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 1-3 of 3 results.

A145904 Square array read by antidiagonals: Hilbert transform of the Narayana numbers A001263.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 6, 5, 1, 1, 10, 16, 7, 1, 1, 15, 40, 31, 9, 1, 1, 21, 85, 105, 51, 11, 1, 1, 28, 161, 295, 219, 76, 13, 1, 1, 36, 280, 721, 771, 396, 106, 15, 1, 1, 45, 456, 1582, 2331, 1681, 650, 141, 17, 1
Offset: 0

Views

Author

Peter Bala, Oct 31 2008

Keywords

Comments

Refer to A145905 for the definition of the Hilbert transform of a lower triangular array. For the Hilbert transform of A008459, the array of type B Narayana numbers, see A108625.
This seems to be a duplicate of A273350. - Alois P. Heinz, Jun 04 2016. This could probably be proved by showing that the g.f.s are the same. - N. J. A. Sloane, Jul 02 2016

Examples

			The array begins
n\k|..0.....1.....2.....3.....4.....5
=====================================
0..|..1.....1.....1.....1.....1.....1
1..|..1.....3.....5.....7.....9....11
2..|..1.....6....16....31....51....76
3..|..1....10....40...105...219...396
4..|..1....15....85...295...771..1681
5..|..1....21...161...721..2331..6083
...
Row 2: (1 + 3x + x^2)/(1 - x)^3 = 1 + 6x + 16x^2 + 31x^3 + ... .
Row 3: (1 + 6x + 6x^2 + x^3)/(1 - x)^4 = 1 + 10x + 40x^2 + 105x^3 + ... .

Crossrefs

Cf. A001263, A005891 (row 2), A063490 (row 3), A108625 (Hilbert transform of h-vectors of type B associahedra).
Cf. also A273350.

Programs

  • Mathematica
    Table[1/(# + 1)*Sum[Binomial[# + 1, i - 1] Binomial[# + 1, i] Binomial[# + k - i + 1, k + 1 - i], {i, 0, k + 1}] &[m - k], {m, 0, 9}, {k, 0, m}] // Flatten (* Michael De Vlieger, Jan 15 2018 *)
  • Maxima
    taylor(((y-1)*sqrt(((x^2+2*x+1)*y-x^2+2*x-1)/(y-1))+(-x-1)*y-x+1)/(2*x^2*y),x,0,10,y,0,10);
T(n,m,k):=1/(n+1)*sum(binomial(n+1,i-1)*binomial(n+1,i)*binomial(n+m-i+1,m+1-i),i,0,m+1); /* Vladimir Kruchinin, Jan 15 2018 */

Formula

Row n generating function: 1/(n+1) * 1/(1-x) * Jacobi_P(n,1,1,(1+x)/(1-x)) = N_n(x)/(1-x)^n where N_n(x) denotes the shifted Narayana polynomial N_n(x) = sum{k = 1..n} A001263(k)*x^(k-1) of degree n-1.
Conjectural column n generating function: N_n(x^2)/(1-x)^(2n+1).
The entries in row n are given by the values of a polynomial function p_n(x) at x = 0,1,2,... . The first few are p_1(x) = 2x + 1, p_2(x) = (5x^2 + 5x + 2)/2, p_3(x) = (2x + 1)*(7x^2 + 7x + 6)/6 and p_4(x) = (7x^4 + 14x^3 + 21x^2 + 14x + 4)/4. These polynomials appear to have their zeros on the line Re x = -1/2; that is, the polynomials p_n(-x) appear to satisfy a Riemann hypothesis. The corresponding result for A108625 is true (see A142995 for details).
Contribution from Paul Barry, Jan 06 2009: (Start)
The g.f. for the corresponding number triangle is:
1/(1-x-xy-x^2y/(1-x-x^2y/(1-x-xy-x^2y/(1-x-x^2y/(1-x-xy-x^2y.... (a continued fraction). (End)
This g.f. satisfies x^2*y*g^2 - (1-x-x*y)*g + 1 = 0. - R. J. Mathar, Jun 16 2016
G.f.: ((y-1)*sqrt(((x^2+2*x+1)*y-x^2+2*x-1)/(y-1))+(-x-1)*y-x+1)/(2*x^2*y). - Vladimir Kruchinin, Jan 15 2018
T(n,m) = 1/(n+1)*Sum_{i=0..m+1} C(n+1,i-1)*C(n+1,i)*C(n+m-i+1,m+1-i). - Vladimir Kruchinin, Jan 15 2018

A273351 Number of up steps in all bargraphs of semiperimeter n (n>=2).

