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.

A211789 Row sums of A211788.

Original entry on oeis.org

1, 2, 9, 50, 310, 2056, 14273, 102410, 753390, 5651948, 43074218, 332553252, 2595442616, 20443630100, 162308182577, 1297503030106, 10435055801110, 84371602316812, 685424273207630, 5592040955107420, 45798007929729828
Offset: 1

Views

Author

Peter Bala, Aug 02 2012

Keywords

Crossrefs

Cf. A211788.

Programs

  • Mathematica
    Rest[CoefficientList[InverseSeries[Series[x*(2*x-1)^2/(x-1)^2, {x, 0, 30}], x], x]] (* Vaclav Kotesovec, Nov 05 2017 *)
  • PARI
    N=21; x='x+O('x^(N+1)); Vec(serreverse(x*((1-2*x)/(1-x))^2)) \\ Gheorghe Coserea, Nov 05 2017

Formula

a(n) = Sum_{k = 1..n} A211788(n,k).
G.f. A(x) satisfies: A(x) = x*((1-A(x))/(1-2*A(x)))^2, a(n) = (Sum_{i=0..n-1} 2^i*(-1)^(n-i-1)*binomial(2*n,n-i-1)*binomial(2*n+i-1,2*n-1))/n for n > 0, a(0)=0. [Vladimir Kruchinin, Feb 08 2013]
From Vaclav Kotesovec, Nov 05 2017: (Start)
Recurrence: 4*n*(2*n - 1)*(17*n - 27)*a(n) = (1207*n^3 - 4331*n^2 + 4818*n - 1584)*a(n-1) - 2*(n-3)*(2*n - 3)*(17*n - 10)*a(n-2).
a(n) ~ sqrt(21/sqrt(17)-5) * ((71 + 17*sqrt(17))/16)^n / (sqrt(8*Pi) * n^(3/2)). (End)
a(n+1) = (1/(n+1)) * Sum_{k=0..n} binomial(2*n+k+1,k) * binomial(n-1,n-k). - Seiichi Manyama, Jan 12 2024

A193091 Augmentation of the triangular array A158405. See Comments.

Original entry on oeis.org

1, 1, 3, 1, 6, 14, 1, 9, 37, 79, 1, 12, 69, 242, 494, 1, 15, 110, 516, 1658, 3294, 1, 18, 160, 928, 3870, 11764, 22952, 1, 21, 219, 1505, 7589, 29307, 85741, 165127, 1, 24, 287, 2274, 13355, 61332, 224357, 638250, 1217270, 1, 27, 364, 3262, 21789, 115003
Offset: 0

Views

Author

Clark Kimberling, Jul 30 2011

Keywords

Comments

Suppose that P is an infinite triangular array of numbers:
p(0,0)
p(1,0)...p(1,1)
p(2,0)...p(2,1)...p(2,2)
p(3,0)...p(3,1)...p(3,2)...p(3,3)...
...
Let w(0,0)=1, w(1,0)=p(1,0), w(1,1)=p(1,1), and define
W(n)=(w(n,0), w(n,1), w(n,2),...w(n,n-1), w(n,n)) recursively by W(n)=W(n-1)*PP(n), where PP(n) is the n X (n+1) matrix given by
...
row 0 ... p(n,0) ... p(n,1) ...... p(n,n-1) ... p(n,n)
row 1 ... 0 ..... p(n-1,0) ..... p(n-1,n-2) .. p(n-1,n-1)
row 2 ... 0 ..... 0 ............ p(n-2,n-3) .. p(n-2,n-2)
...
row n-1 . 0 ..... 0 ............. p(2,1) ..... p(2,2)
row n ... 0 ..... 0 ............. p(1,0) ..... p(1,1)
...
The augmentation of P is here introduced as the triangular array whose n-th row is W(n), for n>=0. The array P may be represented as a sequence of polynomials; viz., row n is then the vector of coefficients: p(n,0), p(n,1),...,p(n,n), from p(n,0)*x^n+p(n,1)*x^(n-1)+...+p(n,n). For example, (C(n,k)) is represented by ((x+1)^n); using this choice of P (that is, Pascal's triangle), the augmentation of P is calculated one row at a time, either by the above matrix products or by polynomial substitutions in the following manner:
...
row 0 of W: 1, by decree
row 1 of W: 1 augments to 1,1
...polynomial version: 1 -> x+1
row 2 of W: 1,1 augments to 1,3,2
...polynomial version: x+1 -> (x^2+2x+1)+(x+1)=x^2+3x+2
row 3 to W: 1,3,2 augments to 1,6,11,6
...polynomial version:
x^2+3x+2 -> (x+1)^3+3(x+1)^2+2(x+1)=(x+1)(x+2)(x+3)
...
Examples of augmented triangular arrays:
(p(n,k)=1) augments to A009766, Catalan triangle.
Catalan triangle augments to A193560.
Pascal triangle augments to A094638, Stirling triangle.
A002260=((k+1)) augments to A023531.
A154325 augments to A033878.
A158405 augments to A193091.
((k!)) augments to A193092.
A094727 augments to A193093.
A130296 augments to A193094.
A004736 augments to A193561.
...
Regarding the specific augmentation W=A193091: w(n,n)=A003169.
From Peter Bala, Aug 02 2012: (Start)
This is the table of g(n,k) in the notation of Carlitz (p. 124). The triangle enumerates two-line arrays of positive integers
............a_1 a_2 ... a_n..........
............b_1 b_2 ... b_n..........
such that
1) max(a_i, b_i) <= min(a_(i+1), b_(i+1)) for 1 <= i <= n-1
2) max(a_i, b_i) <= i for 1 <= i <= n
3) max(a_n, b_n) = k.
See A071948 and A211788 for other two-line array enumerations.
(End)

