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A336633 Triangle read by rows: T(n,k) is the number of generalized permutations related to the degenerate Eulerian numbers with exactly k ascents (0 <= k <= max(0,n-1)).

%I A336633 #39 Jan 11 2021 23:20:43
%S A336633 1,1,2,2,6,16,6,24,116,116,24,120,888,1624,888,120,720,7416,20984,
%T A336633 20984,7416,720,5040,67968,270432,419680,270432,67968,5040,40320,
%U A336633 682272,3587904,7861664,7861664,3587904,682272,40320,362880,7467840,49701024,144570624,204403264,144570624
%N A336633 Triangle read by rows: T(n,k) is the number of generalized permutations related to the degenerate Eulerian numbers with exactly k ascents (0 <= k <= max(0,n-1)).
%H A336633 Orli Herscovici, <a href="https://arxiv.org/abs/2007.13205">Generalized permutations related to the degenerate Eulerian numbers</a>, arXiv preprint arXiv:2007.13205 [math.CO], 2020.
%F A336633 T(n,k) = (n+k)*T(n-1,k) + (2*n-k-1)*T(n-1,k-1) for positive integers n and 0 <= k < n; T(0,0)=1 (or T(1,0)=1); otherwise T(n,k)=0.
%F A336633 From _Peter Bala_, Jan 08 2021: (Start)
%F A336633 The following remarks are all conjectures:
%F A336633 The e.g.f. (without the initial 1) A(x,t) = x + (2 + 2*t)*x^2/2! + (6 + 16*t + 6*t^2)*x^3/3! + ... satisfies the autonomous differential equation dA/dx = (1 + A)^2*(1 + t*A)^2.
%F A336633 The series reversion of A(x,t) with respect to x equals Integral_{u = 0..x} 1/((1 + u)^2*(1 + t*u)^2) du.
%F A336633 Let f(x,t) = (1 + x)^2*(1 + t*x)^2 and let D be the operator f(x,t)*d/dx. Then the (n+1)-th row polynomial = D^n(f(x,t)) evaluated at x = 0. (End)
%e A336633 Triangle T(n,k) (with rows n >= 0 and columns k = 0..max(0,n-1)) begins:
%e A336633      1;
%e A336633      1;
%e A336633      2,     2;
%e A336633      6,    16,      6;
%e A336633     24,   116,    116,     24;
%e A336633    120,   888,   1624,    888,    120;
%e A336633    720,  7416,  20984,  20984,   7416,   720;
%e A336633   5040, 67968, 270432, 419680, 270432, 67968, 5040;
%e A336633   ...
%p A336633 Tnk[0, 0] := 1; for n to N do
%p A336633     for k from 0 to n do if 0 < k and k < n then Tnk[n, k] := (n + k)*Tnk[n - 1, k] + (2*n - k - 1)*Tnk[n - 1, k - 1]; else if k = 0 then Tnk[n, k] := (n + k)*Tnk[n - 1, k]; else Tnk[n, k] := 0; end if; end if; end do;
%p A336633 end do
%Y A336633 Columns k = 0..1 give: A000142, A288964. Row sums give A007559.
%Y A336633 Cf. A008292, A008517.
%K A336633 easy,nonn,tabf
%O A336633 0,3
%A A336633 _Orli Herscovici_, Jul 28 2020