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A332010 Irregular triangle of denominators of the average value of the first letter over all derangements of {1, 2, ..., n} with k descents.

%I A332010 #17 Mar 30 2024 23:08:33
%S A332010 1,2,4,1,1,8,3,4,16,104,40,24,1,32,392,896,480,54,64,152,308,1496,63,
%T A332010 108,1,128,2276,30384,14410,4315,3024,14,256,7340,153400,235252,24766,
%U A332010 180416,4984,448,1,512,23172,365520,1713160,5794944,3739512,881152,63840,453
%N A332010 Irregular triangle of denominators of the average value of the first letter over all derangements of {1, 2, ..., n} with k descents.
%C A332010 Even-indexed rows have length n - 1; odd-indexed rows have length n - 2.
%C A332010 Conjecture: T(n, 1) = 2^(n-2).
%C A332010 T(2n, 2n-1) = 1, since there is only one derangement of 2n letters with 2n-1 descents.
%C A332010 The analogous sequence for permutations is T'(n, k) = 1. That is, the expected value of the first letter of a permutation with k descents is an integer (namely k + 1).
%e A332010 Triangle begins:
%e A332010     1;
%e A332010     2;
%e A332010     4,    1,     1;
%e A332010     8,    3,     4;
%e A332010    16,  104,    40,    24,    1;
%e A332010    32,  392,   896,   480,   54;
%e A332010    64,  152,   308,  1496,   63,  108,  1;
%e A332010   128, 2276, 30384, 14410, 4315, 3024, 14.
%e A332010 T(4,1) = 4 because the derangements of four letters with one descent are
%e A332010 [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 4 as its denominator.
%Y A332010 Cf. A000166, A083329, A219836.
%Y A332010 Numerators are given by A332009.
%K A332010 nonn,tabf
%O A332010 2,2
%A A332010 _Peter Kagey_, Feb 04 2020