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.

A351583 Triangle read by rows: T(n,k) = A(k,n-k), 1 <= k < n, 2 <= n, where A(m,n) is the number of distinct strings consisting of one X, 2*m-1 Y's and 2*n-1 Z's in which the X lies to the right of at least m Y's and at least n Z's.

Original entry on oeis.org

2, 7, 7, 15, 52, 15, 26, 192, 192, 26, 40, 510, 1086, 510, 40, 57, 1115, 4098, 4098, 1115, 57, 77, 2142, 12075, 20840, 12075, 2142, 77, 100, 3752, 30072, 79600, 79600, 30072, 3752, 100, 126, 6132, 66276, 249408, 382510, 249408, 66276, 6132, 126
Offset: 2

Views

Author

Christopher J. Fewster, Feb 14 2022

Keywords

Comments

The general string enumeration problem of counting strings with k+k'-1 X's, m+m' Y's and n+n' Z's in which the k'th X is placed after at least m of the Y's and n of the Z's may be expressed in terms of an integral of incomplete Beta functions and evaluated in terms of Kampe de Feriet functions (see Connor & Fewster, 2022). Other special cases include A351584 and A351585.

Examples

			Triangle starts:
   2;
   7,   7;
  15,  52,   15;
  26, 192,  192,  26;
  40, 510, 1086, 510, 40;
  ...
		

Crossrefs

Cf. A003506 (1/Beta), A005449 (column k=1), A351584, A351585.

Programs

  • Maple
    T:=(n,k) -> 1/(2*Beta(2*k, 2*n - 2*k)) - binomial(n, k)/(2*Beta(k, n - k)); [seq(seq(T(n,k),k=1..n-1),n=2..10)];
  • Mathematica
    t[n_,k_]:=1/(2*Beta[2*k,2*n-2*k])-Binomial[n,k]/(2*Beta[k,n-k]); Table[t[n,k],{n,2,10},{k,1,n-1}]

Formula

T(n+1,1) = A(1,n) = 1/2*n*(3*n+1) = A005449(n), the n-th second pentagonal number.
T(n,k) = 1/(2*Beta(2*k, 2*n - 2*k)) - binomial(n, k)/(2*Beta(k, n - k)), where Beta(x,y) = Gamma(x)*Gamma(y)/Gamma(x+y) is the Beta-function (see A003506). [Connor and Fewster]

A351585 Triangle read by rows: T(n,k) = A(k,n-k), 1 <= k < n, 2 <= n, where A(m,n) is the number of distinct strings consisting of one X, m+n-1 Y's and m+n-1 Z's in which the X lies to the right of at least m Y's and at least m Z's.

Original entry on oeis.org

2, 16, 6, 90, 52, 20, 448, 306, 180, 70, 2100, 1568, 1086, 644, 252, 9504, 7500, 5664, 3948, 2352, 924, 42042, 34452, 27450, 20840, 14580, 8712, 3432, 183040, 154154, 127380, 101950, 77640, 54450, 32604, 12870, 787644, 677248, 574574, 476652, 382510, 291896, 205062, 122980, 48620
Offset: 2

Views

Author

Christopher J. Fewster, Feb 14 2022

Keywords

Comments

The general string enumeration problem of counting strings with k+k'-1 X's, m+m' Y's and n+n' Z's in which the k'th X is placed after at least m of the Y's and n of the Z's may be expressed in terms of an integral of incomplete Beta functions and evaluated in terms of Kampe de Feriet functions (see Connor & Fewster, 2022). Other special cases include A351583 and A351584.

Examples

			Triangle starts:
     2;
    16,    6;
    90,   52,   20;
   448,  306,  180,  70;
  2100, 1568, 1086, 644, 252;
  ...
		

Crossrefs

Programs

  • Maple
    T:=(n,k)->(n - k)*binomial(2*n - 1, n) - 2*k*(n - k)*binomial(2*k - 1, k)*binomial(2*n - 2*k - 1, n - k)/n; [seq(seq(T(n,k),k=1..n-1),n=2..10)];
  • Mathematica
    t[n_,k_]:=(n-k)*Binomial[2*n-1,n]-(2*k*(n-k)/n)*Binomial[2*k-1,k]*Binomial[2*(n-k)-1,n-k]; Table[t[n,k],{n,2,10},{k,1,n-1}]

Formula

T(n+2,1) = A(1,n) = 2*n*binomial(2*n,n-1) = A253487(n-1).
T(m+1,m) = A(m,1) = binomial(2*m,m) = A000984(m) [central binomial coefficients].
T(n,k) = (n - k)*binomial(2*n - 1, n) - 2*k*(n - k)*binomial(2*k - 1, k)*binomial(2*n - 2*k - 1, n - k)/n. See Connor & Fewster (2022).
Showing 1-2 of 2 results.