cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-6 of 6 results.

A126120 Catalan numbers (A000108) interpolated with 0's.

Original entry on oeis.org

1, 0, 1, 0, 2, 0, 5, 0, 14, 0, 42, 0, 132, 0, 429, 0, 1430, 0, 4862, 0, 16796, 0, 58786, 0, 208012, 0, 742900, 0, 2674440, 0, 9694845, 0, 35357670, 0, 129644790, 0, 477638700, 0, 1767263190, 0, 6564120420, 0, 24466267020, 0, 91482563640, 0, 343059613650, 0
Offset: 0

Views

Author

Philippe Deléham, Mar 06 2007

Keywords

Comments

Inverse binomial transform of A001006.
The Hankel transform of this sequence gives A000012 = [1,1,1,1,1,...].
Counts returning walks (excursions) of length n on a 1-d integer lattice with step set {+1,-1} which stay in the chamber x >= 0. - Andrew V. Sutherland, Feb 29 2008
Moment sequence of the trace of a random matrix in G=USp(2)=SU(2). If X=tr(A) is a random variable (A distributed according to the Haar measure on G) then a(n) = E[X^n]. - Andrew V. Sutherland, Feb 29 2008
Essentially the same as A097331. - R. J. Mathar, Jun 15 2008
Number of distinct proper binary trees with n nodes. - Chris R. Sims (chris.r.sims(AT)gmail.com), Jun 30 2010
-a(n-1), with a(-1):=0, n>=0, is the Z-sequence for the Riordan array A049310 (Chebyshev S). For the definition see that triangle. - Wolfdieter Lang, Nov 04 2011
See A180874 (also A238390 and A097610) and A263916 for relations to the general Bell A036040, cycle index A036039, and cumulant expansion polynomials A127671 through the Faber polynomials. - Tom Copeland, Jan 26 2016
A signed version is generated by evaluating polynomials in A126216 that are essentially the face polynomials of the associahedra. This entry's sequence is related to an inversion relation on p. 34 of Mizera, related to Feynman diagrams. - Tom Copeland, Dec 09 2019

Examples

			G.f. = 1 + x^2 + 2*x^4 + 5*x^6 + 14*x^8 + 42*x^10 + 132*x^12 + 429*x^14 + ...
From _Gus Wiseman_, Nov 14 2022: (Start)
The a(0) = 1 through a(8) = 14 ordered binary rooted trees with n + 1 nodes (ranked by A358375):
  o  .  (oo)  .  ((oo)o)  .  (((oo)o)o)  .  ((((oo)o)o)o)
                 (o(oo))     ((o(oo))o)     (((o(oo))o)o)
                             ((oo)(oo))     (((oo)(oo))o)
                             (o((oo)o))     (((oo)o)(oo))
                             (o(o(oo)))     ((o((oo)o))o)
                                            ((o(o(oo)))o)
                                            ((o(oo))(oo))
                                            ((oo)((oo)o))
                                            ((oo)(o(oo)))
                                            (o(((oo)o)o))
                                            (o((o(oo))o))
                                            (o((oo)(oo)))
                                            (o(o((oo)o)))
                                            (o(o(o(oo))))
(End)

References

  • Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Ch. 49, Hemisphere Publishing Corp., 1987.

Links

Crossrefs

Cf. A000108, A036039, A036040, A097610, A127671, A180874, A238390, A263916.
Cf. A126216.
The unordered version is A001190, ranked by A111299.
These trees (ordered binary rooted) are ranked by A358375.
Cf. A000081, A001263, A005043, A032027, A063895, A245824.

