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A317801 G.f. A(x) satisfies: Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(2*n) )^n = 1.

%I A317801 #8 Aug 13 2018 03:56:39
%S A317801 1,2,5,36,465,8504,196900,5448960,174282930,6304990774,254130115519,
%T A317801 11285072891720,547300883009420,28782595694322682,1631647871872693931,
%U A317801 99196878522983084624,6438935162966517263673,444501575692314862825620,32520919561355194120182078,2513781290973908970634293260,204713526722520414595009119193
%N A317801 G.f. A(x) satisfies: Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(2*n) )^n  =  1.
%H A317801 Paul D. Hanna, <a href="/A317801/b317801.txt">Table of n, a(n) for n = 0..200</a>
%F A317801 G.f. A(x) satisfies:
%F A317801 (1) 1 = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(2*n) )^n.
%F A317801 (2) A(x) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(2*n+2) )^n.
%F A317801 (3) 1 = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(2*n+2) )^n / (1+x)^(2*n+2).
%F A317801 (4) Let B(x) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(2*n+1) )^n ,
%F A317801 then B(x) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(2*n+2) )^n / (1+x)^(n+1).
%F A317801 a(n) ~ 2^(n - log(2)/4 - 5/2) * n^n / (sqrt(1-log(2)) * exp(n) * (log(2))^(2*n+1)). - _Vaclav Kotesovec_, Aug 13 2018
%e A317801 G.f.: A(x) = 1 + 2*x + 5*x^2 + 36*x^3 + 465*x^4 + 8504*x^5 + 196900*x^6 + 5448960*x^7 + 174282930*x^8 + 6304990774*x^9 + 254130115519*x^10 + ...
%e A317801 such that
%e A317801 1 = 1  +  (1/A(x) - 1/(1+x)^2)  +  (1/A(x) - 1/(1+x)^4)^2  +  (1/A(x) - 1/(1+x)^6)^3  +  (1/A(x) - 1/(1+x)^8)^4  +  (1/A(x) - 1/(1+x)^10)^5  +  (1/A(x) - 1/(1+x)^12)^6  +  (1/A(x) - 1/(1+x)^14)^7  +  (1/A(x) - 1/(1+x)^16)^8  + ...
%e A317801 Also,
%e A317801 A(x) = 1  +  (1/A(x) - 1/(1+x)^4)  +  (1/A(x) - 1/(1+x)^6)^2  +  (1/A(x) - 1/(1+x)^8)^3  +  (1/A(x) - 1/(1+x)^10)^4  +  (1/A(x) - 1/(1+x)^12)^5  +  (1/A(x) - 1/(1+x)^14)^6  +  (1/A(x) - 1/(1+x)^16)^7  +  (1/A(x) - 1/(1+x)^18)^8  + ...
%e A317801 RELATED SERIES.
%e A317801 The series B(x) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(2*n+1) )^n begins
%e A317801 B(x) = 1 + x + 2*x^2 + 15*x^3 + 193*x^4 + 3523*x^5 + 81497*x^6 + 2254695*x^7 + 72114516*x^8 + 2609080565*x^9 + 105173092838*x^10 + ...
%e A317801 restated,
%e A317801 B(x) = 1  +  (1/A(x) - 1/(1+x)^3)  +  (1/A(x) - 1/(1+x)^5)^2  +  (1/A(x) - 1/(1+x)^7)^3  +  (1/A(x) - 1/(1+x)^9)^4  +  (1/A(x) - 1/(1+x)^11)^5  +  (1/A(x) - 1/(1+x)^13)^6  +  (1/A(x) - 1/(1+x)^15)^7  +  (1/A(x) - 1/(1+x)^17)^8  + ...
%e A317801 which can also be written
%e A317801 B(x) = 1/(1+x)  +  (1/A(x) - 1/(1+x)^4)/(1+x)^2  +  (1/A(x) - 1/(1+x)^6)^2/(1+x)^3  +  (1/A(x) - 1/(1+x)^8)^3/(1+x)^4  +  (1/A(x) - 1/(1+x)^10)^4/(1+x)^5  +  (1/A(x) - 1/(1+x)^12)^5/(1+x)^6  +  (1/A(x) - 1/(1+x)^14)^6/(1+x)^7  +  (1/A(x) - 1/(1+x)^16)^7/(1+x)^8  +  (1/A(x) - 1/(1+x)^18)^8/(1+x)^9  + ...
%e A317801 Compare the above to
%e A317801 1 = 1/(1+x)^2  +  (1/A(x) - 1/(1+x)^4)/(1+x)^4  +  (1/A(x) - 1/(1+x)^6)^2/(1+x)^6  +  (1/A(x) - 1/(1+x)^8)^3/(1+x)^8  +  (1/A(x) - 1/(1+x)^10)^4/(1+x)^10  +  (1/A(x) - 1/(1+x)^12)^5/(1+x)^12  +  (1/A(x) - 1/(1+x)^14)^6/(1+x)^14  +  (1/A(x) - 1/(1+x)^16)^7/(1+x)^16  +  (1/A(x) - 1/(1+x)^18)^8/(1+x)^18  + ...
%o A317801 (PARI) {a(n) = my(A=[1]); for(i=1, n, A=concat(A, 0); A[#A] = Vec( sum(m=0, #A, ( 1/Ser(A) - 1/(1+x +x*O(x^#A))^(2*m+2) )^m ) )[#A]/2 ); A[n+1]}
%o A317801 for(n=0, 25, print1(a(n), ", "))
%Y A317801 Cf. A317339, A317802, A317803, A317666.
%K A317801 nonn
%O A317801 0,2
%A A317801 _Paul D. Hanna_, Aug 12 2018