynomial of this entry from the lower order polynomials and the coefficients of the cycle index polynomials of A036039 is presented in the blog entry "Formal group laws and binomial Sheffer sequences". - Tom Copeland, Feb 06 2018
A formula for computing the polynomials of each row of this matrix is presented as T_{n,1} on p. 196 of the Ihara reference in A139605. - Tom Copeland, Mar 25 2020
Indeterminate substitutions as illustrated in A356145 lead to [E] = [L][P] = [P][E]^(-1)[P] = [P][RT] and [E]^(-1) = [P][L] = [P][E][P] = [RT][P], where [E] contains the refined Eulerian partition polynomials of this entry; [E]^(-1), A356145, the inverse set to [E]; [P], the permutahedra polynomials of A133314; [L], the classic Lagrange inversion polynomials of A134685; and [RT], the reciprocal tangent polynomials of A356144. Since [L]^2 = [P]^2 = [RT]^2 = [I], the substitutional identity, [L] = [E][P] = [P][E]^(-1) = [RT][P], [RT] = [E]^(-1)[P] = [P][L][P] = [P][E], and [P] = [L][E] = [E][RT] = [E]^(-1)[L] = [RT][E]^(-1). - Tom Copeland, Oct 05 2022

Extensions

Title amplified by Tom Copeland, Mar 17 2014
R^5 and R^6 formulas and terms a(19)-a(29) added by Tom Copeland, Jul 11 2016
More terms from Peter Luschny, Jul 20 2016

A139605 Weights for expansion of iterated derivatives, powers of a Lie derivative, or infinitesimal generator in vector form, (f(x)D_x)^n; coefficients of A-polynomials of Comtet; Scherk partition polynomials.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 1, 4, 1, 4, 7, 6, 1, 1, 7, 4, 11, 1, 5, 30, 15, 10, 25, 10, 1, 1, 11, 15, 32, 34, 26, 1, 6, 57, 34, 146, 31, 15, 120, 90, 20, 65, 15, 1, 1, 16, 26, 15, 76, 192, 34, 122, 180, 57, 1, 7, 98, 140, 406, 462, 588, 63, 21, 252, 154, 896, 301
Offset: 1

Views

Author

Tom Copeland, Jun 12 2008

Keywords

Comments

This entry and the references differ slightly among themselves in the order of coefficients for higher order terms. Table on p. 167 of Comtet has at least one index error.
Let F[FI(x)] = FI[F(x)] = x (i.e., F and FI are a compositional inverse pair) about x=0 with F(0)=FI(0)=0. Define f(x) = 1/[dFI(x)/dx], then for w(x) analytic about the origin, exp[t f(x)d/dx] w(x) = w{F[t+FI(x)]} = q(t,x) with q{t,F[s+FI(x)]} = q(t+s,x). See A145271 for w(x)=x and note that A145271 is embedded in A139605. E.g.f. of the binomial Sheffer sequence associated to F(x) is exp[x f(z)d/dz] exp(t*z)= exp{t*F[x+FI(z)]} evaluated at z=0. - Tom Copeland, Oct 19 2011
dq(t,x)/dt - f(x)dq(t,x)/dx = 0, so (1,-f(x)) gives the components of a vector orthogonal to the gradient of q and therefore tangent to the contour of q at (t,x). - Tom Copeland, Oct 26 2011
The formula exp[t f(x)d/dx] w(x)= w{F[t+FI(x)]} above is implicit in the chain rule formulas on pages 10 and 12 of Mathemagical Forests. Another derivation is alluded to in the Dattoli reference in A080635 (repeated below). - Tom Copeland, Nov 28 2011
Let f(x) and g(x) be two infinitely differential functions. Denote f_0 = f(x), f_1 = df_0/dx, f_2 = df_1/dx, and so on. Same with g_0 = g(x). Define the linear operator L(u(x)) = g(x) * du(x)/dx. Denote F_1 = L(f(x)), F_2 = L(F_1), and so on. When n>0, F_n is a linear combination of f_1, ..., f_n where each f_k is multiplied by a homogeneous polynomial (P(n,k)) of degree n in g_0, ..., g_{n-1}. The triangle of the sum of the coefficients of P(n,k) is A130534. - Michael Somos, Mar 23 2014
Triangle with row n length A000070(n+1) and row n consists of the coefficients: P(n,1), ..., P(n,n). The order of coefficients in P(n,k) is Abramowitz and Stegun order for partitions of n-k with parts g_1, ..., g_{n-k}. - Michael Somos, Mar 23 2014
A130534(n,k) gives the number of rooted trees with (k+1) trunks that are associated with D^(k+1) in the forest of "naturally grown" rooted trees with (n+2) nodes, or vertices, that are associated with R^(n+1) in the example below. Cf. MF link. - Tom Copeland, Mar 23 2014
These partition polynomials appeared in 1823 in a dissertation by Heinrich Scherk. See p. 76 of Blasiak and Flajolet. - Tom Copeland, Jul 14 2021
Schröder made use of iterated derivatives, or iterated infinitesimal generators (IGs), ((1/f') D)^n in his investigations of functional iteration, or iterated functional composition, related to extensions of Newton's method of finding zeros of an equation. He constructs the series, in terms of the IGs, for finv[t+f(z)] evaluated at t = -f(z), giving z_1 = finv(0) although he doesn't present his analysis this way. - Tom Copeland, Jul 19 2021

