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A064315
Triangle of number of permutations by length of shortest ascending run.
14
1, 1, 1, 5, 0, 1, 18, 5, 0, 1, 101, 18, 0, 0, 1, 611, 89, 19, 0, 0, 1, 4452, 519, 68, 0, 0, 0, 1, 36287, 3853, 110, 69, 0, 0, 0, 1, 333395, 27555, 1679, 250, 0, 0, 0, 0, 1, 3382758, 233431, 11941, 418, 251, 0, 0, 0, 0, 1, 37688597, 2167152, 59470, 658, 922, 0, 0, 0, 0, 0, 1
OFFSET
1,4
FORMULA
T(2*n,n) = binomial(2*n,n)-1 = A030662(n).
Sum_{k=1..n} k * T(n,k) = A064316(n).
EXAMPLE
Sequence (1, 3, 2, 5, 4) has ascending runs (1, 3), (2, 5), (4), the shortest is length 1. Of all permutations of (1, 2, 3, 4, 5), T(5,1) = 101 have shortest ascending run of length 1.
Triangle T(n,k) begins:
1;
1, 1;
5, 0, 1;
18, 5, 0, 1;
101, 18, 0, 0, 1;
611, 89, 19, 0, 0, 1;
4452, 519, 68, 0, 0, 0, 1,
36287, 3853, 110, 69, 0, 0, 0, 1;
...
MAPLE
A:= proc(n, k) option remember; local b; b:=
proc(u, o, t) option remember; `if`(t+o<=k, (u+o)!,
add(b(u+i-1, o-i, min(k, t)+1), i=1..o)+
`if`(t<=k, u*(u+o-1)!, add(b(u-i, o+i-1, 1), i=1..u)))
end: forget(b):
add(b(j-1, n-j, 1), j=1..n)
end:
T:= (n, k)-> A(n, k) -A(n, k-1):
seq(seq(T(n, k), k=1..n), n=1..12); # Alois P. Heinz, Aug 29 2013
MATHEMATICA
A[n_, k_] := A[n, k] = Module[{b}, b[u_, o_, t_] := b[u, o, t] = If[t+o <= k, (u+o)!, Sum[b[u+i-1, o-i, Min[k, t]+1], {i, 1, o}] + If[t <= k, u*(u+o-1)!, Sum[ b[u-i, o+i-1, 1], {i, 1, u}]]]; Sum[b[j-1, n-j, 1], {j, 1, n}]]; T[n_, k_] := A[n, k] - A[n, k-1]; Table[Table[T[n, k], {k, 1, n}], {n, 1, 12}] // Flatten (* Jean-François Alcover, Jan 28 2015, after Alois P. Heinz *)
CROSSREFS
Row sums give: A000142.
Sequence in context: A222061 A378981 A345453 * A371994 A227322 A216718
KEYWORD
nonn,tabl
AUTHOR
David W. Wilson, Sep 07 2001
STATUS
approved