%I #30 Oct 06 2019 18:20:15
%S 1,1,1,1,2,1,1,4,4,1,1,6,10,6,1,1,10,23,22,9,1,1,14,44,61,41,12,1,1,
%T 22,87,158,148,71,16,1,1,30,151,352,436,301,114,20,1,1,46,280,791,
%U 1210,1092,589,175,25,1,1,62,464,1592,2969,3317,2408,1038,256,30,1
%N Triangle read by rows: T(n,k) = number of permutations without overlaps having k increasing runs.
%C The sums s(n) = Sum_k k*T(n,k) give A259700.
%C Albert Sade in Sur les Chevauchements des Permutation (published by the author in French in 1949) gave the following example for determining the number of increasing runs in a permutation: 176852943 has 3 runs: 123 (left to right), 34567 (right to left), 789 (right to left).
%H Albert Sade, <a href="/A000108/a000108_17.pdf">Sur les Chevauchements des Permutations</a>, published by the author, Marseille, 1949. [Annotated scanned copy]
%e Triangle begins:
%e 1;
%e 1, 1;
%e 1, 2, 1;
%e 1, 4, 4, 1;
%e 1, 6, 10, 6, 1;
%e 1, 10, 23, 22, 9, 1;
%e 1, 14, 44, 61, 41, 12, 1;
%e 1, 22, 87, 158, 148, 71, 16, 1;
%e 1, 30, 151, 352, 436, 301, 114, 20, 1;
%e 1, 46, 280, 791, 1210, 1092, 589, 175, 25, 1;
%e 1, 62, 464, 1592, 2969, 3377, 2408, 1038, 256, 30, 1;
%e ...
%o (PARI)
%o Overlapfree(v)={for(i=1, #v, for(j=i+1, v[i]-1, if(v[j]>v[i], return(0)))); 1}
%o Chords(u)={my(n=2*#u, v=vector(n), s=u[#u]); if(s%2==0, s=n+1-s); for(i=1, #u, my(t=n+1-s); s=u[i]; if(s%2==0, s=n+1-s); v[s]=t; v[t]=s); v}
%o Runs(v)={my(u=vector(#v), s=1); for(i=1, #v, u[v[i]]=i); for(i=2, #u-1, if(sign(u[i]-u[i-1])==sign(u[i]-u[i+1]), s++)); s}
%o row(n)={my(r=vector(n-1)); if(n>=2, forperm(n, v, if(v[1]<>1, break); if(Overlapfree(Chords(v)), r[Runs(v)]++))); r}
%o for(n=2, 8, print(row(n))) \\ _Andrew Howroyd_, Dec 07 2018
%Y Row sums give A000682.
%Y Cf. A259700.
%K nonn,tabl
%O 2,5
%A _N. J. A. Sloane_, Jul 05 2015
%E Corrected and extended by _Roger Ford_, Jul 06 2016