A371277 - OEIS

%I #28 Jun 12 2024 04:10:25

%S 1,64,1,8000,280,1,1728000,104040,792,1,592704000,54996480,681408,

%T 1792,1,303464448000,40685137920,736404480,3066560,3520,1,

%U 221225582592000,40988602368000,1020839500800,6035420160,10800000,6264,1,221225582592000000,54777055334400000,1804999259750400,14280657592320,35670620160,31941000,10360,1

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

%C 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).

%C The (3, 2)-Lah numbers T(n, k) are the (m, r)-Lah numbers for m=3 and r=2.

%C 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).

%H A. Žigon Tankosič, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL26/Tankosic/tank2.html">The (l, r)-Lah Numbers</a>, Journal of Integer Sequences, Article 23.2.6, vol. 26 (2023).

%F Recurrence relation: T(n, k) = T(n-1, k-1) + (n+k-1)^3*T(n-1, k).

%F 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.

%F Special cases:

%F T(n, k) = 0 for n < k or k < 2.

%F T(n, n) = 1.

%F T(n, 2) = (A143497(n,2))^3 = ((n+1)!)^3/216.

%F T(n, n-1) = 2^3 * Sum_{j=2..n-1} j^3.

%e Triangle begins:

%e               1;

%e              64,              1;

%e            8000,            280,             1;

%e         1728000,         104040,           792,          1;

%e       592704000,       54996480,        681408,       1792,        1;

%e    303464448000,    40685137920,     736404480,    3066560,     3520,    1;

%e 221225582592000, 40988602368000, 1020839500800, 6035420160, 10800000, 6264,  1.

%e  ...

%e An example for T(4, 3). The corresponding partitions are

%e pi(1) = {(1),(2),(3,4)},

%e pi(2) = {(1),(2),(4,3)},

%e pi(3) = {(1),(2,3),(4)},

%e pi(4) = {(1),(3,2),(4)},

%e pi(5) = {(1),(2,4),(3)},

%e pi(6) = {(1),(4,2),(3)},

%e pi(7) = {(1,3),(2),(4)},

%e pi(8) = {(3,1),(2),(4)},

%e pi(9) = {(1,4),(2),(3)},

%e 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}.

%e 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.

%p T:= proc(n, k) option remember; `if`(k<2 or k>n, 0,

%p       `if`(n=k, 1, T(n-1, k-1)+(n+k-1)^3*T(n-1, k)))

%p     end:

%p seq(seq(T(n, k), k=2..n), n=2..10);

%t 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]];

%t Table[A371277[n, k], {n, 2, 10}, {k, 2, n}] (* _Paolo Xausa_, Jun 12 2024 *)

%o (Python)

%o def T_Lah(n, k):

%o     if k < 2 or k > n:

%o         return 0

%o     elif n == k == 2:

%o         return 1

%o     else:

%o         return T_Lah(n-1, k-1) + ((n+k-1)**3) * T_Lah(n-1, k)

%o def print_triangle(rows):

%o     for n in range(rows):

%o         row_values = [T_Lah(n, k) for k in range(n+1)]

%o         print(' '.join(map(str, row_values)).center(rows*10))

%o rows = 10

%o print_triangle(rows)

%Y Cf. A143497, A371081, A371259, A372208.

%K nonn,tabl

%O 2,2

%A _Aleks Zigon Tankosic_, Mar 17 2024