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%I #11 Jan 24 2019 22:31:26
%S 1,1,4,32,424,7696,173442,4619266,141315896,4874012942,186981188532,
%T 7896318230898,364045464940596,18196879341802488,980406767669688312,
%U 56648325010279262864,3494752526532046751322,229295129566323954429582,15944415062268028208782178,1171388932048172852048806000,90667183883120180538001042398
%N G.f. satisfies: A(x) = Sum_{n>=0} ( (1+x)^n - A(x)^(1/2) )^n / ( 2 - (1+x)^n * A(x)^(1/2) )^(n+1).
%C It is remarkable that the g.f. should consist entirely of integer coefficients.
%H Paul D. Hanna, <a href="/A322735/b322735.txt">Table of n, a(n) for n = 0..200</a>
%F G.f. A(x) along with B(x) = A(x)^(1/2) satisfy:
%F (1) A(x) = Sum_{n>=0} ( (1+x)^n - B(x) )^n / ( 2 - (1+x)^n*B(x) )^(n+1),
%F (2) A(x) = Sum_{n>=0} ( (1+x)^n + B(x) )^n / ( 2 + (1+x)^n*B(x) )^(n+1).
%e G.f.: A(x) = 1 + x + 4*x^2 + 32*x^3 + 424*x^4 + 7696*x^5 + 173442*x^6 + 4619266*x^7 + 141315896*x^8 + 4874012942*x^9 + 186981188532*x^10 + ...
%e such that A(x) and B = A(x)^(1/2) satisfy
%e A(x) = 1/(2 - B) + ((1+x) - B)/(2 - (1+x)*B)^2 + ((1+x)^2 - B)^2/(2 - (1+x)^2*B)^3 + ((1+x)^3 - B)^3/(2 - (1+x)^3*B)^4 + ((1+x)^4 - B)^4/(2 - (1+x)^4*B)^5 + ((1+x)^5 - B)^5/(2 - (1+x)^5*B)^6 + ...
%e also,
%e A(x) = 1/(2 + B) + ((1+x) + B)/(2 + (1+x)*B)^2 + ((1+x)^2 + B)^2/(2 + (1+x)^2*B)^3 + ((1+x)^3 + B)^3/(2 + (1+x)^3*B)^4 + ((1+x)^4 + B)^4/(2 + (1+x)^4*B)^5 + ((1+x)^5 + B)^5/(2 + (1+x)^5*B)^6 + ...
%e Notice that A(x)^(1/2) is not an integer series, but instead begins
%e A(x)^(1/2) = 1 + 2*(x/4) + 30*(x/4)^2 + 964*(x/4)^3 + 51894*(x/4)^4 + 3807644*(x/4)^5 + 345572460*(x/4)^6 + 36985627016*(x/4)^7 + 4541283789862*(x/4)^8 + 628123762214444*(x/4)^9 + 96578670976842436*(x/4)^10 + ...
%e thus, given the definition, it is remarkable that A(x) should be an integer series.
%o (PARI) {a(n) = my(A=[1,1]); for(i=1, n, A=concat(A, 0); A = Vec( sum(m=0, #A, ( (1+x)^m - Ser(A)^(1/2) )^m / (2 - (1+x)^m*Ser(A)^(1/2))^(m+1) ) ) ); A[n+1]}
%o for(n=0, 30, print1(a(n), ", "))
%Y Cf. A317350, A322737.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Jan 24 2019