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.

A108553 Square array, read by antidiagonals, where row n equals the crystal ball sequence for D_n lattice.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 5, 5, 1, 1, 13, 13, 7, 1, 1, 25, 55, 25, 9, 1, 1, 41, 169, 147, 41, 11, 1, 1, 61, 411, 625, 309, 61, 13, 1, 1, 85, 853, 2051, 1681, 561, 85, 15, 1, 1, 113, 1583, 5577, 6981, 3721, 923, 113, 17, 1, 1, 145, 2705, 13203, 23673, 18733, 7225, 1415, 145, 19, 1
Offset: 0

Views

Author

Paul D. Hanna, Jun 10 2005

Keywords

Comments

Rows 0 and 2 are included by extension since they fit the formula. Row 1 equals the odd numbers in order that triangle A108556 maintains that A108556(n,n-1) = (n/2)*A108556(n,n) for all n>=1, where row n of triangle A108556 equals the inverse binomial transform of row n of this square array.

Examples

			Square array begins:
1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...
1,3,5,7,9,11,13,15,17,19,21,23,25,27,...
1,5,13,25,41,61,85,113,145,181,221,265,...
1,13,55,147,309,561,923,1415,2057,2869,...
1,25,169,625,1681,3721,7225,12769,21025,...
1,41,411,2051,6981,18733,42783,86983,...
1,61,853,5577,23673,76389,204205,476113,...
1,85,1583,13203,68853,264825,824083,...
Inverse binomial transform of rows gives
rows of triangle A108556:
1,
1,2,
1,4,4,
1,12,30,20,
1,24,120,192,96,
1,40,330,940,1080,432, ...
Product of the g.f. of row n and (1-x)^(n+1)
generates the symmetric triangle A108558:
1;
1,1;
1,2,1;
1,9,9,1;
1,20,54,20,1;
1,35,180,180,35,1; ...
The row sums of triangle A108558 equals the
main diagonal of triangle A108556.
		

Crossrefs

Cf. A108554 (diagonal), A108555 (antidiagonal sums), A108556, A108558, A001844 (row 2), A005902 (row 3), A007204 (row 4), A008356 (row 5), A008358 (row 6), A008360 (row 7), A008362 (row 8), A008377 (row 9), A008379 (row 10).

Programs

  • PARI
    T(n,k)=if(n<0 || k<0,0,if(n==0 || k==0,1,if(n==1,2*k+1, sum(j=0,k,binomial(n+k-j,k-j)*(binomial(2*n,2*j)-2*n*binomial(n-2,j-1))))))

Formula

T(n, k) = Sum_{j=0..n} C(n+k-j, k-j)*[C(2*n, 2*j) - 2*j*(n-j)*C(n, j)/(n-1)] for n>1, with T(0, k)=1, T(1, k)=2*k+1.

A108554 Main diagonal of square array A108553, in which row n equals the crystal ball sequence for D_n lattice.

Original entry on oeis.org

1, 3, 13, 147, 1681, 18733, 204205, 2195399, 23429185, 249258777, 2650436845, 28209301211, 300745906449, 3212758605189, 34392151363021, 368908132170511, 3964622452867329, 42682164804034609, 460242445806795853
Offset: 0

Views

Author

Paul D. Hanna, Jun 10 2005

Keywords

Comments

Limit a(n+1)/a(n) ~ 11.08...
Limit is equal to A001622^5 = A244593 = 11.090169943749474241... - Vaclav Kotesovec, Aug 31 2025

Crossrefs

Programs

  • Mathematica
    Join[{1, 3}, Table[Sum[Binomial[2*n - j, n - j]*(Binomial[2*n, 2*j] - 2*j*(n - j)*Binomial[n, j]/(n - 1)), {j, 0, n}], {n, 2, 20}]] (* Vaclav Kotesovec, Aug 31 2025 *)
  • PARI
    {a(n)=if(n<0,0,if(n==0,1,if(n==1,3,sum(j=0,n,binomial(2*n-j,n-j)* (binomial(2*n,2*j)-2*j*(n-j)*binomial(n,j)/(n-1))))))}

Formula

a(n) ~ phi^(5*n + 1) / (2*5^(1/4)*sqrt(Pi*n)), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Aug 31 2025

A109000 Antidiagonal sums of square array A108998, in which row n equals the coordination sequence of B_n lattice.

Original entry on oeis.org

1, 1, 3, 11, 37, 133, 479, 1719, 6121, 21609, 75675, 263171, 909899, 3130963, 10730891, 36639987, 124528283, 420319907, 1403656123, 4615627555, 14868713515, 46702912307, 142489152555, 421113970835, 1203581558011
Offset: 0

Views

Author

Paul D. Hanna, Jun 17 2005

Keywords

Comments

Limit a(n+1)/a(n) ~ 3.3829757679..., real root of cubic (1+x+3*x^2-x^3). Compare to antidiagonal sums A108555 of square array A108553, in which row n equals the crystal ball sequence for D_n lattice.

Crossrefs

Programs

  • PARI
    {a(n)=sum(k=0,n,sum(j=0,k, binomial(n-j-1,k-j)*(binomial(2*n-2*k+1,2*j)-2*(n-k)*binomial(n-k-1,j-1))))}

Formula

a(n) = Sum_{k=0..n} Sum_{j=0..k} C(n-j-1, k-j) * (C(2*n-2*k+1, 2*j)-2*(n-k)*C(n-k-1, j-1)).
Showing 1-3 of 3 results.