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G.f. satisfies: A(x) = G(x/A(x))^2 and G(x)^2 = A(x*G(x)^2) where G(x) = Sum_{n>=0} C(2n,n)*C(2n+2,n+1)/(n+2)*x^n is the g.f. of A172392.
+20
2
1, 8, 12, 0, 28, 0, 264, 0, 3720, 0, 63840, 0, 1232432, 0, 25731216, 0, 568130552, 0, 13081215840, 0, 311178567648, 0, 7597974517056, 0, 189518147463232, 0, 4811962763222784, 0, 124028853694440640, 0, 3238304402221646880, 0
FORMULA
G.f.: A(x) = x/Series_Reversion(x*G(x)^2)) where G(x) is the g.f. of A172392(n) = A000108(n+1)* A000984(n).
EXAMPLE
G.f.: A(x) = 1 + 8*x + 12*x^2 + 28*x^4 + 264*x^6 + 3720*x^8 +...
where A(x) = G(x/A(x))^2 where G(x) is the g.f. of A172392: G(x) = 1 + 4*x + 30*x^2 + 280*x^3 + 2940*x^4 + 33264*x^5 +...+ A172392(n)*x^n +... G(x) = 1 + 2*2*x + 5*6*x^2 + 14*20*x^3 + 42*70*x^4 + 132*252*x^5 +...
PROG
(PARI) {a(n)=local(G=sum(m=0, n, binomial(2*m, m)*binomial(2*m+2, m+1)/(m+2)*x^m)+x*O(x^n)); polcoeff(x/serreverse(x*G^2), n)}
G.f. satisfies: A(x) = G(x/A(x)^2) and G(x) = A(x*G(x)^2) = Sum_{n>=0} C(2n,n)*C(2n+2,n+1)/(n+2)*x^n is the g.f. of A172392.
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2
1, 4, -2, 8, -20, 96, -324, 1648, -6348, 33200, -137848, 732640, -3193296, 17148608, -77335400, 418289696, -1934677436, 10518803376, -49611450120, 270796872160, -1297234193744, 7102371571840, -34458382484976, 189117499963840
FORMULA
G.f. satisfies: A(x) = Sum_{n>=0} A000108(n+1)* A000984(n)*x^n/A(x)^(2n), where A000108 is the Catalan numbers and A000984 is the central binomial coefficients.
EXAMPLE
G.f.: A(x) = 1 + 4*x - 2*x^2 + 8*x^3 - 20*x^4 + 96*x^5 - 324*x^6 +...
A(x)^2 = 1 + 8*x + 12*x^2 + 28*x^4 + 264*x^6 + 3720*x^8 +...
where A(x)^2 equals the g.f. of A172391: A172391=[1,8,12,0,28,0,264,0,3720,0,63840,0,1232432,0,25731216,0,...].
Let G(x) = A(x*G(x)^2) = Sum_{n>=0} C(2n+2,n+1)/(n+2)*C(2n,n)*x^n:
G(x) = 1 + 2*2*x + 5*6*x^2 + 14*20*x^3 + 42*70*x^4 + 132*252*x^5 +...
PROG
(PARI) {a(n)=local(G=sum(m=0, n, binomial(2*m, m)*binomial(2*m+2, m+1)/(m+2)*x^m)+x*O(x^n)); polcoeff((x/serreverse(x*G^2))^(1/2), n)}
Triangle read by rows, a Narayana related triangle whose rows are refinements of twice the Catalan numbers (for n >= 2).
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3
1, 0, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 4, 4, 1, 0, 1, 7, 12, 7, 1, 0, 1, 11, 30, 30, 11, 1, 0, 1, 16, 65, 100, 65, 16, 1, 0, 1, 22, 126, 280, 280, 126, 22, 1, 0, 1, 29, 224, 686, 980, 686, 224, 29, 1, 0, 1, 37, 372, 1512, 2940, 2940, 1512, 372, 37, 1
COMMENTS
This is the second triangle in a sequence of Narayana triangles. The first is A090181, whose n-th row is a refinement of Catalan(n), whereas here the n-th row of T is a refinement of 2*Catalan(n-1). We can show that T(n, k) <= A090181(n, k) for all n, k. The third triangle in this sequence is A353279, where also a recurrence for the general case is given. Here we give a recurrence for the row polynomials, which correspond to the recurrence of the classical Narayana polynomials combinatorially proved by Sulanke (see link).
The polynomials have only real zeros and form a Sturm sequence. This follows from the recurrence along the lines given in the Chen et al. paper.
Some interesting sequences turn out to be the evaluation of the polynomial sequence at a fixed point (see the cross-references), for example the reversion of the Jacobsthal numbers A001045 essentially is -(-2)^n*P(n, -1/2). The polynomials can also be represented as the difference between generalized Narayana polynomials, see the formula section.
