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A307496 Expansion of Product_{k>=1} (1 + ((1 - sqrt(1 - 4*x))/2)^k).

%I A307496 #13 Aug 01 2022 08:38:06
%S A307496 1,1,2,6,18,57,187,629,2156,7502,26427,94053,337653,1221260,4445892,
%T A307496 16277089,59893052,221370725,821499759,3059620076,11432831745,
%U A307496 42848889316,161032785057,606710026659,2291156662259,8670805904186,32879697168622,124910667052026,475357627716839,1811931609379926
%N A307496 Expansion of Product_{k>=1}  (1 + ((1 - sqrt(1 - 4*x))/2)^k).
%C A307496 Catalan transform of A000009 (number of partitions into distinct parts).
%C A307496 From _Peter Bala_, Jan 27 2020: (Start)
%C A307496 This sequence is the main diagonal of the lower triangular array formed by putting A000009 in the first column (k = 0) of the array and then completing the triangle using the relation T(n,k) = T(n-1,k) + T(n,k-1) for k >= 1.
%C A307496     1
%C A307496     1    1
%C A307496     1    2    2
%C A307496     2    4    6    6
%C A307496     2    6   12   18   18
%C A307496     3    9   21   39   57   57
%C A307496   ...
%C A307496 Alternatively, the sequence can be obtained by multiplying A000009 by the array A106566.
%C A307496 (End)
%F A307496 G.f.: Product_{k>=1}  1/(1 - ((1 - sqrt(1 - 4*x))/2)^(2*k-1)).
%F A307496 Sum_{n>=0} a(n)*(x*(1 - x))^n = g.f. of A000009.
%F A307496 a(n) = (1/n) * Sum_{k=1..n} binomial(2*n-k-1,n-k)*k*A000009(k) for n > 0.
%F A307496 a(n) ~ c * 4^n / n^(3/2), where c = 1/sqrt(Pi) * Sum_{k>=0} k*A000009(k)/2^(k+1) = 1.12333545392999500455446757207126193339498222754079045166328600452997969... - _Vaclav Kotesovec_, Jan 28 2020, extended Aug 01 2022
%t A307496 nmax = 29; CoefficientList[Series[Product[(1 + ((1 - Sqrt[1 - 4 x])/2)^k), {k, 1, nmax}], {x, 0, nmax}], x]
%t A307496 nmax = 29; CoefficientList[Series[Product[1/(1 - ((1 - Sqrt[1 - 4 x])/2)^(2 k - 1)), {k, 1, nmax}], {x, 0, nmax}], x]
%t A307496 Join[{1}, Table[1/n Sum[Binomial[2n - k - 1, n - k] k PartitionsQ[k], {k, n}], {n, 29}]]
%Y A307496 Cf. A000009, A100100, A286955, A106566, A307495.
%K A307496 nonn
%O A307496 0,3
%A A307496 _Ilya Gutkovskiy_, Apr 10 2019