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%I #10 Aug 30 2017 08:37:17
%S 1,1,0,1,1,0,1,2,1,0,1,3,3,2,0,1,4,6,6,3,0,1,5,10,13,11,5,0,1,6,15,24,
%T 27,20,9,0,1,7,21,40,55,54,38,15,0,1,8,28,62,100,120,109,70,26,0,1,9,
%U 36,91,168,236,258,216,129,45,0,1,10,45,128,266,426,540,544,423,238,78,0,1,11,55,174,402,721,1035,1205,1127,824,437,135,0
%N Square array A(n,k), n>=0, k>=0, read by antidiagonals, where column k is the expansion of k-th power of continued fraction 1/(1 - x/(1 - x^2/(1 - x^3/(1 - x^4/(1 - x^5/(1 - ...)))))).
%H Seiichi Manyama, <a href="/A291652/b291652.txt">Antidiagonals n = 0..139, flattened</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Rogers-RamanujanContinuedFraction.html">Rogers-Ramanujan Continued Fraction</a>
%F G.f. of column k: (1/(1 - x/(1 - x^2/(1 - x^3/(1 - x^4/(1 - x^5/(1 - ...)))))))^k, a continued fraction.
%e G.f. of column k: A_k(x) = 1 + k*x + k*(k + 1)*x^2/2 + k*(k^2 + 3*k + 8)*x^3/6 + k*(k^3 + 6*k^2 + 35*k + 30)*x^4/24 + ...
%e Square array begins:
%e 1, 1, 1, 1, 1, 1, ...
%e 0, 1, 2, 3, 4, 5, ...
%e 0, 1, 3, 6, 10, 15, ...
%e 0, 2, 6, 13, 24, 40, ...
%e 0, 3, 11, 27, 55, 100, ...
%e 0, 5, 20, 54, 120, 236, ...
%t Table[Function[k, SeriesCoefficient[1/(1 + ContinuedFractionK[-x^i, 1, {i, 1, n}])^k, {x, 0, n}]][j - n], {j, 0, 12}, {n, 0, j}] // Flatten
%t Table[Function[k, SeriesCoefficient[((Sum[(-1)^i x^(i (i + 1))/Product[(1 - x^m), {m, 1, i}], {i, 0, n}])/(Sum[(-1)^i x^(i^2)/Product[(1 - x^m), {m, 1, i}], {i, 0, n}]))^k, {x, 0, n}]][j - n], {j, 0, 12}, {n, 0, j}] // Flatten
%Y Columns k=0..1 give A000007, A005169.
%Y Rows n=0..3 give A000012, A001477, A000217, A003600 (with a(0)=0).
%Y Main diagonal gives A291653.
%Y Cf. A286509, A286933.
%K nonn,tabl
%O 0,8
%A _Ilya Gutkovskiy_, Aug 28 2017
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