A003239 Number of rooted planar trees with n non-root nodes: circularly cycling the subtrees at the root gives equivalent trees.

1, 1, 2, 4, 10, 26, 80, 246, 810, 2704, 9252, 32066, 112720, 400024, 1432860, 5170604, 18784170, 68635478, 252088496, 930138522, 3446167860, 12815663844, 47820447028, 178987624514, 671825133648, 2528212128776, 9536895064400, 36054433810102, 136583761444364, 518401146543812

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• nice
• easy

Comments

Also number of necklaces with 2*n beads, n white and n black (to get the correspondence, start at root, walk around outside of tree, use white if move away from the root, black if towards root).

Also number of terms in polynomial expression for permanent of generic circulant matrix of order n.

a(n) is the number of equivalence classes of n-compositions of n under cyclic rotation. (Given a necklace, split it into runs of white followed by a black bead and record the lengths of the white runs. This gives an n-composition of n.) a(n) is the number of n-multisets in Z mod n whose sum is 0. - David Callan, Nov 05 2003

a(n) is the number of cyclic equivalence classes of triangulations of a once-punctured n-gon. - Esther Banaian, May 06 2025

Examples

			As _David Callan_ said, a(n) is the number of n-multisets in Z mod n whose sum is 0. So for n = 4 the a(4)=10 multisets are (0, 0, 0, 0), (1, 1, 1, 1), (0, 1, 1, 2), (0, 0, 2, 2), (2, 2, 2, 2), (0, 0, 1, 3), (1, 2, 2, 3), (1, 1, 3, 3), (0, 2, 3, 3) and (3, 3, 3, 3). - _Boas Bakker_, Apr 21 2025
		

References

• Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 305 (see R(x)).
• F. Harary and E. M. Palmer, Graphical Enumeration, Academic Press, NY, 1973; page 80, Problem 3.13.
• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
• R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 7.112(b).

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..1669 (terms 0..200 from T. D. Noe)
• Michal Bassan, Serte Donderwinkel, and Brett Kolesnik, Graphical sequences and plane trees, arXiv:2406.05110 [math.CO], 2024.
• Bruce M. Boman, Thien-Nam Dinh, Keith Decker, Brooks Emerick, Christopher Raymond, and Gilberto Schleinger, Why do Fibonacci numbers appear in patterns of growth in nature?, Fibonacci Quarterly, 55(5) (2017), 30-41.
• R. Brualdi and M. Newman, An enumeration problem for a congruence equation, J. Res. Nat. Bureau Standards, B74 (1970), 37-40.
• CombOS - Combinatorial Object Server, Generate rooted plane trees.
• Paul Drube and Puttipong Pongtanapaisan, Annular Non-Crossing Matchings, Journal of Integer Sequences, Vol. 19 (2016), #16.2.4.
• A. Elashvili and M. Jibladze, Hermite reciprocity for the regular representations of cyclic groups, Indag. Math. (N.S.) 9(2) (1998), 233--238. MR1691428 (2000c:13006).
• A. Elashvili, M. Jibladze, and D. Pataraia, Combinatorics of necklaces and "Hermite reciprocity", J. Algebraic Combin. 10(2) (1999), 173--188. MR1719140 (2000j:05009). See p. 174. - _N. J. A. Sloane_, Aug 06 2014
• M. L. Fredman, A symmetry relationship for a class of partitions, J. Combin. Theory Ser. A, 18 (1975), 199-202. See Eq. (4), a(n) = S(n,n,0).
• F. Harary and R. W. Robinson, The number of achiral trees, J. Reine Angew. Math., 278 (1975), 322-335.
• F. Harary and R. W. Robinson, The number of achiral trees, J. Reine Angew. Math., 278 (1975), 322-335. (Annotated scanned copy)
• Thomas C. Hull and Tomohiro Tachi, Double-line rigid origami, arXiv:1709.03210 [math.MG], 2017.
• INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 761.
• Benjamin Ruoyu Kan, Polynomial Approximations for Quantum Hamiltonian Complexity, Bachelor's thesis, Harvard Univ., 2023.
• G. Labelle and P. Leroux, Enumeration of (uni- or bicolored) plane trees according to their degree distribution, Disc. Math. 157 (1996), 227-240, Eq. (1.18).
• J. Malenfant, On the Matrix-Element Expansion of a Circulant Determinant, arXiv preprint arXiv:1502.06012 [math.NT], 2015.
• Paul Melotti, Sanjay Ramassamy, and Paul Thévenin, Points and lines configurations for perpendicular bisectors of convex cyclic polygons, arXiv:2003.11006 [math.CO], 2020.
• J. Sawada, Generating rooted and free plane trees, ACM Transactions on Algorithms, 2(1) (2006), 1-13.
• Hugh Thomas, The number of terms in the permanent and the determinant of a generic circulant matrix, arXiv:math/0301048 [math.CO], 2003.
• D. W. Walkup, The number of plane trees, Mathematika, 19(2) (1972), 200-204. - From _N. J. A. Sloane_, Jun 08 2012
• Index entries for sequences related to necklaces
• Index entries for sequences related to rooted trees
• Index entries for sequences related to trees

Crossrefs

Cf. A002995, A057510, A000108, A022553, A082936, A084575, A037306.

Column k=2 of A208183.

Column k=1 of A261494.

Programs

Maple

with(numtheory): A003239 := proc(n) local t1,t2,d; t2 := divisors(n); t1 := 0; for d in t2 do t1 := t1+phi(n/d)*binomial(2*d,d)/(2*n); od; t1; end;
spec := [ C, {B=Union(Z,Prod(B,B)), C=Cycle(B)}, unlabeled ]; [seq(combstruct[count](spec, size=n), n=0..40)];

Mathematica

a[n_] := Sum[ EulerPhi[n/k]*Binomial[2k, k]/(2n), {k, Divisors[n]}]; a[0] = 1; Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Apr 11 2012 *)

PARI

C(n, k)=binomial(n,k);
a(n) = if(n<=0, n==0, sumdiv(n, d, eulerphi(n/d) * C(2*d, d)) / (2*n) );
/* or, second formula: */
/* a(n) = if(n<=0, n==0, sumdiv(n, d, eulerphi(n/d) * C(2*d-1,d)) / n ); */
/* Joerg Arndt, Oct 21 2012 */

SageMath

def A003239(n):
    if n == 0: return 1
    return sum(euler_phi(n/d)*binomial(2*d, d)/(2*n) for d in divisors(n))
print([A003239(n) for n in (0..29)]) # Peter Luschny, Dec 10 2020

Formula

a(n) = Sum_{d|n} (phi(n/d)*binomial(2*d, d))/(2*n) for n > 0.

a(n) = (1/n)*Sum_{d|n} (phi(n/d)*binomial(2*d-1, d)) for n > 0.

a(n) = A047996(2*n, n). - Philippe Deléham, Jul 25 2006

a(n) ~ 2^(2*n-1) / (sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Aug 22 2015

Extensions

Sequence corrected and extended by Roderick J. Fletcher (yylee(AT)mail.ncku.edu.tw), Aug 1997

Additional comments from Michael Somos