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 1-2 of 2 results.

A099602 Triangle, read by rows, such that row n equals the inverse binomial transform of column n of the triangle of trinomial coefficients (A027907), omitting leading zeros.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 2, 5, 4, 1, 1, 5, 8, 5, 1, 3, 13, 22, 18, 7, 1, 1, 9, 26, 35, 24, 8, 1, 4, 26, 70, 101, 84, 40, 10, 1, 1, 14, 61, 131, 160, 116, 49, 11, 1, 5, 45, 171, 363, 476, 400, 215, 71, 13, 1, 1, 20, 120, 363, 654, 752, 565, 275, 83, 14, 1, 6, 71, 356, 1017, 1856, 2282, 1932
Offset: 0

Views

Author

Paul D. Hanna, Oct 25 2004

Keywords

Comments

Row sums form A099603, where A099603(n) = Fibonacci(n+1)*2^floor((n+1)/2). Central coefficients of even-indexed rows form A082759, where A082759(n) = Sum_{k=0..n} binomial(n,k)*trinomial(n,k). Antidiagonal sums form A099604.
Matrix inverse equals triangle A104495, which is generated from self-convolutions of the Catalan sequence (A000108).

Examples

			Rows begin:
  1;
  1,  1;
  1,  2,   1;
  2,  5,   4,   1;
  1,  5,   8,   5,   1;
  3, 13,  22,  18,   7,   1;
  1,  9,  26,  35,  24,   8,   1;
  4, 26,  70, 101,  84,  40,  10,   1;
  1, 14,  61, 131, 160, 116,  49,  11,  1;
  5, 45, 171, 363, 476, 400, 215,  71, 13,  1;
  1, 20, 120, 363, 654, 752, 565, 275, 83, 14, 1;
  ...
The binomial transform of row 2 = column 2 of A027907: BINOMIAL[1,2,1] = [1,3,6,10,15,21,28,36,45,55,...].
The binomial transform of row 3 = column 3 of A027907: BINOMIAL[2,5,4,1] = [2,7,16,30,50,77,112,156,210,...].
The binomial transform of row 4 = column 4 of A027907: BINOMIAL[1,5,8,5,1] = [1,6,19,45,90,161,266,414,615,...].
The binomial transform of row 5 = column 5 of A027907: BINOMIAL[3,13,22,18,7,1] = [3,16,51,126,266,504,882,1452,...].
		

Crossrefs

Programs

  • PARI
    {T(n,k)=polcoeff(polcoeff((1+(y+1)*x-(y+1)*x^2)/(1-(y+1)*(y+2)*x^2+(y+1)^2*x^4)+x*O(x^n),n,x)+y*O(y^k),k,y)}
    
  • PARI
    {T(n,k)=(matrix(n+1,n+1,i,j,if(i>=j,polcoeff(polcoeff( (1+x*y/(1+x))/(1+x-y^2*(1-(1+4*x+O(x^i))^(1/2))^2/4+O(y^j)),i-1,x),j-1,y)))^-1)[n+1,k+1]}

Formula

G.f.: (1 + (y+1)*x - (y+1)*x^2)/(1 - (y+1)*(y+2)*x^2 + (y+1)^2*x^4).

A099606 Row sums of triangle A099605, in which row n equals the inverse Binomial transform of column n of the triangle A034870 of even-indexed rows of Pascal's triangle.

Original entry on oeis.org

1, 4, 10, 48, 116, 560, 1352, 6528, 15760, 76096, 183712, 887040, 2141504, 10340096, 24963200, 120532992, 290992384, 1405035520, 3392055808, 16378294272, 39540700160, 190919389184, 460920178688, 2225519493120, 5372879343616
Offset: 0

Views

Author

Paul D. Hanna, Oct 25 2004

Keywords

Examples

			Sequence begins: {1*1, 2*2, 5*2, 12*4, 29*4, 70*8, 169*8, 408*16, ...}.
		

Crossrefs

Programs

  • PARI
    a(n)=polcoeff((1+4*x-2*x^2)/(1-12*x^2+4*x^4)+x*O(x^n),n)

Formula

a(n) = Pell(n+1)*2^[(n+1)/2]. a(n) = 12*a(n-2) - 4*a(n-4) for n>=4. G.f.: (1+4*x-2*x^2)/(1-12*x^2+4*x^4).
Showing 1-2 of 2 results.