Programs

  • Magma
    &cat [[Catalan(n), 0]: n in [0..30]]; // Vincenzo Librandi, Jul 28 2016
  • Maple
    with(combstruct): grammar := { BB = Sequence(Prod(a,BB,b)), a = Atom, b = Atom }: seq(count([BB,grammar], size=n),n=0..47); # Zerinvary Lajos, Apr 25 2007
BB := {E=Prod(Z,Z), S=Union(Epsilon,Prod(S,S,E))}: ZL:=[S,BB,unlabeled]: seq(count(ZL, size=n), n=0..45); # Zerinvary Lajos, Apr 22 2007
BB := [T,{T=Prod(Z,Z,Z,F,F), F=Sequence(B), B=Prod(F,Z,Z)}, unlabeled]: seq(count(BB, size=n+1), n=0..45); # valid for n> 0. # Zerinvary Lajos, Apr 22 2007
seq(n!*coeff(series(hypergeom([],[2],x^2),x,n+2),x,n),n=0..45); # Peter Luschny, Jan 31 2015
# Using function CompInv from A357588.
CompInv(48, n -> ifelse(irem(n, 2) = 0, 0, (-1)^iquo(n-1, 2))); # Peter Luschny, Oct 07 2022
  • Mathematica
    a[n_?EvenQ] := CatalanNumber[n/2]; a[n_] = 0; Table[a[n], {n, 0, 45}] (* Jean-François Alcover, Sep 10 2012 *)
a[ n_] := If[ n < 0, 0, n! SeriesCoefficient[ BesselI[ 1, 2 x] / x, {x, 0, n}]]; (* Michael Somos, Mar 19 2014 *)
bot[n_]:=If[n==1,{{}},Join@@Table[Tuples[bot/@c],{c,Table[{k,n-k-1},{k,n-1}]}]];
Table[Length[bot[n]],{n,10}] (* Gus Wiseman, Nov 14 2022 *)
Riffle[CatalanNumber[Range[0,50]],0,{2,-1,2}] (* Harvey P. Dale, May 28 2024 *)
  • Python
    from math import comb
def A126120(n): return 0 if n&1 else comb(n,m:=n>>1)//(m+1) # Chai Wah Wu, Apr 22 2024
  • Sage
    def A126120_list(n) :
    D = [0]*(n+2); D[1] = 1
    b = True; h = 2; R = []
    for i in range(2*n-1) :
        if b :
            for k in range(h,0,-1) : D[k] -= D[k-1]
            h += 1; R.append(abs(D[1]))
        else :
            for k in range(1,h, 1) : D[k] += D[k+1]
        b = not b
    return R
A126120_list(46) # Peter Luschny, Jun 03 2012

Formula

a(2*n) = A000108(n), a(2*n+1) = 0.
a(n) = A053121(n,0).
(1/Pi) Integral_{0 .. Pi} (2*cos(x))^n *2*sin^2(x) dx. - Andrew V. Sutherland, Feb 29 2008
G.f.: (1 - sqrt(1 - 4*x^2)) / (2*x^2) = 1/(1-x^2/(1-x^2/(1-x^2/(1-x^2/(1-... (continued fraction). - Philippe Deléham, Nov 24 2009
G.f. A(x) satisfies A(x) = 1 + x^2*A(x)^2. - Vladimir Kruchinin, Feb 18 2011
E.g.f.: I_1(2x)/x Where I_n(x) is the modified Bessel function. - Benjamin Phillabaum, Mar 07 2011
Apart from the first term the e.g.f. is given by x*HyperGeom([1/2],[3/2,2], x^2). - Benjamin Phillabaum, Mar 07 2011
a(n) = Integral_{x=-2..2} x^n*sqrt((2-x)*(2+x))/(2*Pi) dx. - Peter Luschny, Sep 11 2011
E.g.f.: E(0)/(1-x) where E(k) = 1-x/(1-x/(x-(k+1)*(k+2)/E(k+1))); (continued fraction). - Sergei N. Gladkovskii, Apr 05 2013
G.f.: 3/2- sqrt(1-4*x^2)/2 = 1/x^2 + R(0)/x^2, where R(k) = 2*k-1 - x^2*(2*k-1)*(2*k+1)/R(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 28 2013 (warning: this is not the g.f. of this sequence, R. J. Mathar, Sep 23 2021)
G.f.: 1/Q(0), where Q(k) = 2*k+1 + x^2*(1-4*(k+1)^2)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Jan 09 2014
a(n) = n!*[x^n]hypergeom([],[2],x^2). - Peter Luschny, Jan 31 2015
a(n) = 2^n*hypergeom([3/2,-n],[3],2). - Peter Luschny, Feb 03 2015
a(n) = ((-1)^n+1)*2^(2*floor(n/2)-1)*Gamma(floor(n/2)+1/2)/(sqrt(Pi)* Gamma(floor(n/2)+2)). - Ilya Gutkovskiy, Jul 23 2016
D-finite with recurrence (n+2)*a(n) +4*(-n+1)*a(n-2)=0. - R. J. Mathar, Mar 21 2021
From Peter Bala, Feb 03 2024: (Start)
a(n) = 2^n * Sum_{k = 0..n} (-2)^(-k)*binomial(n, k)*Catalan(k+1).
G.f.: 1/(1 + 2*x) * c(x/(1 + 2*x))^2 = 1/(1 - 2*x) * c(-x/(1 - 2*x))^2 = c(x^2), where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the g.f. of the Catalan numbers A000108. (End)

