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.

A068767 Generalized Catalan numbers 5*x*A(x)^2 -A(x) +1 -4*x=0.

Original entry on oeis.org

1, 1, 10, 105, 1150, 13050, 152500, 1825625, 22293750, 276758750, 3483287500, 44352006250, 570333187500, 7396680812500, 96638930625000, 1270796364765625, 16806545339843750, 223400240246093750
Offset: 0

Views

Author

Wolfdieter Lang, Mar 04 2002

Keywords

Comments

a(n) = K(5,5; n)/5 with K(a,b; n) defined in a comment to A068763.

Links

Crossrefs

Cf. A000108, A068764-A068766, A068768-A068772, A025227-A025230.

Programs

  • Mathematica
    CoefficientList[Series[(1-Sqrt[1-20*x*(1-4*x)])/(10*x), {x, 0, 20}], x] (* Vaclav Kotesovec, Mar 04 2014 *)

Formula

a(n) = (5^n) * p(n, -4/5) with the row polynomials p(n, x) defined from array A068763.
a(n+1) = 5*sum(a(k)*a(n-k), k=0..n), n>=1, a(0)=1=a(1).
G.f.: (1-sqrt(1-20*x*(1-4*x)))/(10*x).
(n+1)*a(n) = 80*(2-n)*a(n-2) + 10*(2*n-1)*a(n-1). - Fung Lam, Mar 04 2014
a(n) ~ sqrt(10+10*sqrt(5)) * (10+2*sqrt(5))^n / (10*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Mar 04 2014
Equivalently, a(n) ~ 2^(2*n) * 5^((n-1)/2) * phi^(n + 1/2) / (sqrt(Pi) * n^(3/2)), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Dec 08 2021

A192933 Triangle read by rows: T(n,k) = Sum_{i <= n, j <= k, (i,j) <> (n,k)} T(i,j), starting with T(1,1) = 1, for n >= 1 and 1 <= k <= n.

Original entry on oeis.org

1, 1, 2, 2, 6, 12, 4, 16, 44, 88, 8, 40, 136, 360, 720, 16, 96, 384, 1216, 3152, 6304, 32, 224, 1024, 3712, 11296, 28896, 57792, 64, 512, 2624, 10624, 36416, 108032, 273856, 547712, 128, 1152, 6528, 29056, 109696, 362624, 1056896, 2661504, 5323008, 256, 2560, 15872, 76800, 314880, 1135616, 3659776, 10528768, 26380544, 52761088
Offset: 1

Views

Author

Andrea Raffetti, Jul 13 2011

Keywords

Comments

The outer diagonal is A059435.
The second outer diagonal is A090442.
The third outer diagonal is essentially 2*A068766.
The first column is A011782.
The second column is essentially A057711 (not considering its first two terms).
The second column is essentially A129952 (not considering its first two terms).
The second column is essentially 2*A001792.
The differences between the terms of the second column is essentially 2*A045623.
The third column is essentially 4*A084266.
The cumulative sums of the third column are essentially 4*A176027.
T(n,k) = 0 for n < k. If this overriding constraint is not applied, you get A059576. - Franklin T. Adams-Watters, Jul 24 2011
For n >= 2 and 1 <= k <= n, T(n,k) is the number of bimonotone subdivisions of a 2-row grid with n points on the first row and k points on the second row (with the lower left point of the grid being the origin). A bimonotone subdivision of a convex polygon (the convex hull of the grid) is one where the internal dividing lines have nonnegative (including infinite) slopes. See Robeva and Sun (2020). - Petros Hadjicostas and Michel Marcus, Jul 15 2020

Examples

			Triangle (with rows n >= 1 and columns k = 1..n) begins:
   1;
   1,   2;
   2,   6,   12;
   4,  16,   44,    88;
   8,  40,  136,   360,   720;
  16,  96,  384,  1216,  3152,   6304;
  32, 224, 1024,  3712, 11296,  28896,  57792;
  64, 512, 2624, 10624, 36416, 108032, 273856, 547712;
  ...
Example: T(4,3) = 44 = 1 + 1 + 2 + 2 + 6 + 12 + 4 + 16.
From _Petros Hadjicostas_, Jul 15 2020: (Start)
Consider the following 2-row grid with n = 3 points at the top and k = 2 points at the bottom:
   A  B  C
   *--*--*
   |    /
   |   /
   *--*
   D  E
The sets of the dividing internal lines of the T(3,2) = 6 bimonotone subdivisions of the above 2-row grid are as follows: { }, {DC}, {DB}, {EB}, {DB, DC}, and {DB, EB}. We exclude subdivisions {EA} and {EA, EB} because they have at least one dividing line with a negative slope. (End)

Links

Crossrefs

Cf. A001003, A001792, A006318, A011782, A045623, A057711, A059435, A059576, A068766, A084266, A090442, A129952, A166444, A176027, A336245.

