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%I #6 Mar 30 2012 18:37:26
%S 1,1,2,5,15,49,169,605,2226,8364,31956,123770,484862,1917800,7648470,
%T 30722318,124180334,504720369,2061489396,8457050387,34831589583,
%U 143972841512,597034531410,2483173470124,10356092457386,43298360910159
%N G.f. satisfies: A(x) = Sum_{n>=0} x^n*A(x)^A001951(n), where A001951(n) = [n*sqrt(2)].
%C Compare to the g.f. of A190271, G(x), which satisfies:
%C * G(x) = Sum_{n>=0} x^n*G(x)^A001952(n),
%C where A001952 is the complementary Beatty sequence to A001951.
%F G.f. satisfies: A(x) = G(x/A(x)^2) where A(x*G(x)^2) = G(x) is the g.f. of A190271, which in turn satisfies: G(x) = Sum_{n>=0} x^n*G(x)^[n*(2+sqrt(2))].
%F G.f.: A(x) = sqrt(x/Series_Reversion(x*G(x)^2)) where G(x) is the g.f. of A190271.
%e G.f.: A(x) = 1 + x + 2*x^2 + 5*x^3 + 15*x^4 + 49*x^5 + 169*x^6 +...
%e The g.f. satisfies:
%e A(x) = 1 + x*A(x) + x^2*A(x)^2 + x^3*A(x)^4 + x^4*A(x)^5 + x^5*A(x)^7 + x^6*A(x)^8 + x^7*A(x)^9 + x^8*A(x)^11 +...+ x^n*A(x)^A001951(n) +...
%e The g.f. of A190271, G(x) = A(x*G(x)^2), satisfies:
%e G(x) = 1 + x*G(x)^3 + x^2*G(x)^6 + x^3*G(x)^10 + x^4*G(x)^13 + x^5*G(x)^17 + x^6*G(x)^20 + x^7*G(x)^23 +...+ x^n*G(x)^A001952(n) +...
%e and begins:
%e G(x) = 1 + x + 4*x^2 + 22*x^3 + 141*x^4 + 986*x^5 + 7295*x^6 +...
%e Since A(x) = G(x/A(x)^2), then:
%e A(x) = 1 + x/A(x)^2 + 4*x^2/A(x)^4 + 22*x^3/A(x)^6 + 141*x^4/A(x)^8 +...
%o (PARI) {a(n)=local(A=1+x, t=sqrt(2)-1); for(i=1, n, A=sum(m=0, n, x^m*(A+x*O(x^n))^floor(m+m*t))); polcoeff(A, n)}
%Y Cf. A190271, A001951; variant: A186576.
%K nonn
%O 0,3
%A _Paul D. Hanna_, May 06 2011
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