Extensions

An erroneous comment removed by Tom Copeland, Jul 23 2016

A097610 Triangle read by rows: T(n,k) is number of Motzkin paths of length n and having k horizontal steps.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 0, 3, 0, 1, 2, 0, 6, 0, 1, 0, 10, 0, 10, 0, 1, 5, 0, 30, 0, 15, 0, 1, 0, 35, 0, 70, 0, 21, 0, 1, 14, 0, 140, 0, 140, 0, 28, 0, 1, 0, 126, 0, 420, 0, 252, 0, 36, 0, 1, 42, 0, 630, 0, 1050, 0, 420, 0, 45, 0, 1, 0, 462, 0, 2310, 0, 2310, 0, 660, 0, 55, 0, 1, 132, 0, 2772, 0
Offset: 0

Views

Author

Emeric Deutsch, Aug 30 2004

Keywords

Comments

Row sums are the Motzkin numbers (A001006). Column 0 gives the aerated Catalan numbers (A000108).
Let P_n(x) = Sum_{k=0..n} T(n,k)*x^k. P_0(x) = 1, P_1(x) = x, P_n(x) = x*P_(n-1)(x) + Sum_{j=0..n-2} P_j(x)*P_(n-2-j)(x); essentially the same as A124027. - Philippe Deléham, Oct 03 2007
G. J. Chaitin's numbers of s-expressions of size n are given by the coefficients of polynomials p(k, x) satisfying: p(k, x) = Sum_{j=2..k-1} p(j, x)*p(k-j, x). The coefficients of these polynomials also give (essentially) the triangle shown here. - Roger L. Bagula, Oct 31 2006
Exponential Riordan array [Bessel_I(1,2x)/x,x]. - Paul Barry, Mar 24 2010
Diagonal sums are the aerated large Schroeder numbers. - Paul Barry, Apr 21 2010
Non-vanishing antidiagonals are rows of A060693. - Tom Copeland, Feb 03 2016
These polynomials are related to the Gegenbauer polynomials which in turn are specializations of the Jacobi polynomials. The o.g.f. of the Gegenbauer polynomials is 1 / [1-2tx+x^2]^a. For the generating function Gb(x,h1,h2,a) = [x / (1 + h1 x + h2 x^2)]^a, the compositional inverse in x is Gbinv(x,h1,h2,a) = [(1-h1*y) - sqrt[(1-h1*y)^2-4h2*y^2]]/(2*h2*y) with y = x^(1/a). The polynomials of this entry are generated by Gbinv(x,t,1,1). The Legendre polynomials are related to the o.g.f. Gb(x,-2t,1,1/2). Cf. A121448. - Tom Copeland, Feb 07 2016
The bivariate o.g.f. in Copeland's Jan 29 2016 formulas can be related to conformal mappings of the complex plane and a solution of the dKP hierarchy. Cf. p. 24 of the Takebe et al. paper. - Tom Copeland, May 30 2018

Examples

			Triangle begins:
1;
0,  1;
1,  0,  1;
0,  3,  0,  1;
2,  0,  6,  0,  1;
0, 10,  0, 10,  0,  1;
5,  0, 30,  0, 15,  0,  1;
Row n has n+1 terms.
T(4,2)=6 because we have HHUD, HUDH, UDHH, HUHD, UHDH, UHHD, where U=(1,1), D=(1,-1) and H=(1,0).

References

  • G. J. Chaitin, Algorithmic Information Theory, Cambridge Univ. Press, 1987, page 169.

Links

Crossrefs

Cf. A001006, A000108. A124027 is an essentially identical triangle.
Cf. A011973, A049310, A055151, A060693, A180874, A121448, A125181, A126120, A134264, A145271, A238390.
Cf. A001263.

Programs

  • Maple
    G:=(1-t*z-sqrt(1-2*t*z+t^2*z^2-4*z^2))/2/z^2:
Gser:=simplify(series(G,z=0,16)): P[0]:=1:
for n from 1 to 13 do P[n]:=sort(coeff(Gser,z^n)) od:
seq(seq(coeff(t*P[n],t^k),k=1..n+1),n=0..13);
# Maple program for the triangular array:
T:=proc(n,k) if n-k mod 2 = 0 and k<=n then n!/k!/((n-k)/2)!/((n-k)/2+1)! else 0 fi end: TT:=(n,k)->T(n-1,k-1): matrix(10,10,TT);
  • Mathematica
    T[n_,k_]:=If[n>=k&&EvenQ[n-k],n!/(k!((n-k)/2)!((n-k)/2+1)!),0];
Flatten[Table[T[n,k],{n,0,20},{k,0,n}]] (* Peter J. C. Moses, Apr 06 2013 *)
T[n_,k_] := If[OddQ[n - k], 0, Binomial[n, k] CatalanNumber[(n - k)/2]]; (* Peter Luschny, Jun 06 2018 *)

