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Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of e.g.f. (1 - (k-1)*log(1 + x))/(1 - k*log(1 + x)).
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%I #33 May 01 2021 17:39:45

%S 1,1,1,1,1,-1,1,1,1,2,1,1,3,2,-6,1,1,5,14,4,24,1,1,7,38,86,14,-120,1,

%T 1,9,74,384,664,38,720,1,1,11,122,1042,4854,6136,216,-5040,1,1,13,182,

%U 2204,18344,73614,66240,600,40320,1,1,15,254,4014,49774,387512,1302552,816672,6240,-362880

%N Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of e.g.f. (1 - (k-1)*log(1 + x))/(1 - k*log(1 + x)).

%F T(0,k)=1 and T(n,k) = Sum_{j=0..n} j! * k^(j-1) * Stirling1(n,j) for n > 0.

%e Square array begins:

%e 1, 1, 1, 1, 1, 1, ...

%e 1, 1, 1, 1, 1, 1, ...

%e -1, 1, 3, 5, 7, 9, ...

%e 2, 2, 14, 38, 74, 122, ...

%e -6, 4, 86, 384, 1042, 2204, ...

%e 24, 14, 664, 4854, 18344, 49774, ...

%t T[0, k_] = 1; T[n_, k_] := Sum[If[k == 0 && j <= 1, 1, k^(j - 1)] * j! * StirlingS1[n, j], {j, 0, n}]; Table[T[k, n - k], {n, 0, 10}, {k, 0, n}] // Flatten (* _Amiram Eldar_, May 01 2021 *)

%Y Columns k=1..3 give A006252, A308878, A335530.

%Y Main diagonal gives A335529.

%Y Cf. A320080.

%K sign,tabl

%O 0,10

%A _Seiichi Manyama_, Jun 12 2020