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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 (list; table; graph; refs; listen; history; text; internal format)
OFFSET
1,4
LINKS
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: A291774 A222061 A345453 * A371994 A227322 A216718
KEYWORD
nonn,tabl
AUTHOR
David W. Wilson, Sep 07 2001
STATUS
approved

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Last modified April 24 20:08 EDT 2024. Contains 371963 sequences. (Running on oeis4.)