Formula

G.f.: [1-tz-sqrt(1-2tz+t^2*z^2-4z^2)]/(2z^2).
T(n, k) = n!/[k!((n-k)/2)!((n-k)/2-1)! ] = A055151(n, (n-k)/2) if n-k is a nonnegative even number; otherwise T(n, k) = 0.
T(n, k) = C(n, k)*C((n-k)/2)*(1+(-1)^(n-k))/2 if k <= n, 0 otherwise. - Paul Barry, May 18 2005
T(n,k) = A121448(n,k)/2^k. - Philippe Deléham, Aug 17 2006
Sum_{k=0..n} T(n,k)*2^k = A000108(n+1). - Philippe Deléham, Aug 22 2006
Sum_{k=0..n} T(n,k)*3^k = A002212(n+1). - Philippe Deléham, Oct 03 2007
G.f.: 1/(1-x*y-x^2/(1-x*y-x^2/(1-x*y-x^2/.... (continued fraction). - Paul Barry, Dec 15 2008
Sum_{k=0..n} T(n,k)*4^k = A005572(n). - Philippe Deléham, Dec 03 2009
T(n,k) = A007318(n,k)*A126120(n-k). - Philippe Deléham, Dec 12 2009
From Tom Copeland, Jan 23 2016: (Start)
E.g.f.: M(x,t) = e^(xt) AC(t) = e^(xt) I_1(2t)/t = e(xt) * e.g.f.(A126120(t)) = e^(xt) Sum_{n>=0} t^(2n)/(n!(n+1)!) = exp[t P(.,x)].
The e.g.f. of this Appell sequence of polynomials P(n,x) is e^(xt) times the e.g.f. AC(t) of the aerated Catalan numbers A126120. AC(t) = I_1(2t)/t, where I_n(x) = T_n(d/dx) I_0(x) are the modified Bessel functions of the first kind and T_n, the Chebyshev polynomials of the first kind.
P(n,x) has the lowering and raising operators L = d/dx = D and R = d/dD log{M(x,D)} = x + d/dD log{AC(D)} = x + Sum_{n>=0} c(n) D^(2n+1)/(2n+1)! with c(n) = (-1)^n A180874(n+1), i.e., L P(n,x) = n P(n-1,x) and R P(n,x) = P(n+1,x).
(P(.,x) + y)^n = P(n,x+y) = Sum_{k=0..n} binomial(n,k) P(k,x) y^(n-k) = (b.+x+y)^n, where (b.)^k = b_k = A126120(k).
Exp(b.D) e^(xt) = exp[(x+b.)t] = exp[P(.,x)t] = e^(b.t) e^(xt) = e^(xt) AC(t).
See p. 12 of the Alexeev et al. link and A055151 for a refinement.
Shifted o.g.f: G(x,t) = [1-tx-sqrt[(1-tx)^2-4x^2]] / 2x = x + t x^2 + (1+t) x^3 + ... has the compositional inverse Ginv(x,t) = x / [1 + tx + x^2] = x - t x^2 +(-1+t^2) x^3 + (2t-t^3) x^4 + (1-3t^2+t^4) x^5 + ..., a shifted o.g.f. for the signed Chebyshev polynomials of the second kind of A049310 (cf. also the Fibonacci polynomials of A011973). Then the inversion formula of A134264, involving non-crossing partitions and free probability with their multitude of interpretations (cf. A125181 also), can be used with h_0 = 1, h_1 = t, and h_2 = 1 to interpret the coefficients of the Motzkin polynomials combinatorially.
(End)
From Tom Copeland, Jan 29 2016: (Start)
Provides coefficients of the inverse of f(x) = x / [1 + h1 x + h2 x^2], a bivariate generating function of A049310 (mod signs).
finv(x) = [(1-h1*x) - sqrt[(1-h1*x)^2-4h2*x^2]]/(2*h2*x) = x + h1 x^2 + (h2 + h1^2) x^3 + (3 h1 h2 + h1^3) x^4 + ... is a bivariate o.g.f. for this entry.
The infinitesimal generator for finv(x) is g(x) d/dx with g(x) = 1 /[df(x)/dx] = x^2 / [(f(x))^2 (1 - h2 x^2)] = (1 + h1 x + h2 x^2)^2 / (1 - h2 x^2) so that [g(x)d/dx]^n/n! x evaluated at x = 0 gives the row polynomials FI(n,h1,h2) of the compositional inverse of f(x), i.e., exp[x g(u)d/du] u |_(u=0) = finv(x) = 1 / [1 -x FI(.,h1,h2)]. Cf. A145271. E.g.,
