2,5

The sums s(n) = Sum_k k*T(n,k) give A259700.

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).

Albert Sade, Sur les Chevauchements des Permutations, published by the author, Marseille, 1949. [Annotated scanned copy]

Triangle begins:

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, 22, 87, 158, 148, 71, 16, 1;

1, 30, 151, 352, 436, 301, 114, 20, 1;

1, 46, 280, 791, 1210, 1092, 589, 175, 25, 1;

1, 62, 464, 1592, 2969, 3377, 2408, 1038, 256, 30, 1;

...

(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}

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}

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}

for(n=2, 8, print(row(n))) \\ Andrew Howroyd, Dec 07 2018

Row sums give A000682.

Cf. A259700.

Sequence in context: A132731 A128966 A055907 * A274643 A172991 A203906

Adjacent sequences: A259695 A259696 A259697 * A259699 A259700 A259701

nonn,tabl

N. J. A. Sloane, Jul 05 2015

Corrected and extended by Roger Ford, Jul 06 2016

approved