A064315 Triangle of number of permutations by length of shortest ascending run.
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
Examples
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; ...
Links
- Alois P. Heinz, Rows n = 1..100, flattened
- D. W. Wilson, Extended tables for A008304 and A064315
Crossrefs
Programs
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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
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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 *)