Examples

			The triangle P, at A158405, is given by rows
  1
  1...3
  1...3...5
  1...3...5...7
  1...3...5...7...9...
The augmentation of P is the array W starts with w(0,0)=1, by definition of W.
Successive polynomials (rows of W) arise from P as shown here:
  ...
  1->x+3, so that W has (row 1)=(1,3);
  ...
  x+3->(x^2+3x+5)+3*(x+3), so that W has (row 2)=(1,6,14);
  ...
  x^2+6x+14->(x^3+3x^2+5x+7)+6(x^2+3x+5)+14(x+3), so that (row 3)=(1,9,37,79).
  ...
First 7 rows of W:
  1
  1    3
  1    6    14
  1    9    37    79
  1   12    69   242    494
  1   15   110   516   1658    3294
  1   18   160   928   3870   11764   22952
		

Crossrefs

Programs

  • Mathematica
    p[n_, k_] := 2 k + 1
    Table[p[n, k], {n, 0, 5}, {k, 0, n}] (* A158405 *)
    m[n_] := Table[If[i <= j, p[n + 1 - i, j - i], 0], {i, n}, {j, n + 1}]
    TableForm[m[4]]
    w[0, 0] = 1; w[1, 0] = p[1, 0]; w[1, 1] = p[1, 1];
    v[0] = w[0, 0]; v[1] = {w[1, 0], w[1, 1]};
    v[n_] := v[n - 1].m[n]
    TableForm[Table[v[n], {n, 0, 6}]] (* A193091 *)
    Flatten[Table[v[n], {n, 0, 9}]]

Formula

From Peter Bala, Aug 02 2012: (Start)
T(n,k) = (n-k+1)/n*Sum_{i=0..k} C(n+1,n-k+i+1)*C(2*n+i+1,i) for 0 <= k <= n.
Recurrence equation: T(n,k) = Sum_{i=0..k} (2*k-2*i+1)*T(n-1,i).
(End)

A071948 Triangle read by rows of numbers of paths in a lattice satisfying certain conditions.

Original entry on oeis.org

1, 1, 2, 1, 4, 7, 1, 6, 18, 30, 1, 8, 33, 88, 143, 1, 10, 52, 182, 455, 728, 1, 12, 75, 320, 1020, 2448, 3876, 1, 14, 102, 510, 1938, 5814, 13566, 21318, 1, 16, 133, 760, 3325, 11704, 33649, 76912, 120175, 1, 18, 168, 1078, 5313, 21252, 70840, 197340, 444015
Offset: 0

Views

Author

N. J. A. Sloane, Jun 15 2002

Keywords

Comments

This is the table of h(n,k) in the notation of Carlitz (p.125). The triangle (with an offset of 1 rather than 0) enumerates two-line arrays of positive integers
............1 a_2 ... a_(n-1) a_n..........
............1 b_2 ... b_(n-1) b_n..........
such that a_i <= i (2 <= i <= n) and b_2 <= a_2 <= ... <= b_n <= a_n = k.
See A193091 and A211788 for other two-line array enumerations. - Peter Bala, Aug 02 2012

Examples

			Triangle begins
  1;
  1, 2;
  1, 4,  7;
  1, 6, 18, 30;
  1, 8, 33, 88, 143;
		

Crossrefs

Row sums give A001764.
Rows are the reversals of the rows of A092276.

Programs

  • Maple
    T := proc(n,k) if k<=n then (n-k+1)*binomial(2*n+k+1,k)/(n+1) else 0 fi end: seq(seq(T(n,k),k=0..n),n=0..10);
  • Mathematica
    t[n_, k_] /; k <= n := (n-k+1)*Binomial[2*n+k+1, k]/(n+1); t[, ] = 0; Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jan 14 2014 *)
  • Sage
    # Computes the first n rows of the triangle.
    def A071948_triangle(n) :
        D = [0 for i in (0..n+1)]; D[1] = 1
        for i in (4..2*n+3) :
            h = i//2 - 1
            for k in (1..h) : D[k] += D[k-1]
            if i%2 == 1 : print([D[z] for z in (1..h)])
    A071948_triangle(10)  # Peter Luschny, Apr 01 2012

Formula

T(n, n) = A006013(n).
T(n, k) = (n-k+1)binomial(2n+k+1, k)/(n+1) if k<=n.
Let M = the infinite square production matrix
2, 1;
3, 2, 1;
4, 3, 2, 1;
5, 4, 3, 2, 1;
...
The top row of M^n gives reversed terms of n-th row of triangle A071948; with leftmost terms of each row generating A006013 starting (1, 2, 7, 30, 143, ...). - Gary W. Adamson, Jul 07 2011

Extensions

Edited by Emeric Deutsch, Mar 04 2004
Showing 1-3 of 3 results.