A001263 Triangle of Narayana numbers T(n,k) = C(n-1,k-1)*C(n,k-1)/k with 1 <= k <= n, read by rows. Also called the Catalan triangle.
1, 1, 1, 1, 3, 1, 1, 6, 6, 1, 1, 10, 20, 10, 1, 1, 15, 50, 50, 15, 1, 1, 21, 105, 175, 105, 21, 1, 1, 28, 196, 490, 490, 196, 28, 1, 1, 36, 336, 1176, 1764, 1176, 336, 36, 1, 1, 45, 540, 2520, 5292, 5292, 2520, 540, 45, 1, 1, 55, 825, 4950, 13860, 19404, 13860, 4950, 825
Offset: 1
Examples
The initial rows of the triangle are: [1] 1 [2] 1, 1 [3] 1, 3, 1 [4] 1, 6, 6, 1 [5] 1, 10, 20, 10, 1 [6] 1, 15, 50, 50, 15, 1 [7] 1, 21, 105, 175, 105, 21, 1 [8] 1, 28, 196, 490, 490, 196, 28, 1 [9] 1, 36, 336, 1176, 1764, 1176, 336, 36, 1; ... For all n, 12...n (1 block) and 1|2|3|...|n (n blocks) are noncrossing set partitions. Example of umbral representation: A007318(5,k)=[1,5/1,5*4/(2*1),...,1]=(1,5,10,10,5,1), so A001263(5,k)={1,b(5)/b(1),b(5)*b(4)/[b(2)*b(1)],...,1} = [1,30/2,30*20/(6*2),...,1]=(1,15,50,50,15,1). First = last term = b.(5!)/[b.(0!)*b.(5!)]= 1. - _Tom Copeland_, Sep 21 2011 Row polynomials and diagonal sequences of A103371: n = 4, P(4, x) = 1 + 6*x + 6*x^2 + x^3, and the o.g.f. of fifth diagonal is G(4, x) = 5* P(4, x)/(1 - x)^9, namely [5, 75, 525, ...]. See a comment above. - _Wolfdieter Lang_, Jul 31 2017
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Links
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Crossrefs
Row sums give A000108 (Catalan numbers), n>0.
A008459 (h-vectors type B associahedra), A033282 (f-vectors type A associahedra), A145903 (h-vectors type D associahedra), A145904 (Hilbert transform). - Peter Bala, Oct 27 2008
Cf. A243752.
Programs
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GAP
Flat(List([1..11],n->List([1..n],k->Binomial(n-1,k-1)*Binomial(n,k-1)/k))); # Muniru A Asiru, Jul 12 2018
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Haskell
a001263 n k = a001263_tabl !! (n-1) !! (k-1) a001263_row n = a001263_tabl !! (n-1) a001263_tabl = zipWith dt a007318_tabl (tail a007318_tabl) where dt us vs = zipWith (-) (zipWith (*) us (tail vs)) (zipWith (*) (tail us ++ [0]) (init vs)) -- Reinhard Zumkeller, Oct 10 2013
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Magma
/* triangle */ [[Binomial(n-1,k-1)*Binomial(n,k-1)/k : k in [1..n]]: n in [1.. 15]]; // Vincenzo Librandi, Oct 19 2014
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Maple
A001263 := (n,k)->binomial(n-1,k-1)*binomial(n,k-1)/k; a:=proc(n,k) option remember; local i; if k=1 or k=n then 1 else add(binomial(n+i-1, 2*k-2)*a(k-1,i),i=1..k-1); fi; end: # Alternatively, as a (0,0)-based triangle: R := n -> simplify(hypergeom([-n, -n-1], [2], x)): Trow := n -> seq(coeff(R(n,x),x,j), j=0..n): seq(Trow(n), n=0..9); # Peter Luschny, Mar 19 2018
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Mathematica
T[n_, k_] := If[k==0, 0, Binomial[n-1, k-1] Binomial[n, k-1] / k]; Flatten[Table[Binomial[n-1,k-1] Binomial[n,k-1]/k,{n,15},{k,n}]] (* Harvey P. Dale, Feb 29 2012 *) TRow[n_] := CoefficientList[Hypergeometric2F1[1 - n, -n, 2, x], x]; Table[TRow[n], {n, 1, 11}] // Flatten (* Peter Luschny, Mar 19 2018 *) aot[n_]:=If[n==1,{{}},Join@@Table[Tuples[aot/@c],{c,Join@@Permutations/@IntegerPartitions[n-1]}]]; Table[Length[Select[aot[n],Length[Position[#,{}]]==k&]],{n,2,9},{k,1,n-1}] (* Gus Wiseman, Jan 23 2023 *) T[1, 1] := 1; T[n_, k_]/;1<=k<=n := T[n, k] = (2n/k-1) T[n-1,k-1] + T[n-1, k]; T[n_, k_] := 0; Flatten@Table[T[n, k], {n, 1, 11}, {k, 1, n}] (* Oliver Seipel, Dec 31 2024 *)
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PARI
{a(n, k) = if(k==0, 0, binomial(n-1, k-1) * binomial(n, k-1) / k)};
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PARI
{T(n,k)=polcoeff(polcoeff(exp(sum(m=1,n,sum(j=0,m,binomial(m,j)^2*y^j)*x^m/m) +O(x^(n+1))),n,x),k,y)} \\ Paul D. Hanna, Oct 13 2010
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Sage
@CachedFunction def T(n, k): if k == n or k == 1: return 1 if k <= 0 or k > n: return 0 return binomial(n, 2) * (T(n-1, k)/((n-k)*(n-k+1)) + T(n-1, k-1)/(k*(k-1))) for n in (1..9): print([T(n, k) for k in (1..n)]) # Peter Luschny, Oct 28 2014
Formula
a(n, k) = C(n-1, k-1)*C(n, k-1)/k for k!=0; a(n, 0)=0.
