

A007317


Binomial transform of Catalan numbers.
(Formerly M1480)


108



1, 2, 5, 15, 51, 188, 731, 2950, 12235, 51822, 223191, 974427, 4302645, 19181100, 86211885, 390248055, 1777495635, 8140539950, 37463689775, 173164232965, 803539474345, 3741930523740, 17481709707825, 81912506777200, 384847173838501, 1812610804416698
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OFFSET

1,2


COMMENTS

Partial sums of A002212 (the restricted hexagonal polyominoes with n cells). Number of Schroeder paths (i.e., consisting of steps U=(1,1),D=(1,1),H=(2,0) and never going below the xaxis) from (0,0) to (2n2,0), with no peaks at even level. Example: a(3)=5 because among the six Schroeder paths from (0,0) to (4,0) only UUDD has a peak at an even level.  Emeric Deutsch, Dec 06 2003
Number of binary trees of weight n where leaves have positive integer weights. Noncommutative Nonassociative version of partitions of n.  Michael Somos, May 23 2005
Appears also as the number of Euler trees with total weight n (associated with even switching class of matrices of order 2n).  David Garber, Sep 19 2005
Number of symmetric hex trees with 2n1 edges; also number of symmetric hex trees with 2n2 edges. A hex tree is a rooted tree where each vertex has 0, 1, or 2 children and, when only one child is present, it is either a left child, or a median child, or a right child (name due to an obvious bijection with certain treelike polyhexes; see the HararyRead reference). A hex tree is symmetric if it is identical with its reflection in a bisector through the root.  Emeric Deutsch, Dec 19 2006
The Hankel transform of [1, 2, 5, 15, 51, 188, ...] is [1, 1, 1, 1, 1, ...], see A000012 ; the Hankel transform of [2, 5, 15, 51, 188, 731, ...] is [2, 5, 13, 34, 89, ...], see A001519.  Philippe Deléham, Dec 19 2006
a(n) = number of 321avoiding partitions of [n]. A partition is 321avoiding if the permutation obtained from its canonical form (entries in each block listed in increasing order and blocks listed in increasing order of their first entries) is 321avoiding. For example, the only partition of [5] that fails to be 321avoiding is 15/24/3 because the entries 5,4,3 in the permutation 15243 form a 321 pattern.  David Callan, Jul 22 2008
The sequence 1,1,2,5,15,51,188,... has Hankel transform A001519.  Paul Barry, Jan 13 2009
Equals INVERT transform of A033321: (1, 1, 2, 6, 21, 79, 311, ...).
Equals INVERTi transform of A002212: (1, 3, 10, 36, 137, ...).
Convolved with A026378, (1, 4, 17, 75, 339, ...) = A026376: (1, 6, 30, 144, ...)
(End)
a(n) is the number of vertices of the composihedron CK(n). The composihedra are a sequence of convex polytopes used to define maps of certain homotopy Hspaces. They are cellular quotients of the multiplihedra and cellular covers of the cubes.  Stefan Forcey (sforcey(AT)gmail.com), Dec 17 2009
a(n) is the number of Motzkin paths of length n1 in which the (1,0)steps at level 0 come in 2 colors and those at a higher level come in 3 colors. Example: a(4)=15 because we have 2^3 = 8 paths of shape UHD, 2 paths of shape HUD, 2 paths of shape UDH, and 3 paths of shape UHD; here U=(1,1), H=(1,0), and D=(1,1).  Emeric Deutsch, May 02 2011
REVERT transform of (1, 2, 3, 5, 8, 13, 21, 34, ... ) where the entries are Fibonacci numbers, A000045. Equivalently, coefficients in the series reversion of x(1x)/(1+xx^2). This means that the substitution of the gf (1x(16x+5x^2)^(1/2))/(2(1x)) for x in x(1x)/(1+xx^2) will simplify to x.  David Callan, Nov 11 2012
The number of plane trees with nodes that have positive integer weights and whose total weight is n.  Brad R. Jones, Jun 12 2014
Let P(x) = x/(1+x) with comp. inverse Pinv(x) = x/(1x) = P[x], and C(x)= [1sqrt(14x)]/2, an o.g.f. for the shifted Catalan numbers A000108, with inverse Cinv(x) = x * (1x).
Fin(x) = P[C(x)] = C(x)/[1 + C(x)] is an o.g.f. for the Fine numbers, A000957 with inverse Fin^(1)(x) = Cinv[Pinv(x)] = Cinv[P(x)].
Mot(x) = C[P(x)] = C[Pinv(x)] gives an o.g.f. for shifted A005043, the Motzkin or Riordan numbers with comp. inverse Mot^(1)(x) = Pinv[Cinv(x)] = (x  x^2) / (1  x + x^2) (cf. A057078).
BTC(x) = C[Pinv(x)] gives A007317, a binomial transform of the Catalan numbers, with BTC^(1)(x) = P[Cinv(x)] = (xx^2) / (1 + x  x^2).
Fib(x) = Fin[Cinv(Cinv(x))] = P[Cinv(x)] = x + 2 x^2 + 3 x^3 + 5 x^4 + ... = (x+x^2)/[1xx^2] is an o.g.f. for the shifted Fibonacci sequence A000045, so the comp. inverse is Fib^(1)(x) = C[Pinv(x)] = BTC(x) and Fib(x) = BTC^(1)(x).
Generalizing to P(x,t) = x /(1 + t*x) and Pinv(x,t) = x /(1  t*x) = P(x,t) gives other relations to lattice paths, such as the o.g.f. for A091867, C[P[x,1t]], and that for A104597, Pinv[Cinv(x),t+1].
(End)
Starting with offset 0, a(n) is also the number of Schröder paths of semilength n avoiding UH (an up step directly followed by a long horizontal step). Example: a(2)=5 because among the six possible Schröder paths of semilength 2 only UHD contains UH.  Valerie Roitner, Jul 23 2020


