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 1-4 of 4 results.

A110438 Triangular array giving the number of NSEW unit step lattice paths of length n with terminal height k subject to the following restrictions. The paths start at the origin (0,0) and take unit steps (0,1)=N(north), (0,-1)=S(south), (1,0)=E(east) and (-1,0)=W(west) such that no paths pass below the x-axis, no paths begin with W, all W steps remain on the x-axis and there are no NS steps.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 5, 4, 3, 1, 12, 10, 7, 4, 1, 29, 25, 18, 11, 5, 1, 71, 62, 47, 30, 16, 6, 1, 175, 155, 121, 82, 47, 22, 7, 1, 434, 389, 311, 220, 135, 70, 29, 8, 1, 1082, 979, 799, 584, 378, 212, 100, 37, 9, 1, 2709, 2471, 2051, 1541, 1039, 620, 320, 138, 46, 10, 1
Offset: 0

Views

Author

Asamoah Nkwanta (Nkwanta(AT)jewel.morgan.edu), Aug 10 2005

Keywords

Comments

The row sums are the even-indexed Fibonacci numbers.
Matrix product Q^(-1) * P * Q, where P denotes Pascal's triangle A007318 and Q denotes A061554 (formed from P by sorting the rows into descending order). Cf. A158793. - Peter Bala, Jul 14 2021

Examples

			Triangle starts:
  1;
  1,1;
  2,2,1;
  5,4,3,1;
  12,10,7,4,1;
		

References

  • A. Nkwanta, A Riordan matrix approach to unifying a selected class of combinatorial arrays, Congressus Numerantium, 160 (2003), pp. 33-55.
  • A. Nkwanta, A note on Riordan matrices, Contemporary Mathematics Series, AMS, 252 (1999), pp. 99-107.
  • A. Nkwanta, Lattice paths, generating functions and the Riordan group, Ph.D. Thesis, Howard University, Washington DC, 1997.

Crossrefs

Row sums are A001519(n+1).

Programs

  • Maple
    A110438 := proc (n, k)
        add((-1)^binomial(n-i+1, 2)*binomial(floor((1/2)*n+(1/2)*i), i)*add(binomial(i, j)*binomial(j, floor((1/2)*j-(1/2)*k)), j = k..i), i = 0..n);
    end proc:
    seq(seq(A110438(n, k), k = 0..n), n = 0..10); # Peter Bala, Jul 14 2021
  • PARI
    \\ ColGf gives g.f. of k-th column.
    ColGf(k,n)={my(g=(1 - x + x^2 - sqrt(1 - 2*x - x^2 - 2*x^3 + x^4 + O(x^(n-k+3))))/(2*x^2)); (1 - x)*g/(1 - x*g)*(x*g)^k}
    T(n,k) = {polcoef(ColGf(k,n), n)} \\ Andrew Howroyd, Mar 02 2023

Formula

Recurrence is d(0, 0) = 1, d(1, 0) = 1, d(n+1, 0) = 2*d(n, 0) + Sum_{j>=1} d(n-j, j), n>=1 for leftmost column and d(n+1, k) = d(n, k-1) + d(n, k) + Sum_{j>=1} d(n-j, k+j), n>=2, k>=1 and n>j; Riordan array d(n, k): (((1-z)/(2*z))*(sqrt(1+z+z^2)/sqrt(1-3*z+z^2) - 1), ((1-z+z^2)-sqrt(1-2*z-z^2-2*z^3+z^4))/(2*z)).

Extensions

Terms a(55) and beyond from Andrew Howroyd, Mar 02 2023

A191579 Triangular array related to continued fractions of square root of (N^2 - 1) for N>1, apparently containing A004148 and summing to A091964.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 2, 3, 3, 1, 4, 6, 6, 4, 1, 8, 13, 13, 10, 5, 1, 17, 28, 30, 24, 15, 6, 1, 37, 62, 69, 59, 40, 21, 7, 1, 82, 140, 160, 144, 105, 62, 28, 8, 1, 185, 320, 375, 350, 271, 174, 91, 36, 9, 1, 423, 740, 885, 852, 690, 474, 273, 128, 45, 10, 1
Offset: 1

Views

Author

Kenneth J Ramsey, Jun 07 2011

Keywords

Comments

The row sums of this triangle seems to be A091964 (verified to 12 terms), cf. diagonal sums of the triangle A124428. The 1st column seems to be A004148. The 2nd and 3rd column seems to be A089735, and A098075 (verified to 10 terms).
As each of these sequence is related to enumeration of RNA molecule structures, but was generated independently by reference to square array A192062 (re continued fractions for square roots of n^2-1 for n>1, see comments in the example below), it could be interesting to check this further for a relationship. As Mathar noted, this triangle appears identical to A097724. - edited by Kenneth J Ramsey, Oct 25 2012
Is this (apart from offsets) the same as A097724? - R. J. Mathar, Aug 01 2011

Examples

			The triangle begins
1;
1, 1;
1, 2, 1;
2, 3, 3, 1;
4, 6, 6, 4, 1;
8, 13, 13, 10, 5, 1;
17, 28, 30, 24, 15, 6, 1;
37, 62, 69, 59, 40, 21, 7, 1;
82, 140, 160, 144, 105, 62, 28, 8, 1;
185, 320, 375, 350, 271, 174, 91, 36, 9, 1;
423, 740, 885, 852, 690, 474, 273, 128, 45, 10, 1;
...
The 4th row is 2,3,3,1 because the 2nd,4th,6th and 8th terms of columns j = 1-5 of square array T(i,j) A192062  form the 4*5 matrix {{1,3,8,21},{1,4,15,56},{1,5,24,115},{1,6,35,204},{1,7,48,329}}. Solving the resulting system of linear equations results in the identities:
2*1 + 3*3 + 3*8 + 1*21 = 56 = T(8,2) of A192062
2*1 + 3*4 + 3*15+ 1*56 = 115 = T(8,3) of A192062
2*1 + 3*5 + 3*24 + 1*115 = 204 = T(8,4) of A192062
2*1 + 3*6 + 3*35 + 1*204 = 329 = T(8,5) of A192062
		

