login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

Number of ways of partitioning n points on a circle into subsets only of sizes 2 and 3.
(Formerly M1353 N0520)
11

%I M1353 N0520 #111 Sep 22 2023 07:13:22

%S 1,0,1,1,2,5,8,21,42,96,222,495,1177,2717,6435,15288,36374,87516,

%T 210494,509694,1237736,3014882,7370860,18059899,44379535,109298070,

%U 269766655,667224480,1653266565,4103910930,10203669285,25408828065,63364046190,158229645720,395632288590,990419552730

%N Number of ways of partitioning n points on a circle into subsets only of sizes 2 and 3.

%C a(n) is also the number of rooted trees on n nodes such that each node has 0, 2, or 3 children. - _Patrick Devlin_, Mar 04 2012

%C a(n) is the number of Motzkin paths that have no flatsteps (F) at ground level and avoid a factor of the form FMF with M a Motzkin path (possibly empty). For example, a(5) = 5 counts UDUFD, UFDUD, UFUDD, UUDFD, UUFDD but not UFFFD. Proof: Such a path can have at most one flatstep at height 1 before the first return to ground level or else the first component will contain an FMF. Hence, with a dot denoting concatenation, such a path is either empty or has the form U.P1.D.P2 or the form U.P1.F.P2.D.P3 where P1, P2, P3 are all paths of the type being counted. Hence the gf F(x) = 1 + x^2 + x^3 + 2*x^4 + ... satisfies F = 1 + x^2*F^2 + x^3*F^3. - _David Callan_, Nov 21 2021

%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).

%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

%H Vincenzo Librandi, <a href="/A001005/b001005.txt">Table of n, a(n) for n = 0..200</a>

%H Paul Barry, <a href="http://dx.doi.org/10.1016/j.laa.2015.10.032">Riordan arrays, generalized Narayana triangles, and series reversion</a>, Linear Algebra and its Applications, 491 (2016) 343-385.

%H F. R. Bernhart & N. J. A. Sloane, <a href="/A006343/a006343.pdf">Emails, April-May 1994</a>

%H Isaac DeJager, Madeleine Naquin, and Frank Seidl, <a href="https://www.valpo.edu/mathematics-statistics/files/2019/08/Drube2019.pdf">Colored Motzkin Paths of Higher Order</a>, VERUM 2019.

%H INRIA Algorithms Project, <a href="http://ecs.inria.fr/services/structure?nbr=396">Encyclopedia of Combinatorial Structures 396</a>

%H T. S. Motzkin, <a href="http://dx.doi.org/10.1090/S0002-9904-1948-09002-4 ">Relations between hypersurface cross ratios and a combinatorial formula for partitions of a polygon, for permanent preponderance and for non-associative products</a>, Bull. Amer. Math. Soc., 54 (1948), 352-360.

%H Len Smiley, <a href="http://www.math.uaa.alaska.edu/~smiley/A001005_7_8.pdf">a(7) and a(8)</a>

%H <a href="/index/Ro#rooted">Index entries for sequences related to rooted trees</a>

%F G.f. for a(n-1), with a(-1) = 0, satisfies A(x)=x*(1+A(x)^2+A(x)^3). - _Christian G. Bower_, Dec 15 1999

%F a(n) = sum(((n)!/(k!*j!*(n-k-j+1)!)*[2*k+3*j=n], k=0..floor(n/2), j=0..floor(n/3)). - _Len Smiley_, Jun 18 2005

%F Recurrence: 2*(n+1)*(2*n+3)*(26*n+1)*a(n) = -(n-1)*(26*n^2 + 53*n + 18)*a(n-1) + 6*(n-1)*(78*n^2 + 42*n - 25)*a(n-2) + 31*(n-2)*(n-1)*(26*n+27)*a(n-3). - _Vaclav Kotesovec_, Aug 14 2013

%F a(n) ~ c*d^n/n^(3/2), where d = ((6371-624*sqrt(78))^(1/3)+(6371+624*sqrt(78))^(1/3)-1)/12 = 2.610718613276039349818649... is the root of the equation 4d^3 + d^2 - 18d - 31 = 0 and c = d^2 / (2*sqrt(Pi)*sqrt(1 + 3*d + sqrt(1 + 3*d))) = 0.559628309722556021604897336422272... - _Vaclav Kotesovec_, Aug 14 2013, updated Jun 27 2018

%F a(n) = Sum_{k=1..floor(n/2)} C(n,k-1)*C(k,n-2k)/k, n > 0. - _Michael D. Weiner_, Mar 02 2015

%F From _Wolfdieter Lang_, Nov 05 2018: (Start)

%F The o.g.f of a(n) is G(x) = F^[-1](x)/x, where F^[-1](x) is the compositional inverse of F(y) = y/(1 + y^2 + y^3), that is F(F^[-1](x)) = x, identically. (Compare this with the g.f. given above, and see the Pari and Mathematica programs below.)

%F a(n) = b(n+1)/(n+1), for n >= 0, where b(n+1) is the coefficient of x^n of (1 + x^2 + x^3)^(n+1). This follows from the Lagrange inversion series for G(x) = F^[-1](x)/x.

%F a(n) = (1/(n+1))*(Sum_{2*e2 + 3*e3 = n} (n+1)!/(n+1 - (e2 + e3))!*e2!*e3!) (from the multinomial formula for (x1 + x2 + x3)^(n+1)). For the solutions of 2*e2 + 3*e3 = n see the array A321201.

%F (End)

%e a(7)=21: 7 rotations of [12][34][567], 7 rotations of [12][45][367], 7 rotations of [12][37][456]. - _Len Smiley_, Jun 18 2005

%e From _Wolfdieter Lang_, Nov 05 2018: (Start)

%e a(7) = b(8)/8, where b(8) = (d^7/dx^7)((1 + x^2 + x^3)^8)/7! evaluated for x = 0, which is 168, and 168/8 = 21.

%e a(7) =(1/8)*8!/((8-(2+1))!*2!*1!) =(1/8)*8!/(5!*2!)= 168/8 = 21, from the only solution [e2, e3] = [2, 1] of 2*e2 + 3*e3 = 7. (End)

%p a:=proc(n::nonnegint) local k,j; a(n):=0; for k from 0 to floor(n/2) do for j from 0 to floor(n/3) do if (2*k+3*j=n) then a(n):=a(n)+(n)!/(k!*j!*(n-k-j+1)!) fi od od; print(a(n)) end proc; seq(a(i),i=0..30); # _Len Smiley_, Jun 18 2005

%p A001005 := n -> ifelse(n=0, 1, add(binomial(n, k-1)*binomial(k, n-2*k)/k, k = 1 + iquo(n-1,3)..iquo(n,2))): seq(A001005(n), n=0..35); # _Peter Luschny_, Oct 18 2022

%t Table[Sum[(n)!/(k!*j!*(n - k - j + 1)!) * KroneckerDelta[2*k + 3*j - n], {k, 0, Floor[n/2]}, {j, 0, Floor[n/3]}], {n, 0, 20}] (* _Ricardo Bittencourt_, Jun 09 2013 *)

%t CoefficientList[ InverseSeries[x/(1+x^2+x^3) + O[x]^66]/x, x] (* _Jean-François Alcover_, Feb 15 2016, after _Joerg Arndt_*)

%o (PARI) Vec(serreverse(x/(1+x^2+x^3)+O(x^66))/x) /* _Joerg Arndt_, Aug 19 2012 */

%Y Cf. A321201.

%K nonn,eigen

%O 0,5

%A _N. J. A. Sloane_

%E More terms from _Christian G. Bower_, Dec 15 1999