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.

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A114590 Number of peaks at even levels in all hill-free Dyck paths of semilength n+2 (a hill in a Dyck path is a peak at level 1).

Original entry on oeis.org

1, 2, 8, 28, 103, 382, 1432, 5408, 20546, 78436, 300636, 1156188, 4459267, 17241526, 66807856, 259361920, 1008598126, 3928120924, 15319329472, 59817190552, 233826979750, 914962032172, 3583556424208, 14047386554368, 55108441878868
Offset: 0

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Author

Emeric Deutsch, Dec 11 2005

Keywords

Examples

			a(1)=2 because in the 2 (=A000957(4)) hill-free Dyck paths of semilength 3, namely UUUDDD and U(UD)(UD)D (U=(1,1), D=(1,-1)) we have altogether 2 peaks at even level (shown between parentheses).

Links

Crossrefs

Cf. A114588, A114587, A114515.

Programs

  • Maple
    G:=(1+2*z^2-(1+2*z)*sqrt(1-4*z))/2/z^2/(2+z)^2/sqrt(1-4*z): Gser:=series(G,z=0,30): 1, seq(coeff(Gser,z^n),n=1..25);
  • Mathematica
    CoefficientList[Series[(1+2*x^2-(1+2*x)*Sqrt[1-4*x])/2/x^2/(2+x)^2/Sqrt[1-4*x], {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 12 2014 *)

Formula

G.f.: (1+2*z^2-(1+2*z)*sqrt(1-4*z))/(2*z^2*(2+z)^2*sqrt(1-4*z)).
a(n) = sum(k*A114588(n+2,k),k=0..n+1).
a(n)=sum{k=0..n, sum{j=0..n-k, C(n-j,k-j)*C(n-j,k)*(j+1)}}; - Paul Barry, Nov 03 2006
Conjecture: 2*(n+2)*a(n) +(-7*n-9)*a(n-1) -18*a(n-2) +2*(-7*n+19)*a(n-3) +4*(-2*n+3)*a(n-4)=0. - R. J. Mathar, Nov 15 2012
Recurrence: 2*n*(n+2)*(3*n+1)*a(n) = (21*n^3 + 34*n^2 + n - 8)*a(n-1) + 2*(n+1)*(2*n+1)*(3*n+4)*a(n-2). - Vaclav Kotesovec, Feb 12 2014
a(n) ~ 4^(n+2) / (9*sqrt(Pi*n)). - Vaclav Kotesovec, Feb 12 2014

A114586 Triangle read by rows: T(n,k) is the number of hill-free Dyck paths of semilength n and having k peaks at odd levels (0<=k<=n-2; n>=2). A hill in a Dyck path is a peak at level 1.

Original entry on oeis.org

1, 1, 1, 3, 2, 1, 6, 8, 3, 1, 15, 22, 15, 4, 1, 36, 68, 52, 24, 5, 1, 91, 198, 191, 100, 35, 6, 1, 232, 586, 651, 425, 170, 48, 7, 1, 603, 1718, 2203, 1656, 820, 266, 63, 8, 1, 1585, 5047, 7285, 6299, 3591, 1435, 392, 80, 9, 1, 4213, 14808, 23832, 23164, 15155, 6972, 2338
Offset: 2

Views

Author

Emeric Deutsch, Dec 11 2005

Keywords

Comments

Row sums are the Fine numbers (A000957). Column 0 yield the Riordan numbers (A005043). Sum(k*T(n,k),k=0..n-2)=A114587(n).

Examples

			T(5,2)=3 because we have UU(UD)DU(UD)DD, UUDU(UD)(UD)DD and UU(UD)(UD)DUDD, where U=(1,1), D=(1,-1) (the peaks at odd levels are shown between parentheses).
Triangle begins:
1;
1,1;
3,2,1;
6,8,3,1;
15,22,15,4,1;

Crossrefs

Cf. A000957, A005043, A114587, A114588, A100754.

Programs

  • Maple
    G:=(t*z+z+1-sqrt(z^2*t^2+2*z^2*t-2*z*t-3*z^2-2*z+1))/2/z/(1+t+z)-1: Gser:=simplify(series(G,z=0,15)): for n from 2 to 12 do P[n]:=coeff(Gser,z^n) od: for n from 2 to 12 do seq(coeff(t*P[n],t^j),j=1..n-1) od; # yields sequence in triangular form

Formula

G.f.=G-1, where G=G(t, z) satisfies z(1+t+z)G^2-(1+z+tz)G+1=0.

A114589 Number of hill-free Dyck paths of semilength n+3 and having no peaks at even levels (a hill in a Dyck path is a peak at level 1).

Original entry on oeis.org

1, 1, 3, 7, 17, 43, 110, 286, 753, 2003, 5376, 14540, 39589, 108427, 298512, 825664, 2293271, 6393539, 17885835, 50191175, 141247519, 398537101, 1127203038, 3195229662, 9076078057, 25830193513, 73643406563, 210312889095
Offset: 0

Views

Author

Emeric Deutsch, Dec 11 2005

Keywords

Comments

Column 0 of A114588. The number of hill-free Dyck paths having no peaks at odd level are given by the Riordan numbers (A005043).
From Paul Barry, Jul 05 2009: (Start)
The sequence 1,0,0,1,1,3,7,...
has g.f. ((1+x)*(1+2*x)-sqrt((1+x)*(1-3*x)))/(2*x*(2+2*x+x^2)).
It is the inverse binomial transform of A035929(n+1). (End)

Examples

			a(2)=3 because we have UUUDDUUDDD, UUUDUDUDDD and UUUUUDDDDD, where U=(1,1), D=(1,-1).

Links

Crossrefs

Cf. A114588, A005043.

Programs

  • Maple
    G:=(1-z-2*z^2-2*z^3-sqrt(1-3*z^2-2*z))/2/z^4/(2+2*z+z^2): Gser:=series(G,z=0,35): 1, seq(coeff(Gser,z^n),n=1..30);
  • Mathematica
    CoefficientList[Series[(1-x-2*x^2-2*x^3-Sqrt[1-3*x^2-2*x])/2/x^4 /(2+2*x+x^2), {x, 0, 20}], x] (* Vaclav Kotesovec, Mar 20 2014 *)
  • PARI
    x='x+('x^50); Vec((1-x-2*x^2-2*x^3-sqrt(1-3*x^2-2*x))/(2*x^4*(2+2*x+x^2))) \\ G. C. Greubel, Mar 17 2017

Formula

G.f.: (1 -z -2*z^2 -2*z^3 -sqrt(1-3*z^2-2*z))/(2*z^4*(2+2*z+z^2)).
a(n) ~ 3^(n+11/2) / (50*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Mar 20 2014
Conjecture: 2*(n+4)*a(n) +2*(-n-1)*a(n-1) +3*(-3*n-4)*a(n-2) +(-8*n-11)*a(n-3) +3*(-n-1)*a(n-4)=0. - R. J. Mathar, Jul 02 2018
Showing 1-3 of 3 results.

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