A371081 Triangle read by rows, (2, 2)-Lah numbers.
1, 16, 1, 400, 52, 1, 14400, 2948, 116, 1, 705600, 203072, 12344, 216, 1, 45158400, 17154432, 1437472, 38480, 360, 1, 3657830400, 1760601600, 191088544, 6978592, 99320, 556, 1, 365783040000, 216690624000, 29277351936, 1370470592, 26445312, 224420, 812, 1
Offset: 2
Examples
Triangle begins: 1; 16, 1; 400, 52, 1; 14400, 2984, 116, 1; 705600, 203072, 12344, 216, 1; 45158400, 1715443, 1437472, 38480, 360, 1; 3657830400, 1760601600, 191088544, 6978592, 99320, 556, 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 2-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) = 6^2 + 4^2 = 52.
Links
- A. Žigon Tankosič, The (l, r)-Lah Numbers, Journal of Integer Sequences, Article 23.2.6, vol. 26 (2023).
Crossrefs
Programs
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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)^2*T(n-1, k))) end: seq(seq(T(n, k), k=2..n), n=2..10); # Alois P. Heinz, Mar 11 2024
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Mathematica
A371081[n_, k_] := A371081[n, k] = Which[n < k || k < 2, 0, n == k, 1, True, A371081[n-1, k-1] + (n+k-1)^2*A371081[n-1, k]]; Table[A371081[n, k], {n, 2, 10}, {k, 2, n}] (* Paolo Xausa, Jun 12 2024 *)
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Python
A371081 = lambda n, k: 0 if (k < 2 or k > n) else (1 if (n == 2 and k == 2) else (A371081(n-1, k-1) + ((n + k - 1)**2) * A371081(n-1, k))) print([A371081(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)^2*T(n-1, k).
Explicit formula: T(n, k) = Sum_{3 <= 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 < 2,
T(n, n) = 1,
T(n, n-1) = 2^2 * Sum_{j=2..n-1} j^2.
Comments