OFFSET
2,1
COMMENTS
Even-indexed rows have length n - 1; odd-indexed rows have length n - 2.
Conjecture: T(n, 1) = A083329(n - 1).
T(2n, 2n-1) = 2n, since the only derangement of 2n letters with 2n-1 descents is [2n, 2n-1,...,3,2,1].
The analogous sequence for permutations is T'(n, k) = k + 1.
EXAMPLE
Triangle begins:
2;
5;
11, 3, 4;
23, 10, 17;
47, 367, 173, 121, 6;
95, 1431, 3973, 2513, 331;
191, 567, 1395, 8011, 391, 761, 8.
T(4,1) = 11 because the derangements of four letters with one descent are
[2,3,4,1], [2,4,1,3], [3,4,1,2], and [4,1,2,3], and the expected value of the first letter is (2+2+3+4)/4 = 11/4, which has 11 as its numerator.
PROG
(PARI)
descents(p)={sum(i=2, #p, p[i]<p[i-1])}
isderange(p)={for(i=1, #p, if(p[i]==i, return(0))); 1}
qrow(n, f=numerator)={my(s=vector(n-n%2-1), d=vector(#s)); forperm(n, p, if(isderange(p), my(k=descents(p)); d[k]++; s[k]+=p[1])); vector(#s, k, f(s[k]/d[k]))}
{ for(n=2, 8, print(qrow(n, numerator))) } \\ Andrew Howroyd, Feb 02 2022
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Peter Kagey, Feb 04 2020
EXTENSIONS
Terms a(27) and beyond from Andrew Howroyd, Feb 02 2022
STATUS
approved