FORMULA
Explicit formula (additive form):
T(n, n) = 1, T(n > 0, 0) = 0 and otherwise T(n, k) = binomial(n, k)*binomial(n - 1, k - 1)/(n - k + 1) - 2*binomial(n - 1, k)*binomial(n - 1, k - 2)/(n - 1).
Multiplicative formula with the same boundary conditions:
T(n, k) = binomial(n, k)^2*(k*(2*k^2 + (n + 1)*(n - 2*k)))/(n^2*(n-1)*(n- k + 1)).
Bivariate generating function:
T(n, k) = [x^n] [y^k](1 - x + (1+y)*(1-x*(y-1) - sqrt((x*y+x-1)^2 - 4*x^2*y))/2).
Recursion based on polynomials:
T(n, k) = [x^k] (((2*n - 3)*(x + 1)*P(n - 1, x) - (n - 3)*(x - 1)^2*P(n - 2, x)) / n) with P(0, x) = 1, P(1, x) = x, and P(2, x) = x + x^2.
Recursion based on rows (see the second Python program):
T(n, k) = (((B(k) + B(k-1)) * (2*n - 3) - (A(k) - 2*A(k-1) + A(k-2))*(n-3))/n), where A(k) = T(n-2, k) and B(k) = T(n-1, k), for n >= 3.
Hypergeometric representation:
T(n, k) = [x^k] x*(x + 1)*hypergeom([1 - n, 2 - n], [2], x) for n >= 2.
Row sums:
Sum_{k=0..n} T(n, k) = (2/n)*binomial(2*(n - 1), n - 1) = A068875(n-1) for n >= 2. A generalization of the Narayana polynomials is given by
N{n, k}(x) = Sum_{j=0..n-2*k}(((k + 1)/(n - k)) * binomial(n - k, j - 1) * binomial(n - k, j + k) * x^(j + k)).
N{n, 0}(x) are the classical Narayana polynomials A001263 and N{n, 1}(x) is a shifted version of A145596 based in (3, 2). Our polynomials are the difference P(n, x) = N{n, 0}(x) - N{n, 1}(x) for n >= 1. Let RS(T, n) denote the row sum of the n-th row of T, then RS(T, n) - RS( A090181, n) = -4*binomial(2*n - 3, n - 3)/(n + 1) = A115143(n + 1) for n >= 3.
EXAMPLE
Triangle starts:
[0] 1;
[1] 0, 1;
[2] 0, 1, 1;
[3] 0, 1, 2, 1;
[4] 0, 1, 4, 4, 1;
[5] 0, 1, 7, 12, 7, 1;
[6] 0, 1, 11, 30, 30, 11, 1;
[7] 0, 1, 16, 65, 100, 65, 16, 1;
[8] 0, 1, 22, 126, 280, 280, 126, 22, 1;
[9] 0, 1, 29, 224, 686, 980, 686, 224, 29, 1;
MAPLE
T := (n, k) -> if n = k then 1 elif k = 0 then 0 else
binomial(n, k)^2*(k*(2*k^2 + (n + 1)*(n - 2*k))) / (n^2*(n - 1)*(n - k + 1)) fi:
seq(seq(T(n, k), k = 0..n), n = 0..10);
# Alternative:
gf := 1 - x + (1 + y)*(1 - x*(y - 1) - sqrt((x*y + x - 1)^2 - 4*x^2*y))/2:
serx := expand(series(gf, x, 16)): coeffy := n -> coeff(serx, x, n):
seq(seq(coeff(coeffy(n), y, k), k = 0..n), n = 0..10);
# Using polynomial recurrence:
P := proc(n, x) option remember; if n < 3 then [1, x, x + x^2] [n + 1] else
((2*n - 3)*(x + 1)*P(n - 1, x) - (n - 3)*(x - 1)^2*P(n - 2, x)) / n fi end:
Trow := n -> seq(coeff(P(n, x), x, k), k = 0..n): seq(Trow(n), n = 0..10);
# Represented by generalized Narayana polynomials:
N := (n, k, x) -> add(((k+1)/(n-k))*binomial(n-k, j-1)*binomial(n-k, j+k)*x^(j+k), j=0..n-2*k): seq(print(ifelse(n=0, 1, expand(N(n, 0, x) - N(n, 1, x)))), n=0..7);
MATHEMATICA
H[0, _] := 1; H[1, x_] := x;
H[n_, x_] := x*(x + 1)*Hypergeometric2F1[1 - n, 2 - n, 2, x];
Hrow[n_] := CoefficientList[H[n, x], x]; Table[Hrow[n], {n, 0, 9}] // TableForm
PROG
(Python)
from math import comb as binomial
def T(n, k):
if k == n: return 1
if k == 0: return 0
return ((binomial(n, k)**2 * (k * (2 * k**2 + (n + 1) * (n - 2 * k))))
// (n**2 * (n - 1) * (n - k + 1)))
def Trow(n): return [T(n, k) for k in range(n + 1)]
for n in range(10): print(Trow(n))
(Python) # The recursion with cache is (much) faster:
from functools import cache
@cache
def T_row(n):
if n < 3: return ([1], [0, 1], [0, 1, 1])[n]
A = T_row(n - 2) + [0, 0]
B = T_row(n - 1) + [1]
for k in range(n - 1, 1, -1):
B[k] = (((B[k] + B[k - 1]) * (2 * n - 3)
- (A[k] - 2 * A[k - 1] + A[k - 2]) * (n - 3)) // n)
return B
for n in range(10): print(T_row(n))
Expansion of 3F2( (1/2, 3/2, 5/2); (3, 5))(64 x)
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1, 8, 140, 3360, 97020, 3171168, 113369256, 4338459840, 175165316040, 7385525026880, 322747443674656, 14534919841012480, 671591162296782000, 31725844951938480000, 1527939354203180010000, 74847268228930016688000, 3722092276301165621547000
COMMENTS
Generalization of formula for A172392. Combinatorial interpretation welcome.
