A371259 Triangle read by rows, (2, 3)-Lah numbers.
1, 36, 1, 1764, 100, 1, 112896, 9864, 200, 1, 9144576, 1099296, 34064, 344, 1, 914457600, 142159392, 6004512, 92200, 540, 1, 110649369600, 21385410048, 1156921920, 24075712, 213700, 796, 1, 15933509222400, 3724783667712, 248142106368, 6573957120, 78782912, 443744, 1120, 1
Offset: 3
Examples
Triangle begins: 1; 36, 1; 1764, 100, 1; 112896, 9864, 200, 1; 9144576, 1099296, 34064, 344, 1; 914457600, 142159392, 6004512, 92200, 540, 1; 110649369600, 21385410048, 1156921920, 24075712, 213700, 796, 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),(3),(2,4)}, pi(4) = {(1),(3),(4,2)}, pi(5) = {(1,4),(2),(3)}, pi(6) = {(4,1),(2),(3)}, since A143498 for n=4, k=3 equals 6. Sets of their block leaders are bl(pi(1)) = bl(pi(2)) = bl(pi(3)) = bl(pi(4)) = bl(pi(5)) = bl(pi(6)) = {1,2,3}. Compute the number of ordered 2-tuples (i.e., ordered pairs) of partitions pi(1), pi(2), ..., pi(6) 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}, T(4, 3) = 6^2 = 36.
Links
- A. Žigon Tankosič, The (l, r)-Lah Numbers, Journal of Integer Sequences, Article 23.2.6, vol. 26 (2023).
Programs
-
Maple
T:= proc(n, k) option remember; `if`(k<3 or k>n, 0, `if`(n=k, 1, T(n-1, k-1)+(n+k-1)^2*T(n-1, k))) end: seq(seq(T(n, k), k=3..n), n=3..10);
-
Mathematica
A371259[n_, k_] := A371259[n, k] = Which[n < k || k < 3, 0, n == k, 1, True, A371259[n-1, k-1] + (n+k-1)^2*A371259[n-1, k]]; Table[A371259[n, k], {n, 3, 10},{k, 3, n}] (* Paolo Xausa, Jun 11 2024 *)
-
Python
A371259 = lambda n, k: 0 if (k < 3 or k > n) else (1 if (n == 3 and k == 3) else (A371259(n-1, k-1) + ((n + k - 1)**2) * A371259(n-1, k))) print([A371259(n, k) for n in range(3, 11) for k in range(3, n+1)])
Formula
Recurrence relation: T(n, k) = T(n-1, k-1) + (n+k-1)^2*T(n-1, k).
Explicit formula: T(n, k) = Sum_{4 <= j(1) < j(2) < ... < j(n-k) <= n} (2j(1)-2)^2 * (2j(2)-3)^2 * ... * (2j(n-k)-(n-k+1))^2.
Special cases:
T(n, k) = 0 for n < k or k < 3, [corrected by Paolo Xausa, Jun 11 2024]
T(n, n) = 1,
T(n, 3) = (A143498(n, 3))^2 = ((n+2)!)^2/14400,
T(n, n-1) = 2^2 * Sum_{j=3..n-1} j^2.
Extensions
More terms from Michel Marcus, Jun 12 2025
Comments