A052750 a(n) = (2*n + 1)^(n - 1).

1, 1, 5, 49, 729, 14641, 371293, 11390625, 410338673, 16983563041, 794280046581, 41426511213649, 2384185791015625, 150094635296999121, 10260628712958602189, 756943935220796320321, 59938945498865420543457

Offset: 0

Author

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Keywords

• easy
• nonn
• changed

Comments

a(n+1) is the number of labeled incomplete ternary trees on n vertices in which each left child has a larger label than its parent. - Brian Drake, Jul 28 2008

Put a(0) = 1. For n > 0, let x(n,k) = 2*cos((2*k-1)*Pi/(2*n+1)), k=1..n. Define the recurrences S(n;0,x(n,k)) = 1, S(n;1,x(n,k)) = x(n,k), S(n;r,x(n,k)) = x(n,k)*S(n;r-1,x(n,k)) - S(n;r-2,x(n,k)), r > 1 an integer, k=1..n. CONJECTURE: For n > 0, a(n) = Product_{k=1..n} (Sum_{m=0..n-1} S(n;2*m,x(n,k))^2). - L. Edson Jeffery, Sep 11 2013

From Wolfdieter Lang, Dec 16 2013: (Start)

Discriminants of the first difference of Chebyshev S-polynomials.

The coefficient table for the first difference polynomials P(n, x) = S(n, x) - S(n-1, x), n >= 0, S(-1, x) = 0, with the Chebyshev S polynomials (see A049310), is given in A130777.

For the discriminant of a polynomial in terms of the square of a determinant of a Vandermonde matrix build from the zeros of the polynomial see, e.g., A127670.

For the proof that D(n) := discriminant(P(n,x)) = (2*n + 1)^(n - 1), n >= 1, use the formula given e.g., in the Rivlin reference, p. 218, Theorem 5.13, eq. (5.3), namely D(n) = (-1)^(n*(n-1)/2)*Product_{j=1..n} P'(n, x(n,j)), with the zeros x(n,j) = -2*cos(2*Pi*j/(2*n+1)) of P(n, x) (see A130777). P'(n, x(n,j)) = (2*n+1)*P(n-1, x(n,j))/(2*sin(Pi*j/(2*n+1))*2*cos(Pi*j/(2*n+1)))^2. P(n-1, x(n,j)) = (-1)^(n+j)*2*cos(Pi*j/(2*n+1)). Product_{j=1..n} 2*sin(Pi*j/(2*n+1)) = 2*n+1 (see the Oct 10 2013 formula in A005408. Product_{j=1..n} 2*cos(Pi*j/(2*n+1)) = 1, because S(2*n, 0) = (-1)^n.

(End)

a(n) is the number of labeled 2-trees with n+2 vertices, rooted at a given edge. - Nikos Apostolakis, Nov 30 2018

a(n) is also the number of 2-trees with n labeled triangles and with a distinguished oriented edge. - Nikos Apostolakis, Dec 14 2018

Examples

			Discriminant: n=4: P(4, x) = 1 + 2*x - 3*x^2 - x^3 + x^4 with the zeros x[1] = -2*cos((2/9)*Pi), x[2] = -2*cos((4/9)*Pi), x[3] = 1, x[4] = 2*cos((1/9)*Pi). D(4) = (Det(Vandermonde(4,[x[1],x[2],x[3],x[4]])))^2 = 729 = a(4). - _Wolfdieter Lang_, Dec 16 2013
		

References

• L. W. Beineke, and J. W. Moon, Several proofs of the number of labelled 2-dimensional trees, In "Proof Techniques in Graph Theory" (F. Harary editor). Academic Press, New York, 1969, pp. 11-20.
• Theodore J. Rivlin, Chebyshev polynomials: from approximation theory to algebra and number theory, 2. ed., Wiley, New York, 1990.

Links

• G. C. Greubel, Table of n, a(n) for n = 0..350
• T. Fowler, I. Gessel, G. Labelle, and P. Leroux, The specification of 2-trees, Adv. Appl. Math. 28 (2) (2002) 145-168, eq. (9).
• INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 706
• Henri Muehle and Philippe Nadeau, A Poset Structure on the Alternating Group Generated by 3-Cycles, arXiv:1803.00540 [math.CO], 2018.
• J.-C. Novelli and J.-Y. Thibon, Hopf Algebras of m-permutations,(m+1)-ary trees, and m-parking functions, arXiv preprint arXiv:1403.5962 [math.CO], 2014.
• Index entries for sequences related to Chebyshev polynomials.

Crossrefs

Cf. A127670, A130777, A000169, A052752, A052774, A052782, A000272, A001705.

