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.

A371277 Triangle read by rows, (3, 2)-Lah numbers.

Original entry on oeis.org

1, 64, 1, 8000, 280, 1, 1728000, 104040, 792, 1, 592704000, 54996480, 681408, 1792, 1, 303464448000, 40685137920, 736404480, 3066560, 3520, 1, 221225582592000, 40988602368000, 1020839500800, 6035420160, 10800000, 6264, 1, 221225582592000000, 54777055334400000, 1804999259750400, 14280657592320, 35670620160, 31941000, 10360, 1
Offset: 2

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Author

Aleks Zigon Tankosic, Mar 17 2024

Keywords

Comments

The (3, 2)-Lah numbers T(n, k) count ordered 3-tuples (pi(1), pi(2), pi(3)) of partitions of the set {1, ..., n} into k linearly ordered blocks (lists, for short) such that the numbers 1, 2 are in distinct lists, and bl(pi(1)) = bl(pi(2))= bl(pi(3)) where for i = {1, 2, 3} and pi(i) = b(1)^i, b(2)^i, ..., b(k)^i, where b(1)^i, b(2)^i, ..., b(k)^i are the blocks of partition pi(i), bl(pi(i)) = {min(b(1))^i, min(b(2))^i, ..., min(b(k))^i} is the set of block leaders, i.e., of minima of the lists in partition pi(i).
The (3, 2)-Lah numbers T(n, k) are the (m, r)-Lah numbers for m=3 and r=2.
More generally, the (m, r)-Lah numbers count ordered m-tuples (pi(1), pi(2), ..., pi(m)) of partitions of the set {1, 2, ..., n} into k linearly ordered blocks (lists, for short) such that the numbers 1, 2, ..., r are in distinct lists, and bl(pi(1)) = bl(pi(2)) = ... = bl(pi(m)) where for i = {1, 2, ..., m} and pi(i) = {b(1)^i, b(2)^i, ..., b(k)^i}, where b(1)^i, b(2)^i, ..., b(k)^i are the blocks of partition pi(i), bl(pi(i)) = {min(b(1))^i, min(b(2))^i, ..., min (b(k))^i} is the set of block leaders, i.e., of minima of the lists in partition pi(i).

Examples

			Triangle begins:
              1;
             64,              1;
           8000,            280,             1;
        1728000,         104040,           792,          1;
      592704000,       54996480,        681408,       1792,        1;
   303464448000,    40685137920,     736404480,    3066560,     3520,    1;
221225582592000, 40988602368000, 1020839500800, 6035420160, 10800000, 6264,  1.
 ...
An example for T(4, 3). The corresponding partitions are
pi(1) = {(1),(2),(3,4)},
pi(2) = {(1),(2),(4,3)},
pi(3) = {(1),(2,3),(4)},
pi(4) = {(1),(3,2),(4)},
pi(5) = {(1),(2,4),(3)},
pi(6) = {(1),(4,2),(3)},
pi(7) = {(1,3),(2),(4)},
pi(8) = {(3,1),(2),(4)},
pi(9) = {(1,4),(2),(3)},
pi(10) = {(4,1),(2),(3)}, since A143497 for n=4, k=3 equals 10. Sets of their block leaders are bl(pi(1)) = bl(pi(2)) = bl(pi(5)) = bl(pi(6)) = bl(pi(9)) = bl(pi(10)) = {1,2,3} and bl(pi(3)) = bl(pi(4)) = bl(pi(7)) = bl(pi(8)) = {1,2,4}.
Compute the number of ordered 3-tuples (i.e., ordered pairs) of partitions pi(1), pi(2), ..., pi(10) such that partitions in the same pair share the same set of block leaders. As there are six partitions with the set of block leaders equal to {1,2,3}, and four partitions with the set of block leaders equal to {1,2,4}, T(4, 3) = 6^3 + 4^3 = 280.
		

Crossrefs

Programs

  • Maple
    T:= proc(n, k) option remember; `if`(k<2 or k>n, 0,
          `if`(n=k, 1, T(n-1, k-1)+(n+k-1)^3*T(n-1, k)))
        end:
    seq(seq(T(n, k), k=2..n), n=2..10);
  • Mathematica
    A371277[n_, k_] := A371277[n, k] = Which[n < k || k < 2, 0, n == k, 1, True, A371277[n-1, k-1] + (n+k-1)^3*A371277[n-1, k]];
    Table[A371277[n, k], {n, 2, 10}, {k, 2, n}] (* Paolo Xausa, Jun 12 2024 *)
  • Python
    A371277 = lambda n, k: 0 if (k < 2 or k > n) else (1 if (n == 2 and k == 2) else (A371277(n-1, k-1) + ((n + k - 1)**3) * A371277(n-1, k)))
    print([A371277(n, k) for n in range(2, 10) for k in range(2, n+1)])

Formula

Recurrence relation: T(n, k) = T(n-1, k-1) + (n+k-1)^3*T(n-1, k).
Explicit formula: T(n, k) = Sum_{3 <= j(1) < j(2) < ... < j(n-k) <= n} (2j(1)-2)^3 * (2j(2)-3)^3 * ... * (2j(n-k)-(n-k+1))^3.
Special cases:
T(n, k) = 0 for n < k or k < 2.
T(n, n) = 1.
T(n, 2) = (A143497(n,2))^3 = ((n+1)!)^3/216.
T(n, n-1) = 2^3 * Sum_{j=2..n-1} j^3.