Examples

			Let R = f(x)d/dx = f(x)D and (j,k) = [(d/dx)^j f(x)]^k ; then
R^0  = 1.
R^1  = (0,1)D.
R^2  = (0,1)(1,1)D + (0,2)D^2.
R^3  = [(0,1)(1,2) + (0,2)(2,1)]D + 3 (0,2)(1,1)D^2 + (0,3)D^3.
R^4  = [(0,1)(1,3) + 4 (0,2)(1,1)(2,1) + (0,3)(3,1)]D +
       [7 (0,2)(1,2) + 4 (0,3)(2,1)]D^2 + 6 (0,3)(1,1)D^3 + (0,4)D^4. - _Tom Copeland_, Jun 12 2008
R^5  = [(0,1)(1,4) + 11 (0,2)(1,2)(2,2) + 4 (0,3)(2,2) + (0,4)(4,1) + 7 (0,3)(1,1)(3,1)]D + [15 (0,2)(1,3) + 30 (0,3)(1,1)(2,1) + 5 (0,4)(1,3)]D^2 + [25 (0,3)(1,2) + 10 (0,4)(2,1) + 25 (0,3)(1,2)]D^3  + 10 (0,4)(1,1)D^4 + (0,5)D^5. - _Tom Copeland_, Jul 17 2016
R^6  = [(0,1)(1,5) + 26 (0,2)(1,3)(2,1) + 34 (0,3)(1,1)(2,2) + 32 (0,3)(1,2)(3,1) + 11 (0,4)(1,1)(4,1) + 15 (0,4)(2,1)(3,1) + (0,5)(1,5)]D + [31 (0,2)(1,4) + 146 (0,3)(1,2)(2,1) + 57 (0,4)(1,1)(3,1) + 34 (0,4)(2,2) + 6 (0,5)(4,1)]D^2 + [90 (0,3)(1,3) + 120 (0,4)(1,1)(2,1) + 15 (0,5)(3,1)]D^3 + [65 (0,4)(1,2) + 20 (0,5)(2,1)]D^4 + 15 (0,5)(1,1)D^5 + (0,6)D^6. - _Tom Copeland_, Jul 17 2016
------------
F_1 = (1*g_0) * f_1, F_2 = (1*g_0*g_1) * f_1 + (1*g_0^2) * f_2, F_3 = (1*g_0*g_1^2 + 1*g_0^2*g_2) * f_1 + (3*g_0^2*g_1) * f_2 + (1*g_0^3) * f_3. - _Michael Somos_, Mar 23 2014
P(4,2) = 4*g0^3*g2 + 7*g0^2*g1^2. P(5,2) = 5*g0^4*g3 + 30*g0^3*g1*g2 + 15*g0^2*g1^3. - _Michael Somos_, Mar 23 2014
1
1 , 1
1 1 , 3 , 1
1 4 1 , 4 7 , 6 , 1
1 7 4 11 1, 5 30 15 , 10 25 , 10 , 1
1 11 15 32 34 26 1 , 6 57 34 146 31 , 15 120 90 , 20 65 , 15 , 1
		

References

  • F. Bergeron, G. Labelle and P. Leroux, Combinatorial Species and Tree-like Structures, (1997), Cambridge University Press, p. 386.
  • H. Davis, The Theory of Linear Operators, (1936), The Principia Press, p. 13.
  • T. Mansour and M. Schork, Commutation Relations, Normal Ordering, and Stirling Numbers, Chapman and Hall/CRC, 2015.