FI(0,h1,h2) = 0
FI(1,h1,h2) = 1
FI(2,h1,h2) = 1 h1
FI(3,h1,h2) = 1 h2 + 1 h1^2
FI(4,h1,h2) = 3 h2 h1 + 1 h1^3
FI(5,h1,h2) = 2 h2^2 + 6 h2 h1^2 + 1 h1^4
FI(6,h1,h2) = 10 h2^2 h1 + 10 h2 h1^3 + 1 h1^5.
And with D = d/dh1, FI(n+1, h1,h2) = MT(n,h1,h2) = (b.y + h1)^n = Sum_{k=0..n} binomial(n,k) b(k) y^k h1^(n-k) = exp[(b.y D] (h1)^n = AC(y D) (h1)^n, where b(k) = A126120(k), y = sqrt(h2), and AC(t) is defined in my Jan 23 formulas above. Equivalently, AC(y D) e^(x h1) = exp[x MT(.,h1,h2)].
The MT polynomials comprise an Appell sequence in h1 with e.g.f. e^(h1*x) AC(xy) = exp[x MT(.,h1,h2)] with lowering operator L = d/dh1 = D, i.e., L MT(n,h1,h2) = dMT(n,h1,h2)/dh1 = n MT(n-1,h1,h2) and raising operator R = h1 + dlog{AC(y L)}/dL = h1 + Sum_{n>=0} c(n) h2^(n+1) D^(2n+1)/(2n+1)! = h1 + h2 d/dh1 - h2^2 (d/dh1)^3/3! + 5 h2^3 (d/dh1)^5/5! - ... with c(n) = (-1)^n A180874(n+1) (consistent with the raising operator in the Jan 23 formulas).
The compositional inverse finv(x) is also obtained from the non-crossing partitions of A134264 (or A125181) with h_0 = 1, h_1 = h1, h_2 = h2, and h_n = 0 for all other n.
See A238390 for the umbral compositional inverse in h1 of MT(n,h1,h2) and inverse matrix.
(End)
From Tom Copeland, Feb 13 2016: (Start)
z1(x,h1,h2) = finv(x), the bivariate o.g.f. above for this entry, is the zero that vanishes for x=0 for the quadratic polynomial Q(z;z1(x,h1,h2),z2(x,h1,h2)) = (z-z1)(z-z2) = z^2 - (z1+z2) z + (z1*z2) = z^2 - e1 z + e2 = z^2 - [(1-h1*x)/(h2*x)] z + 1/h2, where e1 and e2 are the elementary symmetric polynomials for two indeterminates.
The other zero is given by z2(x,h1,h2) = (1 - h1*x)/(h2*x) - z1(x,h1,h2) = [1 - h1*x + sqrt[(1-h1*x)^2 - 4 h2*x^2]] / (2h2*x).
The two are the nontrivial zeros of the elliptic curve in Legendre normal form y^2 = z (z-z1)(z-z2), (see Landweber et al., p. 14, Ellingsrud, and A121448), and the zeros for the Riccati equation z' = (z - z1)(z - z2), associated to soliton solutions of the KdV equation (see Copeland link).
(End)
Comparing the shifted o.g.f. S(x) = x / (1 - h_1 x + h_2 x^2) for the bivariate Chebyshev polynomials S_n(h_1,h_2) of A049310 with the shifted o.g.f. H(x) = x / ((1 - a x)(1 - b x)) for the complete homogeneous symmetric polynomials H_n(a,b) = (a^(n+1)-b^(n+1)) / (a - b) shows that S_n(h_1,h_2) = H_n(a,b) for h_1 = a + b and h_2 = ab and, conversely, a = (h_1 + sqrt(h_1^2 - 4 h_2)) / 2 and b = (h_1 - sqrt(h_1^2 - 4 h_2)) / 2. The compositional inverse about the origin of S(x) gives a bivariate o.g.f. for signed Motzkin polynomials M_n(h_1,h_2) of this entry, and that of H(x) gives one for signed Narayana polynomials N_n(a,b) of A001263, thereby relating the bivariate Motzkin and Narayana polynomials by the indeterminate transformations. E.g., M_2(h_1,h_2) = h_2 + h_1^2 = ab + (a + b)^2 = a^2 + 3 ab + b^2 = N_2(a,b). - Tom Copeland, Jan 27 2024