Programs

  • PARI
    lista(nn) = {my(T=matrix(nn, nn)); T[1,1] = 1; for (n=2, nn, for (k=1, n, T[n,k] = sum(i=1, n, sum(j=1, k, if ((i!=n) || (j!=k), T[i,j]))););); vector(nn, k, vector(k, i, T[k, i]));} \\ Michel Marcus, Mar 18 2020

Formula

T(n,1) = 2^(n-2) for n >= 2.
T(n,2) = n*2^(n-2) for n >= 2.
T(n,3) = 2^(n-2)*((n-k+1)^2 + 7*(n-k+1) + 4)/2 = 2^(n-3)*(n^2 + 3*n - 6) for k = 3 and n >= 3.
In general: For 1 <= k <= n with (n,k) <> 1,
T(n,k) = 2^(n-2)*Sum_{i=0..k-1} c(k,i)*(n-k+1)^(k-1-i)/(k-1)! and
T(n,k) = 2^(n-2)*Sum_{j=0..k-1} c(k,k-1-j)*(n-k+1)^j/(k-1)!
with c(k,i) being specific coefficients. Below are the first values for c(k,i):
1;
1, 1;
1, 7, 4;
1, 18, 77, 36;
1, 34, 359, 1238, 528,
1, 55, 1065, 8705, 26654, 10800;
... [Formula corrected by Petros Hadjicostas, Jul 15 2020]
The diagonal of this triangle for c(k,i) divided by (k-1)! (except for the first term) is equal to the Shroeder number sequence A006318(k-1).
From Petros Hadjicostas and Michel Marcus, Jul 15 2020: (Start)
T(n,1) = 2^(n-2) for n >= 2; T(n,k) = 2*(T(n,k-1) + T(n-1,k) - T(n-1,k-1)) for n > k >= 2; T(n,n) = 2*T(n,n-1) for n = k >= 2; and T(n,k) = 0 for 1 <= n < k. [Robeva and Sun (2020)] (They do not specify T(1,1) explicitly since they do not care about subdivisions of a degenerate polygon with only one side.)
T(n,k) = (2^(n-2)/(k-1)!) * P_k(n) = (2^(n-2)/(k-1)!) * Sum_{j=1..k} A336245(k,j)*n^(k-j) for n >= k >= 1 with (n,k) <> (1,1), where P_k(n) is some polynomial with integer coefficients of degree k-1. [Robeva and Sun (2020)]
A336245(k,j) = Sum_{s=0..j-1} c(k,s) * binomial(k-1-s, k-j) * (1-k)^(j-1-s) for 1 <= j <= k, in terms of the above coefficients c(k,i).
So c(k,s) = Sum_{j=1..s+1} A336245(k,j) * binomial(k-j, k-s-1) * (k-1)^(s+1-j) for k >= 1 and 0 <= s <= k-1, obtained by inverting the binomial transform.
Bivariate o.g.f.: x*y*(1 - x)*(1 - 2*y*g(2*x*y))/(1 - 2*x - 2*y + 2*x*y), where g(w) = 2/(1 + w + sqrt(1 - 6*w + w^2)) = g.f. of A001003.
Letting y = 1 in the above joint o.g.f., we get the o.g.f. of the row sums: x*(1-x)*(2*g(2*x) - 1). It can then be easily proved that
Sum_{k=1..n} T(n,k) = 2^n*A001003(n-1) - 2^(n-1)*A001003(n-2) for n >= 3. (End)

Extensions

Offset changed by Andrew Howroyd, Dec 31 2017
Name edited by Petros Hadjicostas, Jul 15 2020

A068765 Generalized Catalan numbers 3*x*A(x)^2 -A(x) +1 -2*x = 0.

Original entry on oeis.org

1, 1, 6, 39, 270, 1962, 14796, 114831, 911574, 7368894, 60457428, 502162902, 4214515212, 35686162548, 304491863448, 2615468845311, 22598114065254, 196269877811574, 1712578870493316, 15005719955119698
Offset: 0

Views

Author

Wolfdieter Lang, Mar 04 2002

Keywords

Comments

a(n)=K(3,3; n)/3 with K(a,b; n) defined in a comment to A068763.

Links

Crossrefs

Cf. A000108, A068764, A068766-72, A025227-30.

Programs

  • Mathematica
    CoefficientList[Series[(1-Sqrt[1-12*x*(1-2*x)])/(6*x), {x, 0, 20}], x] (* Vaclav Kotesovec, Mar 04 2014 *)

Formula

a(n)=(3^n)*p(n, -2/3) with the row polynomials p(n, x) defined from array A068763.
a(n+1)= 3*sum(a(k)*a(n-k), k=0..n), n>=1, a(0)=1=a(1).
G.f.: (1-sqrt(1-12*x*(1-2*x)))/(6*x).
Recurrence: (n+1)*a(n) = 24*(2-n)*a(n-2) + 6*(2*n-1)*a(n-1). - Fung Lam, Mar 04 2014
a(n) ~ sqrt(6+6*sqrt(3)) * (6+2*sqrt(3))^n / (6*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Mar 04 2014
Showing 1-3 of 3 results.

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