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%I #22 May 30 2024 06:59:30
%S 1,2,6,24,106,496,2428,12288,63762,337392,1813628,9876096,54365876,
%T 302037408,1691327224,9536234496,54093070626,308474110000,
%U 1767481876540,10170367611008,58746459504884,340513035730944,1979964903739992,11546094361266176,67509252360531940
%N Expansion of g.f. A(x) satisfies A(x) = x * Product_{n>=0} (1 + 2*A(x)^(2^n)).
%C From _Paul D. Hanna_, May 29 2024: (Start)
%C Compare to the following identities of the Catalan function C(x) = x + C(x)^2 (A000108):
%C (1) C(x)^2 = C( x*C(x)*(1 + C(x)) ),
%C (2) C(x)^4 = C( x*C(x)^3*(1 + C(x))*(1 + C(x)^2) ),
%C (3) C(x)^(2^n) = C( x*C(x)^(2^n-1)*Product_{k=0..n-1} (1 + C(x)^(2^k)) ) for n > 0. (End)
%H Paul D. Hanna, <a href="/A356782/b356782.txt">Table of n, a(n) for n = 1..600</a>
%F G.f. A(x) = Sum_{n>=1} a(n)*x^n satisfies:
%F (1) A(x)^2 = A( x*A(x) + 2*x*A(x)^2 ).
%F (2) A(x)^4 = A( x*A(x)^3*(1 + 2*A(x))*(1 + 2*A(x)^2) ).
%F (3) A(x) = x * Product_{n>=0} (1 + 2*A(x)^(2^n)).
%F (4) A(x) = x * Sum_{n>=0} 2^A000120(n) * A(x)^n, where A000120(n) = number of 1's in binary expansion of n.
%F (5) A(x) = x * Sum_{n>=0} A001316(n) * A(x)^n, where A001316(n) = number of odd entries in row n of Pascal's triangle.
%F (6) A(x) = Series_Reversion( x / Product_{n>=0} (1 + 2*x^(2^n)) ).
%F (7) A(x)^(2^n) = A( x*A(x)^(2^n-1)*Product_{k=0..n-1} (1 + 2*A(x)^(2^k)) ) for n > 0. - _Paul D. Hanna_, May 29 2024
%F a(n) ~ c * d^n / n^(3/2), where d = 6.21914802593104186425760427502624741814921... and c = 0.11969150889542130656096151211011193088986347... - _Vaclav Kotesovec_, Apr 05 2024
%F The radius of convergence r of g.f. A(x) and A(r) satisfy 1 = Sum_{n>=0} 2^(n+1) * A(r)^(2^n) / (1 + 2*A(r)^(2^n)) and r = A(r) / Product_{n>=0} (1 + 2*A(r)^(2^n)), where r = 0.160793728631389877... = 1/d (d is given above) and A(r) = 0.368704226036154413721793821323276205873691236089885... - _Paul D. Hanna_, May 04 2024
%e G.f.: A(x) = x + 2*x^2 + 6*x^3 + 24*x^4 + 106*x^5 + 496*x^6 + 2428*x^7 + 12288*x^8 + 63762*x^9 + 337392*x^10 + 1813628*x^11 + 9876096*x^12 + ...
%e such that
%e A(x) = x * (1 + 2*A(x)) * (1 + 2*A(x)^2) * (1 + 2*A(x)^4) * (1 + 2*A(x)^8) * (1 + 2*A(x)^16) * ... * (1 + 2*A(x)^(2^n)) * ...
%e which is equivalent to
%e A(x) = x * (1 + 2*A(x) + 2*A(x)^2 + 4*A(x)^3 + 2*A(x)^4 + 4*A(x)^5 + 4*A(x)^6 + 8*A(x)^7 + 2*A(x)^8 + 4*A(x)^9 + 4*A(x)^10 + 8*A(x)^11 + 4*A(x)^12 + 8*A(x)^13 + 8*A(x)^14 + 16*A(x)^15 + ... + 2^A000120(n)*A(x)^n + ...).
%e Also, g.f. A(x) satisfies
%e A(x)^2 = A( x*A(x) + 2*x*A(x)^2 )
%e where
%e A(x)^2 = x^2 + 4*x^3 + 16*x^4 + 72*x^5 + 344*x^6 + 1704*x^7 + 8688*x^8 + 45328*x^9 + 240856*x^10 + 1298984*x^11 + 7092544*x^12 + ...
%o (PARI) {a(n) = my(A = serreverse( x/prod(k=0,#binary(n),1 + 2*x^(2^k) +x*O(x^n)) ));
%o polcoeff(A,n)}
%o for(n=1,30,print1(a(n),", "))
%Y Cf. A372534, A000120, A001316.
%K nonn
%O 1,2
%A _Paul D. Hanna_, Sep 01 2022
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