Triangle equals [0, 1, 0, 1, 0, 1, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, ...] where DELTA is Deléham's operator defined in A084938.
0Mike Zabrocki, Aug 26 2004
T(n, k) = C(n, k)*C(n-1, k-1) - C(n, k-1)*C(n-1, k) (determinant of a 2 X 2 subarray of Pascal's triangle A007318). - Gerald McGarvey, Feb 24 2005
T(n, k) = binomial(n-1, k-1)^2 - binomial(n-1, k)*binomial(n-1, k-2). - David Callan, Nov 02 2005
a(n,k) = C(n,2) (a(n-1,k)/((n-k)*(n-k+1)) + a(n-1,k-1)/(k*(k-1))) a(n,k) = C(n,k)*C(n,k-1)/n. - Mitch Harris, Jul 06 2006
Central column = A000891, (2n)!*(2n+1)! / (n!*(n+1)!)^2. - Zerinvary Lajos, Oct 29 2006
G.f.: (1-x*(1+y)-sqrt((1-x*(1+y))^2-4*y*x^2))/(2*x) = Sum_{n>0, k>0} a(n, k)*x^n*y^k.
From Peter Bala, Oct 22 2008: (Start)
Relation with Jacobi polynomials of parameter (1,1):
Row n+1 generating polynomial equals 1/(n+1)*x*(1-x)^n*Jacobi_P(n,1,1,(1+x)/(1-x)). It follows that the zeros of the Narayana polynomials are all real and nonpositive, as noted above. O.g.f for column k+2: 1/(k+1) * y^(k+2)/(1-y)^(k+3) * Jacobi_P(k,1,1,(1+y)/(1-y)). Cf. A008459.
T(n+1,k) is the number of walks of n unit steps on the square lattice (i.e., each step in the direction either up (U), down (D), right (R) or left (L)) starting from the origin and finishing at lattice points on the x axis and which remain in the upper half-plane y >= 0 [Guy]. For example, T(4,3) = 6 counts the six walks RRL, LRR, RLR, UDL, URD and RUD, from the origin to the lattice point (1,0), each of 3 steps. Compare with tables A145596 - A145599.
Define a functional I on formal power series of the form f(x) = 1 + ax + bx^2 + ... by the following iterative process. Define inductively f^(1)(x) = f(x) and f^(n+1)(x) = f(x*f^(n)(x)) for n >= 1. Then set I(f(x)) = lim_{n -> infinity} f^(n)(x) in the x-adic topology on the ring of formal power series; the operator I may also be defined by I(f(x)) := 1/x*series reversion of x/f(x).