REFERENCES

J. Brunvoll et al., Studies of some chemically relevant polygonal systems: monoqpolyhexes, ACH Models in Chem., 133 (3) (1996), 277298, Eq. 15.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).


LINKS



FORMULA

(n+2)*a(n+2) = (6n+4)*a(n+1)  5n*a(n).
G.f.: 3/2(1/2)*sqrt((15*x)/(1x)) [GesselKim].  N. J. A. Sloane, Jul 05 2014
G.f. for sequence doubled: (1/(2*x))*(1+x(1x)^(1)*(1x^2)^(1/2)*(15*x^2)^(1/2)).
a(n) = hypergeom([1/2, n], [2], 4), n=0, 1, 2...; Integral representation as nth moment of a positive function on a finite interval of the positive halfaxis: a(n)=int(x^n*sqrt((5x)/(x1))/(2*Pi), x=1..5), n=0, 1, 2... This representation is unique.  Karol A. Penson, Sep 24 2001
a(1)=1, a(n)=1+sum(i=1, n1, a(i)*a(ni)).  Benoit Cloitre, Mar 16 2004
a(n) = Sum_{k=0..n} (1)^k*3^(nk)*binomial(n, k)*binomial(k, floor(k/2)) [offset 0].  Paul Barry, Jan 27 2005
G.f. A(x) satisfies 0=f(x, A(x)) where f(x, y)=x(1x)(yy^2).  Michael Somos, May 23 2005
G.f. A(x) satisfies 0=f(x, A(x), A(A(x))) where f(x, y, z)=x(zz^2)+(x1)y^2 .  Michael Somos, May 23 2005
G.f. (for offset 0): (1+x+(16*x+5*x^2)^(1/2))/(2*(x+x^2)).
G.f. =z*c(z/(1z))/(1z) = 1/2  (1/2)sqrt(14z/(1z)), where c(z)=(1sqrt(14z))/(2z) is the Catalan function (follows from Michael Somos' first comment).  Emeric Deutsch, Aug 12 2007
G.f.: 1/(12xx^2/(13xx^2/(13xx^2/(13xx^2/(13xx^2/(1.... (continued fraction).  Paul Barry, Apr 19 2009
E.g.f.: exp(3x)*(I_0(2x)I_1(2x)), where I_k(x) is a modified Bessel function of the first kind.  Emanuele Munarini, Apr 15 2011
If we prefix sequence with an additional term a(0)=1, g.f. is (33*xsqrt(16*x+5*x^2))/(2*(1x)). [See Kim, 2011]  N. J. A. Sloane, May 13 2011
a(n) = upper left term in M^(n1), M = an infinite square production matrix as follows:
2, 1, 0, 0, 0, 0, ...
1, 2, 1, 0, 0, 0, ...
1, 1, 2, 1, 0, 0, ...
1, 1, 1, 2, 1, 0, ...
1, 1, 1, 1, 2, 1, ...
1, 1, 1, 1, 1, 2, ...
... (End)
G.f. satisfies: A(x) = Sum_{n>=0} x^n * (1  A(x)^(n+1))/(1  A(x)); offset=0.  Paul D. Hanna, Nov 07 2011
G.f.: 1/x  1/x/Q(0), where Q(k)= 1 + (4*k+1)*x/((1x)*(k+1)  x*(1x)*(2*k+2)*(4*k+3)/(x*(8*k+6)+(2*k+3)*(1x)/Q(k+1))); (continued fraction).  Sergei N. Gladkovskii, May 14 2013
G.f.: (1x  (15*x)*G(0))/(2*x*(1x)), where G(k)= 1 + 4*x*(4*k+1)/( (4*k+2)*(1x)  2*x*(1x)*(2*k+1)*(4*k+3)/(x*(4*k+3) + (1x)*(k+1)/G(k+1))); (continued fraction).  Sergei N. Gladkovskii, Jun 25 2013
Asymptotics (for offset 0): a(n) ~ 5^(n+3/2)/(8*sqrt(Pi)*n^(3/2)).  Vaclav Kotesovec, Jun 28 2013
G.f.: G(0)/(1x), where G(k) = 1 + (4*k+1)*x/((k+1)*(1x)  2*x*(1x)*(k+1)*(4*k+3)/(2*x*(4*k+3) + (2*k+3)*(1x)/G(k+1) )); (continued fraction).  Sergei N. Gladkovskii, Jan 29 2014
0 = +a(n)*(+25*a(n+1) 50*a(n+2) +15*a(n+3)) +a(n+1)*(10*a(n+1) +31*a(n+2) 14*a(n+3)) +a(n+2)*(+2*a(n+2) +a(n+3)) for all n in Z.  Michael Somos, Jan 17 2018
a(n+1) = (2/Pi) * Integral_{x = 1..1} (m + 4*x^2)^n*sqrt(1  x^2) dx at m = 1. In general, the integral, qua sequence in n, gives the mth binomial transform of the Catalan numbers.  Peter Bala, Jan 26 2020