A115369 Decimal expansion of first zero of BesselJ(1,z).

Original entry on oeis.org

3, 8, 3, 1, 7, 0, 5, 9, 7, 0, 2, 0, 7, 5, 1, 2, 3, 1, 5, 6, 1, 4, 4, 3, 5, 8, 8, 6, 3, 0, 8, 1, 6, 0, 7, 6, 6, 5, 6, 4, 5, 4, 5, 2, 7, 4, 2, 8, 7, 8, 0, 1, 9, 2, 8, 7, 6, 2, 2, 9, 8, 9, 8, 9, 9, 1, 8, 8, 3, 9, 3, 0, 9, 5, 1, 9, 0, 1, 1, 4, 7, 0, 2, 1, 4, 1, 1, 2, 8, 7, 4, 7, 5, 7, 4, 2, 3, 1, 2, 6, 7, 2, 4, 4, 7
Offset: 1

Views

Author

Eric W. Weisstein, Jan 21 2006

Keywords

Comments

Also the first root of the sinc(2,x) function, that is, the radial component of the 2D Fourier transform of a 2-dimensional unit disc. - Stanislav Sykora, Nov 14 2013
Also the first root of the derivative of BesselJ_0. - Jean-François Alcover, Jul 01 2015

Examples

			3.8317059702075123156...

Links

Crossrefs

Cf. A115368, A115370, A115371, A115372, A115373.
Cf. A103365, A103366, A238390, A180874.

Programs

A180874 Lassalle's sequence connected with Catalan numbers and Narayana polynomials.

Original entry on oeis.org

1, 1, 5, 56, 1092, 32670, 1387815, 79389310, 5882844968, 548129834616, 62720089624920, 8646340208462880, 1413380381699497200, 270316008395632253340, 59800308109377016336155, 15151722444639718679892150, 4359147487054262623576455600
Offset: 1

Views

Author

Jonathan Vos Post, Sep 22 2010

Keywords

Comments

Defined by the recurrence formula in Theorem 1, page 2 of Lasalle.
From Tom Copeland, Jan 26 2016: (Start)
Let G(t) = Sum_{n>=0} t^(2n)/(n!(n+1)!) = exp(c.t) be the e.g.f. of the aerated Catalan numbers c_n of A126120.
R = x + H(D) = x + d/dD log[G(D)] = x + D - D^3/3! + 5 D^5/5! - 56 D^7/7! + ... = x + e^(r. D) generates a signed, aerated version of this entry's sequence a(n), (r.)^(2n+1) = r(2n+1) = (-1)^n a(n+1) for n>=0 and r(0) = a(0) = 0, and is, with D = d/dx, the raising operator for the Appell polynomials P(n,x) of A097610, where P(n,x) = (c. + x)^n = Sum{k=0 to n} binomial(n,k) c_k x^(n-k) with c_k = A126120(k), i.e., R P(n,x) = P(n+1,x).
d/dt log[G(t)] = e^(r.t) = e^(q.t) / e^(c.t) = Ev[c. e^(c.t)] / Ev[e^(c.t)] = e^(q.t) e^(d.t) = [Sum_{n>=0} 2n t^(2n-1)/(n!(n+1)!)] / [Sum_{n>=0} t^(2n)/(n!(n+1)!)] with Ev[..] denoting umbral evaluation, so q(n) = c(n+1) = A126120(n+1) and d(2n) = (-1)^n A238390(n) and vanishes otherwise. Then (r. + c.)^n = q(n) = Sum_{k=0..n} binomial(n,k) r(k) c(n-k) and (q. + d.)^n = r(n), relating A180874, A126120 (A000108), and A238390 through binomial convolutions.
The sequence can also be represented in terms of the Faber polynomials of A263916 as a(n) = |(2n-1)! F(2n,0,b(2),0,b(4),0,..)| = |h(2n)| where b(2n) = 1/(n!(n + 1)!) = A126120(2n)/(2n)! = A000108(n)/(2n)!, giving h(0) = 1, h(1) = 0, h(2) = 1, h(3) = 0, h(4) = -1, h(5) = 0, h(6) = 5, h(7) = 0, h(8) = -56, ..., implying, among other relations, that A000108(n/2)= A126120(n) = Bell(n,0,h(2),0,h(4),...), the Bell polynomials of A036040 which reduce to A257490 in this case.
(End)
From Colin Defant, Sep 06 2018: (Start)
a(n) is the number of pairs (rho,r), where rho is a matching on [2n] and r is an acyclic orientation of the crossing graph of rho in which the block containing 1 is the only source (see the Josuat-Verges paper or the Defant-Engen-Miller paper for definitions).
a(n) is the number of permutations of [2n-1] that have exactly 1 preimage under West's stack-sorting map.
a(n) is the number of valid hook configurations of permutations of [2n-1] that have n-1 hooks (see the paper by Defant, Engen, and Miller for definitions).
Say a binary tree is full if every vertex has either 0 or 2 children. If u is a left child in such a tree, then we can start at the sibling of u and travel down left edges until reaching a leaf v. Call v the leftmost nephew of u. A decreasing binary plane tree on [m] is a binary plane tree labeled with the elements of [m] in which every nonroot vertex has a label that is smaller than the label of its parent. a(n) is the number of full decreasing binary plane trees on [2n-1] in which every left child has a label that is larger than the label of its leftmost nephew.
(End)

Links

Crossrefs

Cf. A000108, A001263, A188664, A115369.
Cf. A036040, A097610, A126120, A238390, A263916, A257490.