FORMULA
D-finite with recurrence +n*(n+4)*(n+2)*a(n) -8*(2*n+3)*(2*n+1)*(2*n-1)*a(n-1)=0. - R. J. Mathar, Jul 27 2022 a(n) = 16 * (2*n+3) * (2*n+1)^2 * (2*n)!^3 / (n!^4 * (n+2)! * (n+4)!).
a(n) ~ 2^(6*n + 7) / (Pi^(3/2) * n^(9/2)). (End)
MATHEMATICA
CoefficientList[Series[HypergeometricPFQ[{1/2, 3/2, 5/2}, {3, 5}, 64 x], {x, 0, 20}], x]
Table[16 * (2*n+3) * (2*n+1)^2 * (2*n)!^3 / (n!^4 * (n+2)! * (n+4)!), {n, 0, 20}] (* Vaclav Kotesovec, Feb 17 2024 *)
Expansion of 2F1( 1/2, 3/2; 4; 16*x ).
+10
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1, 3, 18, 140, 1260, 12474, 132132, 1472328, 17065620, 204155380, 2506399896, 31443925968, 401783498480, 5215458874500, 68633685693000, 914099013896400, 12304253831789700, 167193096184907100, 2291164651422801000, 31637804708163654000, 439903041116118980400
COMMENTS
Combinatorial interpretation welcome.
Could involve planar maps, lattice walks, interpretations of catalan numbers.
FORMULA
D-finite with recurrence n*(n+3)*a(n) -4*(2*n-1)*(2*n+1)*a(n-1)=0. - R. J. Mathar, Jun 17 2016
MATHEMATICA
CoefficientList[
Series[HypergeometricPFQ[{1/2, 3/2}, {4}, 16*x], {x, 0, 20}], x]
Table[3 CatalanNumber[n] CatalanNumber[n+1] * (n+1) / (n+3), {n, 0, 20}] (* Indranil Ghosh, Mar 05 2017 *)
PROG
(PARI)
c(n) = binomial(2*n, n) / (n+1);
a(n) = 3 * c(n) * c(n+1) *(n+1) / (n+3); \\ Indranil Ghosh, Mar 05 2017 (Python)
import math
f=math.factorial
def C(n, r): return f(n) / f(r) / f(n-r)
def Catalan(n): return C(2*n, n) / (n+1)
a(n) = 16^n*[x^n]hypergeometric([3/2, -2*n], [3], -x).
+10
1
1, 16, 480, 17920, 752640, 34062336, 1623638016, 80408739840, 4100845731840, 214072431738880, 11388653368508416, 615465127495335936, 33704042696173158400, 1866685441634205696000, 104401050057113075712000, 5889038054986331298201600, 334693662791723162114457600
FORMULA
a(n) = 4^n*C(2*n,n)*C(2*n+2,n+1)/(n+2).
a(n) = (2^(6*n+2)*Gamma(n+1/2)*Gamma(n+3/2))/(Pi*Gamma(n+1)*Gamma(n+3)).
a(n) = [x^n]hypergeom([1/2, 3/2], [3], 64*x).
a(n) = a(n-1)*( 16*(4*n^2-1)/(n*(n+2)) ) for n >= 1.
MAPLE
a := n -> 16^n*coeff(simplify(hypergeom([3/2, -2*n], [3], -x)), x, n):
seq(a(n), n=0..16);
a_list := len -> seq(coeff(series(hypergeom([1/2, 3/2], [3], 64*x), x, len+1), x, n), n=0..len);
a_list(16);
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