OFFSET
2,4
COMMENTS
The 12th row of the triangle (as given in the reference) is definitely wrong, since the first column of this triangle is known (it is A000560). The row sums are also known - see A000682.
From Roger Ford, Jul 06 2016: (Start)
To determine the first increasing run of the permutation 176852943 start on the left and move to the right counting the consecutive integers.
(1)7685(2)94(3). This permutation a has a first run of (3-1)=2. The permutation 123465 has a first run of (5-1)=4. (1)(2)(3)(4)6(5). (End)
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]
EXAMPLE
Triangle begins:
1;
1, 1;
2, 1, 1;
5, 2, 2, 1;
12, 5, 4, 2, 1;
33, 13, 12, 4, 3, 1;
87, 35, 30, 12, 6, 3, 1;
252, 98, 90, 32, 21, 6, 4, 1;
703, 278, 243, 94, 54, 21, 8, 4, 1;
2105, 812, 745, 270, 175, 57, 32, 8, 5, 1;
6099, 2385, 2108, 808, 485, 181, 84, 32, 10, 5, 1;
...
PROG
(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(Overlapfree(Chords(v)), r[FirstRunLen(v)]++))); r}
for(n=2, 8, print(row(n))) \\ Andrew Howroyd, Dec 07 2018
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
N. J. A. Sloane, Jul 05 2015
EXTENSIONS
Corrected and extended by Roger Ford, Jul 06 2016
STATUS
approved