EXAMPLE

a(3)=5 since {3, (1+2), (1+(1+1)), (2+1), ((1+1)+1)} are the five weighted binary trees of weight 3.
G.f. = x + 2*x^2 + 5*x^3 + 15*x^4 + 51*x^5 + 188*x^6 + 731*x^7 + 2950*x^8 + 12235*x^9 + ... Michael Somos, Jan 17 2018


MAPLE

G := (1sqrt(14*z/(1z)))*1/2: Gser := series(G, z = 0, 30): seq(coeff(Gser, z, n), n = 1 .. 26); # Emeric Deutsch, Aug 12 2007
seq(round(evalf(JacobiP(n1, 1, n1/2, 9)/n, 99)), n=1..25); # Peter Luschny, Sep 23 2014


MATHEMATICA

Rest@ CoefficientList[ InverseSeries[ Series[(y  y^2)/(1 + y  y^2), {y, 0, 26}], x], x] (* then A(x)=y(x); note that InverseSeries[Series[yy^2, {y, 0, 24}], x] produces A000108(x) *) (* Len Smiley, Apr 10 2000 *)
Range[0, 25]! CoefficientList[ Series[ Exp[ 3x] (BesselI[0, 2x]  BesselI[1, 2x]), {x, 0, 25}], x] (* Robert G. Wilson v, Apr 15 2011 *)
a[n_] := Sum[ Binomial[n, k]*CatalanNumber[k], {k, 0, n}]; Table[a[n], {n, 0, 25}] (* JeanFrançois Alcover, Aug 07 2012 *)
Rest[CoefficientList[Series[3/2  (1/2) Sqrt[(1  5 x)/(1  x)], {x, 0, 40}], x]] (* Vincenzo Librandi, Nov 03 2014 *)
Table[Hypergeometric2F1[1/2, n+1, 2, 4], {n, 1, 30}] (* Vaclav Kotesovec, May 12 2022 *)


PROG

(PARI) {a(n) = my(A); if( n<2, n>0, A=vector(n); for(j=1, n, A[j] = 1 + sum(k=1, j1, A[k]*A[jk])); A[n])}; /* Michael Somos, May 23 2005 */
(PARI) {a(n) = if( n<1, 0, polcoeff( serreverse( (x  x^2) / (1 + x  x^2) + x * O(x^n)), n))}; /* Michael Somos, May 23 2005 */
(PARI) /* Offset = 0: */ {a(n)=local(A=1+x); for(i=1, n, A=sum(m=0, n, x^m*sum(k=0, m, A^k)+x*O(x^n))); polcoeff(A, n)} \\ Paul D. Hanna


CROSSREFS

First column of triangle A104259. Row sums of absolute values of A091699.
Number of vertices of multiplihedron A121988.
Cf. A055879, A033321, A026376, A026378, A059346, A000045, A000957, A057078, A091867, A104597, A249548, A162326.


KEYWORD

easy,nonn,nice


AUTHOR



STATUS

approved



