A259701 Triangle read by rows: T(n,k) = number of permutations without overlaps in which the first increasing run has length k and the second element is not 2.
0, 1, 0, 2, 0, 0, 5, 0, 1, 0, 12, 0, 2, 0, 0, 33, 1, 7, 0, 1, 0, 87, 2, 17, 0, 2, 0, 0, 252, 11, 55, 2, 9, 0, 1, 0, 703, 26, 145, 4, 22, 0, 2, 0, 0, 2105, 109, 467, 27, 81, 3, 11, 0, 1, 0, 6099, 280, 1296, 63, 215, 6, 27, 0, 2, 0, 0
Offset: 2
Examples
Triangle begins:
0;
1, 0;
2, 0, 0;
5, 0, 1, 0;
12, 0, 2, 0, 0;
33, 1, 7, 0, 1, 0;
87, 2, 17, 0, 2, 0, 0;
252, 11, 55, 2, 9, 0, 1, 0;
703, 26, 145, 4, 22, 0, 2, 0, 0;
2105, 109, 467, 27, 81, 3, 11, 0, 1, 0;
...
References
- A. Sade, Sur les Chevauchements des Permutations, published by the author, Marseille, 1949
Links
- Albert Sade, Sur les Chevauchements des Permutations, published by the author, Marseille, 1949. [Annotated scanned copy]
Programs
-
PARI
Overlapfree(v)={for(i=1, #v, for(j=i+1, v[i]-1, if(v[j]>v[i], return(0)))); 1} 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} FirstRunLen(v)={my(e=1); for(i=1, #v, if(v[i]==e, e++)); e-2} row(n)={my(r=vector(n-1)); if(n>=2, forperm(n, v, if(v[1]<>1, break); if(v[2]<>2 && Overlapfree(Chords(v)), r[FirstRunLen(v)]++))); r} for(n=2, 8, print(row(n))) \\ Andrew Howroyd, Dec 07 2018
Extensions
a(49) corrected and a(57)-a(67) from Andrew Howroyd, Dec 07 2018
Comments