OFFSET
0,3
COMMENTS
a(n) is the number of noncrossing partial matchings on points 1, 2, ... , 3*n where point 1 is unmatched if n>0 and only points congruent modulo 3 can be matched. See Example 57 on p. 47 of the Burstein-Shapiro reference. - Alexander Burstein, Jun 03 2022
LINKS
Seiichi Manyama, Table of n, a(n) for n = 0..1000
Alexander Burstein and Louis W. Shapiro, Pseudo-Involutions in the Riordan Group, J. Integer Sequences 25 (2022), Article 22.3.6.
FORMULA
G.f. A(x) satisfies [from Paul D. Hanna, Mar 21 2016]: (Start)
(1) A(x)^2 = 1 + x*(A(x)^2 + A(x)^5).
(2) A(x)^3 = 1 + x*(A(x)^2 + A(x)^4 + A(x)^6).
Let F(x) = (1+x - sqrt(1 - 2*x - 3*x^2)) / (2*x), then g.f. A(x) satisfies:
(3) A(x) = sqrt( (1/x)*Series_Reversion(x/F(x)^2) ),
(4) A(x) = F(x*A(x)^2) and F(x) = A(x/F(x)^2),
where F(x) = 1 + x*M(x) such that M(x) = 1 + x*M(x) + x^2*M(x)^2 is the g.f. of the Motzkin numbers (A001006).
Let G(x) = 1 + x*G(x)/(1 - x*G(x)^2), then g.f. A(x) satisfies:
(5) A(x) = (1/x)*Series_Reversion(x/G(x)),
(6) A(x) = G(x*A(x)) and G(x) = A(x/G(x)).
where G(x) is the g.f. of A106228. (End)
Recurrence: 3*n*(3*n-1)*(3*n+1)*(5*n-11)*(5*n-8)*(5*n-6)*a(n) = 6*(5*n-11)*(900*n^5 - 3870*n^4 + 6033*n^3 - 4165*n^2 + 1238*n - 120)*a(n-1) - 2*(n-2)*(5*n-1)*(950*n^4 - 5510*n^3 + 11199*n^2 - 9207*n + 2430)*a(n-2) + 6*(n-3)*(n-2)*(2*n-5)*(5*n-6)*(5*n-3)*(5*n-1)*a(n-3). - Vaclav Kotesovec, Aug 19 2013
a(n) ~ sqrt(300+75*10^(2/3)+30*10^(1/3))/90 * (5/9*10^(2/3)+10/9*10^(1/3)+8/3)^n / (sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Aug 19 2013
Recurrence: 18*n*(2*n+1)*(n+2)*(n+1)*a(n)-(6*(n+1))*(n+2)*(46*n^2+120*n+81)*a(n+1)+(2*(n+2))*(470*n^3+2605*n^2+4748*n+2829)*a(n+2)-(n+3)*(845*n^3+6465*n^2+14764*n+8712)*a(n+3)-(2*(n+4))*(130*n^3+2080*n^2+10407*n+16641)*a(n+4)+(3*(n+5))*(153*n^3+2493*n^2+13520*n+24404)*a(n+5)-(6*(n+5))*(3*n+17)*(3*n+19)*(n+6)*a(n+6) = 0. - Robert Israel, Feb 25 2018
G.f. A(x) satisfies: A(-x*A(x)^5) = 1/A(x). - Alexander Burstein, Jun 03 2022
a(n) = (1/n) * Sum_{k=0..n-1} binomial(n,k) * binomial(2*n+k,n-1-k) for n > 0. - Seiichi Manyama, Aug 05 2023
a(n) = (1/n) * Sum_{k=0..n-1} (-1)^k * binomial(n,k) * binomial(4*n-2*k,n-1-k) for n > 0. - Seiichi Manyama, Aug 06 2023
G.f.: A(x) = sqrt(B(x)) where B(x) is the g.f. of A366400. - Seiichi Manyama, Mar 31 2024
a(n) = (1/n) * Sum_{k=0..floor(n-1)/2} binomial(n,k) * binomial(3*n-k,n-1-2*k) for n > 0. - Seiichi Manyama, Apr 01 2024
a(n) = Sum_{k=0..n} binomial(n,k) * binomial(n+3*k/2+1/2,n)/(2*n+3*k+1). - Seiichi Manyama, Apr 04 2024
EXAMPLE
G.f.: A(x) = 1 + x + 3*x^2 + 13*x^3 + 66*x^4 + 366*x^5 + 2148*x^6 +...