Programs

  • Maple
    A000108 := proc(n) binomial(2*n,n)/(1+n) ; end proc:
A180874 := proc(n) option remember; if n = 1 then 1; else A000108(n)+add((-1)^j*binomial(2*n-1,2*j-1)*procname(j)*A000108(n-j),j=1..n-1) ;   %*(-1)^(n-1) ; end if; end proc: # R. J. Mathar, Apr 16 2011
  • Mathematica
    nmax=20; a = ConstantArray[0,nmax]; a[[1]]=1; Do[a[[n]] = (-1)^(n-1)*(Binomial[2*n,n]/(n+1) + Sum[(-1)^j*Binomial[2n-1,2j-1]*a[[j]]* Binomial[2*(n-j),n-j]/(n-j+1),{j,1,n-1}]),{n,2,nmax}]; a (* Vaclav Kotesovec, Feb 28 2014 *)

Formula

a(n) = (-1)^(n-1) * (C(n)+Sum_{j=1..n-1} (-1)^j *binomial(2n-1,2j-1) * a(j) *C(n-j)), where C() = A000108(). - R. J. Mathar, Apr 17 2011, corrected by Vaclav Kotesovec, Feb 28 2014
E.g.f.: Sum_{k>=0} a(k)*x^(2*k+2)/(2*k+2)! = log(x/BesselJ(1,2*x)). - Sergei N. Gladkovskii, Dec 28 2011
a(n) ~ (n!)^2 / (sqrt(Pi) * n^(3/2) * r^n), where r = BesselJZero[1, 1]^2/16 = 0.917623165132743328576236110539381686855099186384686... - Vaclav Kotesovec, added Feb 28 2014, updated Mar 01 2014
Define E(m,n) by E(1,1) = 1, E(n,n) = 0 for n > 1, and E(m,n) = Sum_{j=1..m} Sum_{i=1..n-m-1} binomial(n-m-1,i-1) * F_j(i+j-1) * F_{m-j}(n-j-i) for 0 <= m < n, where F_m(n) = Sum_{j=m..n} E_j(n). Then a(n) = F_0(2n-1). - Colin Defant, Sep 06 2018

A103365 First column of triangle A103364, which equals the matrix inverse of the Narayana triangle (A001263).

Original entry on oeis.org

1, -1, 2, -7, 39, -321, 3681, -56197, 1102571, -27036487, 810263398, -29139230033, 1238451463261, -61408179368043, 3513348386222286, -229724924077987509, 17023649385410772579, -1419220037471837658603, 132236541042728184852942, -13690229149108218523467549
Offset: 1

Views

Author

Paul D. Hanna, Feb 02 2005

Keywords

Examples

			From _Paul D. Hanna_, Jan 31 2009: (Start)
G.f.: A(x) = 1 - x + 2*x^2/3 - 7*x^3/18 + 39*x^4/180 - 321*x^5/2700 +...
G.f.: A(x) = 1/B(x) where:
B(x) = 1 + x + x^2/3 + x^3/18 + x^4/180 + x^5/2700 +...+ x^n/[n!*(n+1)!/2^n] +... (End)

Links

Crossrefs

Cf. A001263, A103364, A103366, A103367.
Cf. A155926, A155927. [Paul D. Hanna, Jan 31 2009]
Cf. A238390, A180874, A115369.

Programs

  • Mathematica
    Table[(-1)^((n-1)/2) * (CoefficientList[Series[x/BesselJ[1,2*x],{x,0,40}],x])[[n]] * ((n+1)/2)! * ((n-1)/2)!,{n,1,41,2}] (* Vaclav Kotesovec, Mar 01 2014 *)
  • PARI
    a(n)=if(n<1,0,(matrix(n,n,m,j,binomial(m-1,j-1)*binomial(m,j-1)/j)^-1)[n,1])
  • PARI
    {a(n)=local(B=sum(k=0,n,x^k/(k!*(k+1)!/2^k))+x*O(x^n));polcoeff(1/B,n)*n!*(n+1)!/2^n} \\ Paul D. Hanna, Jan 31 2009

