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A264227 G.f. A(x) satisfies: A(x)^2 = A( x^2/(1-10*x) ), with A(0) = 0.

%I A264227 #31 Nov 16 2015 18:30:37
%S A264227 1,5,40,350,3220,30500,294625,2886875,28598035,285786575,2876602225,
%T A264227 29131678625,296574083425,3033183585125,31148390740375,
%U A264227 321040368434375,3319845741478030,34433523106882550,358129419509956150,3734203057793066750,39027568927659117700,408777143934160983500,4290195975642644398000,45111124579414224095000
%N A264227 G.f. A(x) satisfies: A(x)^2 = A( x^2/(1-10*x) ), with A(0) = 0.
%C A264227 Radius of convergence is r = 1/11, where r = r^2/(1-10*r), with A(r) = 1.
%C A264227 Compare to a g.f. M(x) of Motzkin numbers: M(x)^2 = M(x^2/(1-2*x)) where M(x) = (1-x - sqrt(1-2*x-3*x^2))/(2*x).
%H A264227 Paul D. Hanna, <a href="/A264227/b264227.txt">Table of n, a(n) for n = 1..300</a>
%F A264227 G.f. also satisfies:
%F A264227 (1) A(x) = -A( -x/(1-10*x) ).
%F A264227 (2) A( x/(1+5*x) ) = -A( -x/(1-5*x) ), an odd function.
%F A264227 (3) A( x/(1+5*x) )^2 = A( x^2/(1-25*x^2) ), an even function.
%F A264227 (4) A(x)^4 = A( x^4/((1-10*x)*(1-10*x-10*x^2)) ).
%F A264227 (5) [x^(2*n+1)] (x/A(x))^(2*n) = 0 for n>=0.
%F A264227 (6) [x^(2^n+k)] (x/A(x))^(2^n) = 0 for k=1..2^n-1, n>=1.
%F A264227 Given g.f. A(x), let F(x) denote the g.f. of A264415, then:
%F A264227 (7) A(x) = F(A(x))^2 * x/(1+25*x),
%F A264227 (8) A(x) = F(A(x)^2) * x/(1-5*x),
%F A264227 (9) A( x/(F(x)^2 - 25*x) ) = x,
%F A264227 (10) A( x/(F(x^2) + 5*x) ) = x,
%F A264227 where F(x)^2 = F(x^2) + 30*x.
%F A264227 Sum_{k=0..n} binomial(n,k) * (-5)^(n-k) * a(k+1) = 0 for odd n.
%F A264227 Sum_{k=0..n} binomial(n,k) *(-10)^(n-k) * a(k+1) = (-1)^n * a(n+1) for n>=0.
%e A264227 G.f.: A(x) = x + 5*x^2 + 40*x^3 + 350*x^4 + 3220*x^5 + 30500*x^6 + 294625*x^7 + 2886875*x^8 + 28598035*x^9 + 285786575*x^10 + 2876602225*x^11 +...
%e A264227 where A(x)^2 = A(x^2/(1-10*x)).
%e A264227 RELATED SERIES.
%e A264227 A(x)^2 = x^2 + 10*x^3 + 105*x^4 + 1100*x^5 + 11540*x^6 + 121200*x^7 + 1274350*x^8 + 13414000*x^9 + 141353220*x^10 + 1491161000*x^11 + 15747360500*x^12 +...
%e A264227 A( x/(1+5*x) ) = x + 15*x^3 + 345*x^5 + 9000*x^7 + 251160*x^9 + 7328475*x^11 + 220880925*x^13 + 6824229750*x^15 + 214969962405*x^17 + 6877343600775*x^19 +...
%e A264227 A( x^2/(1-25*x^2) ) = x^2 + 30*x^4 + 915*x^6 + 28350*x^8 + 891345*x^10 + 28401750*x^12 + 915916500*x^14 + 29852415000*x^16 + 982068551160*x^18 +...
%e A264227 where A( x^2/(1-25*x^2) ) = A( x/(1+5*x) )^2.
%e A264227 Let B(x) = x/Series_Reversion(A(x)), then A(x) = x*B(A(x)), where
%e A264227 B(x) = 1 + 5*x + 15*x^2 - 105*x^4 + 1575*x^6 - 29190*x^8 + 603225*x^10 - 13352850*x^12 + 309605625*x^14 - 7422255645*x^16 +...+ A264415(n)*x^(2*n) +...
%e A264227 such that B(x) = F(x^2) + 5*x = F(x)^2 - 25*x and F(x) is the g.f. of A264415.
%o A264227 (PARI) {a(n) = my(A=x); for(i=1,n, A = sqrt( subst(A,x,x^2/(1-10*x +x*O(x^n))) ) ); polcoeff(A,n)}
%o A264227 for(n=1,30,print1(a(n),", "))
%Y A264227 Cf. A264415, A264224, A264225, A264226.
%K A264227 nonn
%O A264227 1,2
%A A264227 _Paul D. Hanna_, Nov 08 2015