The o.g.f. for this array is I(1 + t*x + t*x^2 + t*x^3 + ...) = 1 + t*x + (t + t^2)*x^2 + (t + 3*t^2 + t^3)*x^3 + ... = 1/(1 - x*t/(1 - x/(1 - x*t/(1 - x/(1 - ...))))) (as a continued fraction). Cf. A108767, A132081 and A141618. (End)
G.f.: 1/(1-x-xy-x^2y/(1-x-xy-x^2y/(1-... (continued fraction). - Paul Barry, Sep 28 2010
E.g.f.: exp((1+y)x)*Bessel_I(1,2*sqrt(y)x)/(sqrt(y)*x). - Paul Barry, Sep 28 2010
G.f.: A(x,y) = exp( Sum_{n>=1} [Sum_{k=0..n} C(n,k)^2*y^k] * x^n/n ). - Paul D. Hanna, Oct 13 2010
With F(x,t) = (1-(1+t)*x-sqrt(1-2*(1+t)*x+((t-1)*x)^2))/(2*x) an o.g.f. in x for the Narayana polynomials in t, G(x,t) = x/(t+(1+t)*x+x^2) is the compositional inverse in x. Consequently, with H(x,t) = 1/ (dG(x,t)/dx) = (t+(1+t)*x+x^2)^2 / (t-x^2), the n-th Narayana polynomial in t is given by (1/n!)*((H(x,t)*D_x)^n)x evaluated at x=0, i.e., F(x,t) = exp(x*H(u,t)*D_u)u, evaluated at u = 0. Also, dF(x,t)/dx = H(F(x,t),t). - Tom Copeland, Sep 04 2011
With offset 0, A001263 = Sum_{j>=0} A132710^j / A010790(j), a normalized Bessel fct. May be represented as the Pascal matrix A007318, n!/[(n-k)!*k!], umbralized with b(n)=A002378(n) for n>0 and b(0)=1: A001263(n,k)= b.(n!)/{b.[(n-k)!]*b.(k!)} where b.(n!) = b(n)*b(n-1)...*b(0), a generalized factorial (see example). - Tom Copeland, Sep 21 2011
With F(x,t) = {1-(1-t)*x-sqrt[1-2*(1+t)*x+[(t-1)*x]^2]}/2 a shifted o.g.f. in x for the Narayana polynomials in t, G(x,t)= x/[t-1+1/(1-x)] is the compositional inverse in x. Therefore, with H(x,t)=1/(dG(x,t)/dx)=[t-1+1/(1-x)]^2/{t-[x/(1-x)]^2}, (see A119900), the (n-1)-th Narayana polynomial in t is given by (1/n!)*((H(x,t)*d/dx)^n)x evaluated at x=0, i.e., F(x,t) = exp(x*H(u,t)*d/du) u, evaluated at u = 0. Also, dF(x,t)/dx = H(F(x,t),t). - Tom Copeland, Sep 30 2011
T(n,k) = binomial(n-1,k-1)*binomial(n+1,k)-binomial(n,k-1)*binomial(n,k). - Philippe Deléham, Nov 05 2011
A166360(n-k) = T(n,k) mod 2. - Reinhard Zumkeller, Oct 10 2013
Damped sum of a column, in leading order: lim_{d->0} d^(2k-1) Sum_{N>=k} T(N,k)(1-d)^N=Catalan(n). - Joachim Wuttke, Sep 11 2014
Multiplying the n-th column by n! generates the revert of the unsigned Lah numbers, A089231. - Tom Copeland, Jan 07 2016
Row polynomials: (x - 1)^(n+1)*(P(n+1,(1 + x)/(x - 1)) - P(n-1,(1 + x)/(x - 1)))/((4*n + 2)), n = 1,2,... and where P(n,x) denotes the n-th Legendre polynomial. - Peter Bala, Mar 03 2017
The coefficients of the row polynomials R(n, x) = hypergeom([-n,-n-1], [2], x) generate the triangle based in (0,0). - Peter Luschny, Mar 19 2018
Multiplying the n-th diagonal by n!, with the main diagonal n=1, generates the Lah matrix A105278. With G equal to the infinitesimal generator of A132710, the Narayana triangle equals Sum_{n >= 0} G^n/((n+1)!*n!) = (sqrt(G))^(-1) * I_1(2*sqrt(G)), where G^0 is the identity matrix and I_1(x) is the modified Bessel function of the first kind of order 1. (cf. Sep 21 2011 formula also.) - Tom Copeland, Sep 23 2020
T(n,k) = T(n,k-1)*C(n-k+2,2)/C(k,2). - Yuchun Ji, Dec 21 2020
From Sergii Voloshyn, Nov 25 2024: (Start)
G.f.: F(x,y) = (1-x*(1+y)-sqrt((1-x*(1+y))^2-4*y*x^2))/(2*x) is the solution of the differential equation x^3 * d^2(x*F(x,y))/dx^2 = y * d^2(x*F(x,y))/dy^2.
Let E be the operator x*D*D, where D denotes the derivative operator d/dx. Then (1/(n! (1 + n)!)) * E^n(x/(1 - x)) = (row n generating polynomial)/(1 - x)^(2*n+1) = Sum_{k >= 0} C(n-1, k-1)*C(n, k-1)/k*x^k. For example, when n = 4 we have (1/4!/5!)*E^3(x/(1 - x)) = x (1 + 6 x + 6 x^2 + x^3)/(1 - x)^9. (End)
Extensions
Deleted certain dangerous or potentially dangerous links. - N. J. A. Sloane, Jan 30 2021
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