A332009 - OEIS

%I #18 Mar 30 2024 23:08:37

%S 2,5,11,3,4,23,10,17,47,367,173,121,6,95,1431,3973,2513,331,191,567,

%T 1395,8011,391,761,8,383,8615,139897,78483,27201,21685,113,767,28073,

%U 715277,1298393,158189,1310499,40613,4049,10,1535,89311,1721203,9555617,37414903,27455053,7254087,582767,4549

%N Irregular triangle of numerators of the average value of the first letter over all derangements of {1, 2, ..., n} with k descents.

%C Even-indexed rows have length n - 1; odd-indexed rows have length n - 2.

%C Conjecture: T(n, 1) = A083329(n - 1).

%C T(2n, 2n-1) = 2n, since the only derangement of 2n letters with 2n-1 descents is [2n, 2n-1,...,3,2,1].

%C The analogous sequence for permutations is T'(n, k) = k + 1.

%e Triangle begins:

%e     2;

%e     5;

%e    11,    3,    4;

%e    23,   10,   17;

%e    47,  367,  173,  121,   6;

%e    95, 1431, 3973, 2513, 331;

%e   191,  567, 1395, 8011, 391, 761, 8.

%e T(4,1) = 11 because the derangements of four letters with one descent are

%e [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.

%o (PARI)

%o descents(p)={sum(i=2, #p, p[i]<p[i-1])}

%o isderange(p)={for(i=1, #p, if(p[i]==i, return(0))); 1}

%o 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]))}

%o { for(n=2, 8, print(qrow(n, numerator))) } \\ _Andrew Howroyd_, Feb 02 2022

%Y Cf. A000166, A083329, A219836.

%Y Denominators are given by A332010.

%K nonn,tabf

%O 2,1

%A _Peter Kagey_, Feb 04 2020

%E Terms a(27) and beyond from _Andrew Howroyd_, Feb 02 2022