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%I #13 Aug 20 2017 10:01:58
%S 1,1,1,1,1,2,1,1,1,5,1,1,1,2,14,1,1,1,1,3,42,1,1,1,1,1,5,132,1,1,1,1,
%T 1,2,9,429,1,1,1,1,1,1,3,15,1430,1,1,1,1,1,1,1,4,26,4862,1,1,1,1,1,1,
%U 1,1,5,45,16796,1,1,1,1,1,1,1,1,1,7,78,58786,1,1,1,1,1,1,1,1,1,2,10,135,208012
%N Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of the continued fraction 1/(1 - x/(1 - x^(2^k)/(1 - x^(3^k)/(1 - x^(4^k)/(1 - x^(5^k)/(1 - ...)))))).
%F G.f. of column k: 1/(1 - x/(1 - x^(2^k)/(1 - x^(3^k)/(1 - x^(4^k)/(1 - x^(5^k)/(1 - ...)))))), a continued fraction.
%e Square array begins:
%e 1, 1, 1, 1, 1, 1, ...
%e 1, 1, 1, 1, 1, 1, ...
%e 2, 1, 1, 1, 1, 1, ...
%e 5, 2, 1, 1, 1, 1, ...
%e 14, 3, 1, 1, 1, 1, ...
%e 42, 5, 2, 1, 1, 1, ...
%t Table[Function[k, SeriesCoefficient[1/(1 + ContinuedFractionK[-x^(i^k), 1, {i, 1, n}]), {x, 0, n}]][j - n], {j, 0, 12}, {n, 0, j}] // Flatten
%Y Columns k = 0..5 give A000108, A005169, A206739, A291146, A291149, A291168.
%K nonn,tabl
%O 0,6
%A _Ilya Gutkovskiy_, Aug 10 2017