Crossrefs

Cf. A000070 (number of distinct terms for each order).
Cf. A130534 (sum of numerical coefficients of the derivatives).

Programs

  • Mathematica
    row[n_] := With[{pn = CoefficientRules[Nest[g[x] D[#, x] &, f[x], n], Derivative[#][f][x] & /@ Range[n]][[;; , 2]] /. {Derivative[k_][g][x] :> h[k], g[x] -> 1}}, Table[Coefficient[pn[[k]], Product[h[x], {x, p}]], {k, n - 1}, {p, Sort[Sort /@ IntegerPartitions[n - k]]}]~Join~{{1}}];
    Table[row[n], {n, 7}] // Flatten (* Andrey Zabolotskiy, Mar 08 2024 *)

Formula

Equivalent matrix computation: Multiply the n-th diagonal (with n=0 the main diagonal) of the lower triangular Pascal matrix by f_n = (d/dx)^n f(x) to obtain the matrix VP with VP(n,k) = binomial(n,k) f_(n-k). Then R^n = (1, 0, 0, 0,..) [VP * S]^n (1, D, D^2, ..)^T, where S is the shift matrix A129185, representing differentiation in the basis x^n/n!. Cf. A145271. - Tom Copeland, Jul 17 2016
A formula for the coefficients of this matrix is presented in Ihara, obtained from Comtet. - Tom Copeland, Mar 25 2020
Elaborating on my 2011 comments: Let exp[x F(t)] = exp[t p.(x)] be the e.g.f. for the binomial Sheffer sequence of polynomials (p.(x))^n = p_n(x). Then, evaluated at x = t = 0, the coefficient p_(n,k) = (D_x^k/k!) p_n(x) = D_t^n [F(t)]^k/k! = (f(x)D_x)^n x^k/k! = R^n x^k/k!, and so p_(n,k) is the coefficient of D^k of the operator R^n below evaluated at x=0. - Tom Copeland, May 14 2020
Per earlier comments, the action of the differentials of this entry on an exponential is exp(x g(u)D_u) e^(ut) = e^(t H^{(-1)}(H(u)+x)) with g(x) = 1/D(H(x)) and H^{(-1)}(x) the compositional inverse of H(x). With H^{(-1)}(x) = -log(1-x), the inverse about x=0 is H(x) = 1-e^(-x), giving g(x) = e^x and the resulting action e^(-t log(1-x)) = (1-x)^(-t) for u=0, an e.g.f. for the unsigned Stirling numbers of the first kind (see A008275, A048994, and A130534). Consequently, summing the coefficients of this entry over each associated derivative gives these Stirling numbers. E.g., the fifth row in the examples reduces to (1+4+1) D + (7+4) D^2 + 6 D^3 + D^4 = 6 D + 11 D^2 + 6 D^3 + D^4. The Connes-Moscovici weights A139002 are a refinement of this entry. - Tom Copeland, Jul 14 2021

Extensions

Title expanded by Tom Copeland, Mar 17 2014
Sequence terms rearranged in Abramowitz and Stegun order by Michael Somos, Mar 23 2014
Title expanded by Tom Copeland, Jul 14 2021

A206496 The Connes-Moscovici weight of the rooted tree with Matula-Goebel number n. It is defined as the number of ways to build up the rooted tree from the one-vertex tree by adding successively edges to the existing vertices.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 1, 3, 4, 1, 6, 3, 4, 10, 1, 1, 15, 1, 10, 10, 5, 3, 10, 10, 15, 15, 10, 4, 60, 1, 1, 15, 5, 20, 45, 6, 5, 45, 20, 3, 60, 4, 15, 105, 18, 10, 15, 10, 70, 15, 45, 1, 105, 35, 20, 15, 24, 1, 210, 15, 6, 105, 1, 105, 105, 1, 15, 63, 140
Offset: 1

