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A125181
Triangle read by rows: T(n,k) is the number of Dyck paths of semilength n whose ascent lengths form the k-th partition of the integer n; the partitions of n are ordered in the "Mathematica" ordering.
14
1, 1, 1, 1, 3, 1, 1, 4, 2, 6, 1, 1, 5, 5, 10, 10, 10, 1, 1, 6, 6, 15, 3, 30, 20, 5, 30, 15, 1, 1, 7, 7, 21, 7, 42, 35, 21, 21, 105, 35, 35, 70, 21, 1, 1, 8, 8, 28, 8, 56, 56, 4, 56, 28, 168, 70, 28, 84, 168, 280, 56, 14, 140, 140, 28, 1, 1, 9, 9, 36, 9, 72, 84, 9, 72, 36, 252, 126, 36
OFFSET
1,5
COMMENTS
Equivalently, T(n,k) is the number of ordered trees with n edges whose node degrees form the k-th partition of the integer n.
Also the number of non-crossing set partitions whose block sizes are the parts of the n-th integer partition in graded Mathematica ordering. - Gus Wiseman, Feb 15 2019
For relations to Lagrange inversion through shifted reciprocals of a function, refined Narayana numbers, non-crossing partitions, trees, and other lattice paths, see A134264 and A091867. - Tom Copeland, Nov 01 2014
REFERENCES
R. P. Stanley, Enumerative Combinatorics Vol. 2, Cambridge University Press, Cambridge, 1999; Theorem 5.3.10.
LINKS
Germain Kreweras, Sur les partitions non croisées d'un cycle, Discrete Math. 1 333-350 (1972).
FORMULA
Row n has A000041(n) terms (equal to the number of partitions of n).
Row sums yield the Catalan numbers (A000108).
Given a partition p = [a(1)^e(1), ..., a(j)^e(j)] into k parts (e(1) +...+ e(j) = k), the number of Dyck paths whose ascent lengths yield the partition p is n!/[(n-k+1)!e(1)!e(2)! ... e(j)! ]. - Franklin T. Adams-Watters
EXAMPLE
Example: T(5,3)=5 because the 3rd partition of 5 is [3,2] and we have (UU)DD(UUU)DDD, (UUU)DDD(UU)DD, (UU)D(UUU)DDDD, (UUU)D(UU)DDDD and (UUU)DD(UU)DDD; here U=(1,1), D=(1,-1) and the ascents are shown between parentheses.
Triangle begins:
1
1 1
1 3 1
1 4 2 6 1
1 5 5 10 10 10 1
1 6 6 15 3 30 20 5 30 15 1
1 7 7 21 7 42 35 21 21 105 35 35 70 21 1
Row 4 counts the following non-crossing set partitions:
{{1234}} {{1}{234}} {{12}{34}} {{1}{2}{34}} {{1}{2}{3}{4}}
{{123}{4}} {{14}{23}} {{1}{23}{4}}
{{124}{3}} {{12}{3}{4}}
{{134}{2}} {{1}{24}{3}}
{{13}{2}{4}}
{{14}{2}{3}}
MAPLE
with(combinat): for n from 1 to 9 do p:=partition(n): for q from 1 to numbpart(n) do m:=convert(p[numbpart(n)+1-q], multiset): k:=nops(p[numbpart(n)+1-q]): s[n, q]:=n!/(n-k+1)!/product(m[j][2]!, j=1..nops(m)) od: od: for n from 1 to 9 do seq(s[n, q], q=1..numbpart(n)) od; # yields sequence in triangular form
# second Maple program:
b:= proc(n, i, k) `if`(n=0, [k!], `if`(i<1, [],
[seq(map(x->x*j!, b(n-i*j, i-1, k-j))[], j=0..n/i)]))
end:
T:= proc(n) local l, m;
l:= b(n, n, n+1); m:=nops(l);
seq(n!/l[m-i], i=0..m-1)
end:
seq(T(n), n=1..10); # Alois P. Heinz, May 25 2013
MATHEMATICA
b[n_, i_, k_] := b[n, i, k] = If[n == 0, {k!}, If[i<1, {}, Flatten @ Table[Map[#*j! &, b[n-i*j, i-1, k-j]], {j, 0, n/i}]]]; T[n_] := Module[{l, m}, l = b[n, n, n+1]; m = Length[l]; Table[n!/l[[m-i]], {i, 0, m-1}]]; Table[T[n], {n, 1, 10}] // Flatten (* Jean-François Alcover, May 26 2015, after Alois P. Heinz *)
Table[Binomial[Total[y], Length[y]-1]*(Length[y]-1)!/Product[Count[y, i]!, {i, Max@@y}], {y, Join@@Table[IntegerPartitions[n], {n, 1, 8}]}] (* Gus Wiseman, Feb 15 2019 *)
PROG
(SageMath)
def C(p):
n = sum(p); l = n - len(p) + 1
def f(x): return factorial(len(list(filter(lambda y: y == x, p))))
return factorial(n) // (factorial(l) * prod(f(x) for x in set(p)))
def row(n): return list(C(p) for p in Partitions(n))
for n in range(1, 9): print(row(n)) # Peter Luschny, Jul 14 2022
CROSSREFS
Other orderings are A134264 and A306438.
Sequence in context: A219842 A351402 A134264 * A157076 A348608 A342805
KEYWORD
nonn,look,tabf
AUTHOR
Emeric Deutsch, Nov 23 2006
STATUS
approved