Related expansions:
A(x)^2 = 1 + 2*x + 7*x^2 + 32*x^3 + 167*x^4 + 942*x^5 + 5593*x^6 +...
A(x)^3 = 1 + 3*x + 12*x^2 + 58*x^3 + 312*x^4 + 1794*x^5 + 10794*x^6 +...
A(x)^4 = 1 + 4*x + 18*x^2 + 92*x^3 + 511*x^4 + 3000*x^5 + 18316*x^6 +...
A(x)^5 = 1 + 5*x + 25*x^2 + 135*x^3 + 775*x^4 + 4651*x^5 + 28845*x^6 +...
A(x)^6 = 1 + 6*x + 33*x^2 + 188*x^3 + 1116*x^4 + 6852*x^5 + 43204*x^6 +...
where A(x) = 1 + x*(A(x)^2 - A(x)^3 + A(x)^4),
and A(x)^2 = 1 + x*(A(x)^2 + A(x)^5),
and A(x)^3 = 1 + x*(A(x)^2 + A(x)^4 + A(x)^6),
and A(x)^4 = 1 + x*(A(x)^2 + A(x)^4 + A(x)^5 + A(x)^7),
and A(x)^5 = 1 + x*(A(x)^2 + A(x)^4 + A(x)^5 + A(x)^6 + A(x)^8), etc.
The g.f. satisfies A(x) = F(x*A(x)^2) and F(x) = A(x/F(x)^2) where
F(x) = 1 + x + x^2 + 2*x^3 + 4*x^4 + 9*x^5 + 21*x^6 + 51*x^7 +...+ A001006(n-1)*x^n +...
is a g.f. of the Motzkin numbers (A001006, shifted right 1 place).
The g.f. satisfies A(x) = G(x*A(x)) and G(x) = A(x/G(x)) where
G(x) = 1 + x + 2*x^2 + 6*x^3 + 21*x^4 + 80*x^5 + 322*x^6 +...+ A106228(n)*x^n +...
satisfies G(x) = 1 + x*G(x)/(1 - x*G(x)^2).
MAPLE
rec := {(36*n^4+126*n^3+126*n^2+36*n)*a(n)+(-276*n^4-1548*n^3-3198*n^2-2898*n-972)*a(n+1)+(940*n^4+7090*n^3+19916*n^2+24650*n+11316)*a(n+2)+(-845*n^4-9000*n^3-34159*n^2-53004*n-26136)*a(n+3)+(-260*n^4-5200*n^3-37454*n^2-116538*n-133128)*a(n+4)+(459*n^4+9774*n^3+77955*n^2+276012*n+366060)*a(n+5)+(-54*n^4-1242*n^3-10686*n^2-40758*n-58140)*a(n+6), a(0) = 1, a(1) = 1, a(2) = 3, a(3) = 13, a(4) = 66, a(5) = 366}:
f:= gfun:-rectoproc(rec, a(n), remember):
map(f, [$0..50]); # Robert Israel, Feb 25 2018
MATHEMATICA
nmax = 23; sol = {a[0] -> 1};
Do[A[x_] = Sum[a[k] x^k, {k, 0, n}] /. sol; eq = CoefficientList[A[x] - (1 + x (A[x]^2 - A[x]^3 + A[x]^4)) + O[x]^(n + 1), x] == 0 /. sol; sol = sol ~Join~ Solve[eq][[1]], {n, 1, nmax}];
sol /. Rule -> Set;
a /@ Range[0, nmax] (* Jean-François Alcover, Nov 01 2019 *)
PROG
(PARI) /* Formula A(x) = 1 + x*(A(x)^2 - A(x)^3 + A(x)^4): */
{a(n)=local(A=1); for(i=1, n, A=1+x*(A^2-A^3+A^4) +x*O(x^n)); polcoeff(A, n)}
for(n=0, 25, print1(a(n), ", "))
(PARI) /* Formula using Series Reversion involving Motzkin numbers: */
{a(n)=local(A=1); A=(1+x-sqrt(1-2*x-3*x^2+x^3*O(x^n)))/(2*x); polcoeff(sqrt(1/x*serreverse(x/A^2)), n)}
for(n=0, 25, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Nov 21 2012
STATUS
approved