Programs

GAP

List([0..20],n->(2*n+1)^(n-1)); # Muniru A Asiru, Dec 05 2018

Magma

[(2*n+1)^(n-1) : n in [0..20]]; // Wesley Ivan Hurt, Jan 20 2017

Maple

spec := [S,{B=Prod(Z,S,S),S=Set(B)},labeled]: seq(combstruct[count](spec,size=n), n=0..20);

Mathematica

max = 16; (Series[Exp[-1/2*ProductLog[-2*x]], {x, 0, max}] // CoefficientList[#, x] & ) * Range[0, max]! (* Jean-François Alcover, Jun 20 2013 *)
Table[(2*n+1)^(n-1),{n,0,20}] (* Harvey P. Dale, Jul 14 2025 *)

PARI

a(n)=(2*n+1)^(n-1) \\ Charles R Greathouse IV, Nov 20 2011

PARI

{a(n)=local(A=1+x);for(i=1,21,A=sqrt(1+2*sum(n=1,21,x^(2*n-1)/(2*n-1)!*A^(4*n-1))+x*O(x^n)));n!*polcoeff(A,n)} \\ Paul D. Hanna, Sep 07 2012

Python

for n in range(0, 20): print((2*n + 1)**(n - 1), end=', ') # Stefano Spezia, Dec 01 2018

Formula

E.g.f.: exp(-1/2*W(-2*x)), where W is Lambert's W function.

E.g.f. satisfies: A(x) = sqrt(1 + 2*Sum_{n>=1} x^(2*n-1)/(2*n-1)! * A(x)^(4*n-1)). - Paul D. Hanna, Sep 07 2012

E.g.f. satisfies: A(x) = 1/A(-x*A(x)^4). - Paul D. Hanna, Sep 07 2012

a(n) = discriminant of P(n,x) = S(n,x) - S(n-1,x), n >= 1, with the Chebyshev S polynomials from A049310. For the proof see the comment above. a(n) is also the discriminant of S(n,x) + S(n-1,x) = (-1)^n*(S(n,-x) - S(n-1,-x)). - Wolfdieter Lang, Dec 16 2013

From Peter Bala, Dec 19 2013: (Start)

The e.g.f. A(x) = 1 + x + 5*x^2/2! + 49*x^3/3! + 729*x^4/4! + ... satisfies:

1) A(x*exp(-2*x)) = exp(x) = 1/A(-x*exp(2*x));

2) A^2(x) = 1/x*series reversion(x*exp(-2*x));

3) A(x^2) = 1/x*series reversion(x*exp(-x^2));

4) A(x) = exp(x*A(x)^2). (End)

E.g.f.: sqrt(-LambertW(-2*x)/(2*x)). - Vaclav Kotesovec, Dec 07 2014

Related to A001705 by Sum_{n >= 1} a(n)*x^n/n! = series reversion( 1/(1 + x)^2*log(1 + x) ) = series reversion(x - 5*x^2/2! + 26*x^3/3! - 154*x^4/4! + ...). Cf. A000272, A052752, A052774, A052782. - Peter Bala, Jun 15 2016

From Peter Bala, Dec 13 2022: (Start)

The e.g.f. A(x) = 1/x * series reversion of x^2/T(x), where the tree function T(x) = Sum_{n >= 1} n^(n-1)*x^n/n!. See A000169.

For c in C, A(x)^c = 1 + Sum_{n >= 1} c*(2*n + c)^(n-1)*x^n/n!.

First derivative A'(x) = A(x)^3/(1 - 2*x*A(x)^2).

Series reversion of (1 - A(-z)) = -log(1 - z)/(1 - z)^2 is the e.g.f. of A001705.

1/z * series reversion of z/A(z) = 1 + z + 7*z^2/2! + (10^2)*z^3/3! + (13^3)*z^4/4! + ... is the e.g.f. of A052752.

1/z * series reversion of z/A(z^2) = 1 + z^2 + 9*z^4/2! + (13^2)*z^6/3! + (17^3)*z^8/4! + ... = Sum_{n >= 0} A052774(n)*z^(2*n)/n!.

1/z * series reversion of z/A(z^3) = 1 + z^3 + 11*z^6/2! + (16^2)*z^9/3! + (21^3)*z^12/4! + ... = Sum_{n >= 0} A052782(n)*z^(3*n)/n!.

1/z * series reversion of z/A(z)^2 = A(2*z) = 2*Sum_{n >= 0} (4*n + 2)^(n-1)*z^n/n!.

1/z * series reversion of z/A(z)^k = k*Sum_{n >= 0} ((k+2)*n + k)^(n-1)*z^n/n!. (End)

a(n) = Sum_{k=1..n} (-1)^(n-k)*(n+k)^(n-1)*binomial(n,k-1), a(0)=1. - Vladimir Kruchinin, Aug 14 2025

Extensions

Better description from Vladeta Jovovic, Sep 02 2003