Crossrefs

Formula

The only way I know to generate this triangle is by reference to the square array A192062. The columns of that array, T(i,j) are such that for any given i>0, each term T(i,2*n) equals the sum as k = 1 to n, T(i-1,2*k)*C_k where C_k is the k th term of the n th row of this triangle. So solving the system of linear equations for each n > 0 gives the n th row of this triangle.

A205810 Irregular triangle read by rows: Whitney numbers c_{n,k} (n >= 0, 0 <= k <= 2n) of Lucas lattices.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 3, 3, 4, 3, 3, 1, 1, 4, 6, 8, 9, 8, 6, 4, 1, 1, 5, 10, 15, 20, 21, 20, 15, 10, 5, 1, 1, 6, 15, 26, 39, 48, 52, 48, 39, 26, 15, 6, 1, 1, 7, 21, 42, 70, 98, 119, 127, 119, 98, 70, 42, 21, 7, 1
Offset: 0

Views

Author

N. J. A. Sloane, Jan 31 2012

Keywords

Examples

			Triangle begins:
  1;
  1, 1,  1;
  1, 2,  1,  2,  1;
  1, 3,  3,  4,  3,  3,   1;
  1, 4,  6,  8,  9,  8,   6,   4,   1;
  1, 5, 10, 15, 20, 21,  20,  15,  10,  5,  1;
  1, 6, 15, 26, 39, 48,  52,  48,  39, 26, 15,  6,  1;
  1, 7, 21, 42, 70, 98, 119, 127, 119, 98, 70, 42, 21, 7, 1;
  ...
		

Crossrefs

Main diagonal is A051292.

Programs

  • Maple
    c:= (n, k)-> `if`(k=2*n, 1, n*add(1/(n-i)*binomial(n-i, n-k+i)*binomial(k-i-1, i), i=0..floor(k/2))): seq(seq(c(n, k), k=0..2*n), n=0..8);  # Leonid Bedratyuk, May 15 2018
  • PARI
    T(n,k) = if (k==2*n, 1, n*sum(i=0, k\2, 1/(n-i)*binomial(n-i,n-k+i)*binomial(k-i-1,i)));
    tabf(nn) = for (n=0, nn, for (k=0, 2*n, print1(T(n,k), ", ")); print); \\ Michel Marcus, May 16 2018

Formula

c(n, k) = n*Sum_{i = 0..floor(k/2)} 1/(n-i)*binomial(n-i, n-k+i)*binomial(k-i-1, i) for 0 <= k <= 2*n-1; c(n, 2*n) = 1. - Leonid Bedratyuk, May 15 2018
From Peter Bala, Jun 26 2025: (Start)
For n >= 1, the n-th row polynomial R(n, x) = x^n * t(n, 1 + x + 1/x), where t(n, x) = 2*Chebyshev_T(n, x/2) (AlSukaiti and Chbili, Proposition 2.1).
Conjecture: for n >= 1, t(n, x + 1) = Sum_{k = 0..n} c(n, n-k)*t(k, x) - c(n, n). Cf. A097724. (End)

A110439 Triangular array formed by the odd-indexed Fibonacci numbers.

Original entry on oeis.org

1, 1, 1, 3, 2, 1, 8, 5, 3, 1, 21, 14, 8, 4, 1, 55, 38, 23, 12, 5, 1, 144, 102, 65, 36, 17, 6, 1, 377, 273, 180, 106, 54, 23, 7, 1, 987, 728, 494, 304, 166, 78, 30, 8, 1, 2584, 1936, 1346, 858, 494, 251, 109, 38, 9, 1
Offset: 0

Views

Author

Asamoah Nkwanta (nkwanta(AT)jewel.morgan.edu), Aug 09 2005

Keywords

Comments

The leftmost column of the array is the odd-indexed Fibonacci numbers plus leading one.

Examples

			Triangle starts:
   1;
   1,  1;
   3,  2,  1;
   8,  5,  3,  1;
  21, 14,  8,  4,  1;
		

References

  • A. Nkwanta, A Riordan matrix approach to unifying a selected class of combinatorial arrays, Congressus Numerantium, 160 (2003), pp. 33-55.
  • A. Nkwanta, A note on Riordan matrices, Contemporary Mathematics Series, AMS, 252 (1999), pp. 99-107.
  • A. Nkwanta, Lattice paths, generating functions and the Riordan group, Ph.D. Thesis, Howard University, Washington DC 1997.

Crossrefs

Cf. A097724.

Formula

Riordan array: ((1-2z+z^2)/(1-3z+z^2), ((1-z+z^2)-sqrt(1-2z-z^2-2z^3+z^4))/2z), R(n, k). Recurrence: R(n+1, 0) = 2R(n, 0) + Sum_{j>=1} R(n-j, 0), leftmost column. For other columns: R(n+1, k) = R(n, k-1) + R(n, k) + Sum_{j>=1} R(n-j, k+j).
Showing 1-4 of 4 results.