Views

Author

Emeric Deutsch, Jul 20 2012

Keywords

Comments

See A206494 for the number of ways to take apart the rooted tree corresponding to the Matula-Goebel number n by sequentially removing terminal edges.
The Matula-Goebel number of a rooted tree can be defined in the following recursive manner: to the one-vertex tree there corresponds the number 1; to a tree T with root degree 1 there corresponds the t-th prime number, where t is the Matula-Goebel number of the tree obtained from T by deleting the edge emanating from the root; to a tree T with root degree m>=2 there corresponds the product of the Matula-Goebel numbers of the m branches of T.

Examples

			a(6)=3 because the rooted tree with Matula-Goebel number 6 is the path ARBC with root at R; starting with R we can obtain the tree ARBC by adding successively edges at the vertices  (i) R, R, A, or at (ii) R, R, B, or at (iii) R, A, R.
a(8)=1 because the rooted tree with Matula-Goebel number 8 is the star tree with 3 edges emanating from the root; obviously, there is only 1 way to build up this tree from the root.
		

References

  • J. C. Butcher, The Numerical Analysis of Ordinary Differential Equations, 1987, Wiley, Chichester.

Crossrefs

A139002 is a permutation of this sequence.

Programs

  • Maple
    with(numtheory): V := proc (n) local r, s: r := proc (n) options operator, arrow: op(1, factorset(n)) end proc: s := proc (n) options operator, arrow: n/r(n) end proc: if n = 1 then 1 elif bigomega(n) = 1 then 1+V(pi(n)) else V(r(n))+V(s(n))-1 end if end proc: TF := proc (n) local r, s: r := proc (n) options operator, arrow: op(1, factorset(n)) end proc: s := proc (n) options operator, arrow; n/r(n) end proc: if n = 1 then 1 elif bigomega(n) = 1 then V(n)*TF(pi(n)) else TF(r(n))*TF(s(n))*V(n)/(V(r(n))*V(s(n))) end if end proc: SF := proc (n) if n = 1 then 1 elif nops(factorset(n)) = 1 then factorial(log[factorset(n)[1]](n))*SF(pi(factorset(n)[1]))^log[factorset(n)[1]](n) else SF(expand(op(1, ifactor(n))))*SF(expand(n/op(1, ifactor(n)))) end if end proc: a := proc (n) options operator, arrow: factorial(V(n))/(TF(n)*SF(n)) end proc: seq(a(n), n = 1 .. 120);
  • Mathematica
    V[n_] := Module[{u, v}, u = FactorInteger[#][[1, 1]]&; v = #/u[#]&; If[n == 1, 1, If[PrimeQ[n], 1 + V[PrimePi[n]], V[u[n]] + V[v[n]] - 1]]];
    r[n_] := FactorInteger[n][[1, 1]];
    s[n_] := n/r[n];
    v[n_] := Which[n == 1, 1, PrimeOmega[n] == 1, 1 + v[PrimePi[n]], True, v[r[n]] + v[s[n]] - 1];
    TF[n_] := Which[n == 1, 1, PrimeOmega[n] == 1, V[n]*TF[PrimePi[n]], True, TF[r[n]]*TF[s[n]]*v[n]/(v[r[n]]*v[s[n]])];
    SF[n_] := SF[n] = If[n == 1, 1, If[PrimeQ[n], SF[PrimePi[n]], Module[{p, e}, Product[{p, e} = pe; e! * SF[p]^e, {pe, FactorInteger[n]}]]]];
    a[n_] := V[n]!/(TF[n]*SF[n]);
    Table[a[n], {n, 1, 120}] (* Jean-François Alcover, Jun 24 2024, after Emeric Deutsch in A061775, A206493 and A206497 *)

Formula

a(n) = V(n)!/[TF(n)*SF(n)], where V denotes "number of vertices" (A061775), TF denotes "tree factorial" (A206493), and SF denotes "symmetry factor" (A206497).
Showing 1-3 of 3 results.