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Square array A(n,k), n>=0, k>=0, read by antidiagonals, where column k is the expansion of continued fraction 1/(1 - x/(1 - k*x/(1 - k^2*x/(1 - k^3*x/(1 - k^4*x/(1 - ...)))))).
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%I #24 Jun 06 2020 12:59:13

%S 1,1,1,1,1,1,1,1,2,1,1,1,3,5,1,1,1,4,17,14,1,1,1,5,43,171,42,1,1,1,6,

%T 89,1252,3113,132,1,1,1,7,161,5885,104098,106419,429,1,1,1,8,265,

%U 20466,1518897,25511272,7035649,1430,1,1,1,9,407,57799,12833546,1558435125,18649337311,915028347,4862,1

%N Square array A(n,k), n>=0, k>=0, read by antidiagonals, where column k is the expansion of continued fraction 1/(1 - x/(1 - k*x/(1 - k^2*x/(1 - k^3*x/(1 - k^4*x/(1 - ...)))))).

%C This is the transpose of the array in A090182.

%H Seiichi Manyama, <a href="/A290759/b290759.txt">Antidiagonals n = 0..55, flattened</a>

%F G.f. of column k: 1/(1 - x/(1 - k*x/(1 - k^2*x/(1 - k^3*x/(1 - k^4*x/(1 - ...)))))), a continued fraction.

%e G.f. of column k: A_k(x) = 1 + x + (k + 1)*x^2 + (k^3 + k^2 + 2*k + 1)*x^3 + (k^6 + k^5 + 2*k^4 + 3*k^3 + 3*k^2 + 3*k + 1)*x^4 + ...

%e Square array begins:

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

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

%e 1, 2, 3, 4, 5, 6, ...

%e 1, 5, 17, 43, 89, 161, ...

%e 1, 14, 171, 1252, 5885, 20466, ...

%e 1, 42, 3113, 104098, 1518897, 12833546, ...

%p A:= proc(n, k) option remember; `if`(n=0, 1, add(

%p A(j, k)*A(n-j-1, k)*k^j, j=0..n-1))

%p end:

%p seq(seq(A(n, d-n), n=0..d), d=0..12); # _Alois P. Heinz_, Aug 10 2017

%t Table[Function[k, SeriesCoefficient[1/(1 - x/(1 + ContinuedFractionK[-k^i x, 1, {i, 1, n}])), {x, 0, n}]][j - n], {j, 0, 10}, {n, 0, j}] // Flatten

%o (Python)

%o from sympy.core.cache import cacheit

%o @cacheit

%o def A(n, k): return 1 if n==0 else sum(A(j, k)*A(n - j - 1, k)*k**j for j in range(n))

%o for n in range(13): print([A(k, n - k) for k in range(n + 1)]) # _Indranil Ghosh_, Aug 10 2017, after Maple code

%Y Columns k=0-11 give: A000012, A000108, A015083, A015084, A015085, A015086, A015089, A015091, A015092, A015093, A015095, A015096.

%Y Main diagonal gives A290777.

%Y Cf. A090182, A290789.

%K nonn,tabl

%O 0,9

%A _Ilya Gutkovskiy_, Aug 09 2017