Formula

From Paul D. Hanna, Jan 31 2009: (Start)
G.f.: A(x) = 1/B(x) where A(x) = Sum_{n>=0} (-1)^n*a(n)*x^n/[n!*(n+1)!/2^n] and B(x) = Sum_{n>=0} x^n/[n!*(n+1)!/2^n].
G.f. satisfies: A(x) = 1/F(x*A(x)) and F(x) = 1/A(x*F(x)) where F(x) = Sum_{n>=0} A155926(n)*x^n/[n!*(n+1)!/2^n].
G.f. satisfies: A(x) = 1/G(x/A(x)) and G(x) = 1/A(x/G(x)) where G(x) = Sum_{n>=0} A155927(n)*x^n/[n!*(n+1)!/2^n]. (End)
a(n) ~ (-1)^(n+1) * c * n! * (n-1)! * d^n, where d = 4/BesselJZero[1, 1]^2 = 0.2724429913055159309179376055957891881897555639652..., and c = 9.11336321311226744479181866135367355200240221549667284076... = BesselJZero[1, 1]^2 / (4*BesselJ[2, BesselJZero[1, 1]]). - Vaclav Kotesovec, Mar 01 2014, updated Apr 01 2018

A280580 Triangle read by rows: T(n,k) = binomial(2*n,2*k)*binomial(2*n-2*k,n-k)/(n+1-k) for 0<=k<=n.

Original entry on oeis.org

1, 1, 1, 2, 6, 1, 5, 30, 15, 1, 14, 140, 140, 28, 1, 42, 630, 1050, 420, 45, 1, 132, 2772, 6930, 4620, 990, 66, 1, 429, 12012, 42042, 42042, 15015, 2002, 91, 1, 1430, 51480, 240240, 336336, 180180, 40040, 3640, 120, 1, 4862, 218790, 1312740, 2450448, 1837836, 612612, 92820, 6120, 153, 1
Offset: 0

Views

Author

Werner Schulte, Jan 05 2017

Keywords

Examples

			Triangle begins:
n\k:     0      1       2       3       4      5     6    7  8  . . .
  0:     1
  1:     1      1
  2:     2      6       1
  3:     5     30      15       1
  4:    14    140     140      28       1
  5:    42    630    1050     420      45      1
  6:   132   2772    6930    4620     990     66     1
  7:   429  12012   42042   42042   15015   2002    91    1
  8:  1430  51480  240240  336336  180180  40040  3640  120  1
  etc.
T(3,2) = binomial(6,4) * binomial(2,1) / (3+1-2) = 15 * 2 / 2 = 15. - _Indranil Ghosh_, Feb 15 2017

Links

Crossrefs

Row sums are A026945.
Triangle related to A000108, A001006, A001263, and A039599.
Cf. A000007, A001003, A005568, A103365, A238390.

Formula

T(n,k) = A001263(n+1,k+1)*A000108(n)/A000108(k) for 0 <= k <= n.
T(n,k) = binomial(2*n,2*k)*A000108(n-k) for 0 <= k <= n.
T(n,k) = A039599(n,k)*binomial(n+1+k,2*k+1)/(n+1-k) for 0 <= k <= n.
Recurrences: T(n,0) = A000108(n) and (1) T(n,k) = T(n,k-1)*(n+1-k)*(n+2-k)/ (2*k*(2*k-1)) for 0 < k <= n, (2) T(n,k) = T(n-1,k-1)*n*(2*n-1)/(k*(2*k-1)).
The row polynomials p(n,x) = Sum_{k=0..n} T(n,k)*x^(2*k) satisfy the recurrence equation p"(n,x) = 2*n*(2*n-1)*p(n-1,x) with initial value p(0,x) = 1 ( n > 0, p" is the second derivative of p ), and Sum_{n>=0} p(n,x)*t^(2*n)/((2*n)!) = cosh(x*t)*(Sum_{n>=0} A000108(n)*t^(2*n)/((2*n)!)).
Conjectures: (1) Sum_{k=0..n} (-1)^k*T(n,k)*A238390(k) = A000007(n);
(2) Antidiagonal sums equal A001003(n);
(3) Matrix inverse equals T(n,k)*A103365(n+1-k).
Sum_{k=0..n} (n+1-k)*T(n,k) = A002426(2*n) = A082758(n).
Sum_{k=0..n} T(n,k)*A000108(k) = A000108(n)*A000108(n+1) = A005568(n).
Matrix product: Sum_{i=0..n} T(n,i)*T(i,k) = T(n,k)*A000108(n+1-k) for 0<=k<=n.
T(n,k) = A097610(2*n,2*k) for 0 <= k <= n.
Sum_{k=0..n} (k+1)*T(n,k)*A000108(k) = binomial(2*n+1,n)*A000108(n).
Showing 1-6 of 6 results.

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