Original entry on oeis.org

1, 3, 10, 32, 102, 326, 1046, 3370, 10899, 35369, 115123, 375705, 1228970, 4028366, 13228516, 43511464, 143329157, 472761015, 1561246112, 5161512902, 17081176912, 56579333508, 187570898065, 622318325281, 2066208751201, 6864800067363, 22821993704857, 75915970992635, 252667993114760
Offset: 2

Views

Author

Emeric Deutsch, Jun 02 2016

Keywords

Examples

			a(4) = 10 because the 5 (=A082582(4)) bargraphs of semiperimeter 4 correspond to the compositions [1,1,1], [1,2], [2,1], [2,2], [3] which, clearly, have 1,2,2,2,3 up steps.

Links

Crossrefs

Cf. A082582, A273350.

Programs

  • Maple
    g := ((1-z-z^2-z^3-(1+z)*sqrt(1-4*z+2*z^2+z^4))*(1/2))/sqrt(1-4*z+2*z^2+z^4): gser := series(g, z = 0, 33): seq(coeff(gser, z, n), n = 2 .. 30);

Formula

G.f.: g(z) = (1-z-z^2-z^3-(1+z)*h)/(2*h), where h = sqrt(1-4*z+2*z^2+z^4).
a(n) = Sum_{k>=1} k*A273350(n,k).
Conjecture: n*(13*n-40)*a(n) +(-55*n^2+211*n-129)*a(n-1) +(38*n^2-212*n+255)*a(n-2) +(-6*n^2+56*n-75)*a(n-3) +(13*n^2-66*n+3)*a(n-4) -(3*n-13)*(n-6)*a(n-5)=0. - R. J. Mathar, Jun 06 2016
Conjecture: n*(n-3)*(n-2)^2*a(n) -(n-3)*(2*n-3)*(2*n^2-6*n+3) *a(n-1) +(2*n^4-16*n^3+41*n^2-36*n+5) *a(n-2) +2*(n-1)*(2*n-5) *a(n-3) +(n-2)*(n-5)*(n-1)^2 *a(n-4)=0. - R. J. Mathar, Jun 06 2016

A274488 Triangle read by rows: T(n,k) is the number of bargraphs of semiperimeter n having least column-height k (n>=2, k>=1).

Original entry on oeis.org

1, 1, 1, 3, 1, 1, 8, 3, 1, 1, 22, 8, 3, 1, 1, 62, 22, 8, 3, 1, 1, 178, 62, 22, 8, 3, 1, 1, 519, 178, 62, 22, 8, 3, 1, 1, 1533, 519, 178, 62, 22, 8, 3, 1, 1, 4578, 1533, 519, 178, 62, 22, 8, 3, 1, 1, 13800, 4578, 1533, 519, 178, 62, 22, 8, 3, 1, 1, 41937, 13800, 4578, 1533, 519, 178, 62, 22, 8, 3, 1, 1
Offset: 2

Views

Author

Emeric Deutsch, Jul 01 2016

Keywords

Comments

T(n,k) = number of bargraphs of semiperimeter n for which the width of the leftmost horizontal segment is k. A horizontal segment is a maximal sequence of adjacent horizontal steps (1,0). Example: T(4,1)=3 because the 5 (=A082582(4)) bargraphs of semiperimeter 4 correspond to the compositions [1,1,1], [1,2], [2,1], [2,2], [3] and the widths of their leftmost horizontal segments are 3, 1, 1, 2, 1.
Number of entries in row n is n-1.

Examples

			Row 4 is 3,1,1 because the 5 (=A082582(4)) bargraphs of semiperimeter 4 correspond to the compositions [1,1,1], [1,2], [2,1], [2,2], [3] and, clearly, their least column-heights are 1,1,1,2,3.
Triangle starts
1;
1,1;
3,1,1;
8,3,1,1;
22,8,3,1,1

Links

Crossrefs

Sum of entries in row n = A082582(n).
T(n,1) = A188464(n-3)(n>=3).
Sum(k*T(n,k),k>=1)= A008909(n).
Cf. A273350, A274490.

Programs

  • Maple
    G:=(1/2)*t*(1-z)*(1-2*z-z^2-sqrt((1-z)*(1-3*z-z^2-z^3)))/(z*(1-t*z)): Gser:= simplify(series(G,z=0,28)):for n from 2 to 20 do P[n]:= sort(coeff(Gser,z,n)) end do: for n from 2 to 15 do seq(coeff(P[n],t,k),k=1..n-1) end do; # yields sequence in triangular form
  • Mathematica
    gf = t(1-z)((1 - 2z - z^2 - Sqrt[(1-z)(1 - 3z - z^2 - z^3)])/(2z(1 - t z)));
Rest[CoefficientList[#, t]]& /@ Drop[CoefficientList[gf + O[z]^14, z], 2] // Flatten (* Jean-François Alcover, Nov 16 2018 *)

Formula

G.f.: t(1-z)(1-2z-z^2-sqrt((1-z)(1-3z-z^2-z^3)))/(2z(1-tz)).
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