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%I #15 Aug 16 2018 21:19:29
%S 1,4,22,308,7877,287224,13293116,735955720,47105160785,3410314286768,
%T 275071315285416,24442342714268592,2371821148074889444,
%U 249559207019813962752,28303003280888905543584,3442273720243525242224992,446977352681757476329452018,61724119095080041604018873868,9033234491867095630258647812994,1396682556807057529868101744945708,227509260041431637641628131782970335
%N G.f. A(x) satisfies: Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(4*n) )^n = 1.
%H Paul D. Hanna, <a href="/A317803/b317803.txt">Table of n, a(n) for n = 0..200</a>
%F G.f. A(x) satisfies:
%F (1) 1 = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(4*n) )^n.
%F (2) A(x) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(4*n+4) )^n.
%F (3) 1 = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(4*n+4) )^n / (1+x)^(4*n+4).
%F (4) Let B(x,p) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(4*n + p) )^n ,
%F then B(x,p) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(4*(n+1)) )^n / (1+x)^((4-p)*(n+1)), where B(x,0) = 1 and B(x,4) = A(x).
%F a(n) ~ 2^(2*n - log(2)/8 - 5/2) * n^n / (sqrt(1-log(2)) * exp(n) * (log(2))^(2*n+1)). - _Vaclav Kotesovec_, Aug 13 2018
%e G.f.: A(x) = 1 + 4*x + 22*x^2 + 308*x^3 + 7877*x^4 + 287224*x^5 + 13293116*x^6 + 735955720*x^7 + 47105160785*x^8 + 3410314286768*x^9 + 275071315285416*x^10 + ...
%e such that
%e 1 = 1 + (1/A(x) - 1/(1+x)^4) + (1/A(x) - 1/(1+x)^8)^2 + (1/A(x) - 1/(1+x)^12)^3 + (1/A(x) - 1/(1+x)^16)^4 + (1/A(x) - 1/(1+x)^20)^5 + (1/A(x) - 1/(1+x)^24)^6 + (1/A(x) - 1/(1+x)^28)^7 + (1/A(x) - 1/(1+x)^32)^8 + ...
%e Also,
%e A(x) = 1 + (1/A(x) - 1/(1+x)^8) + (1/A(x) - 1/(1+x)^12)^2 + (1/A(x) - 1/(1+x)^16)^3 + (1/A(x) - 1/(1+x)^20)^4 + (1/A(x) - 1/(1+x)^24)^5 + (1/A(x) - 1/(1+x)^28)^6 + (1/A(x) - 1/(1+x)^32)^7 + (1/A(x) - 1/(1+x)^36)^8 + ...
%e RELATED SERIES.
%e (1) The series B(x,1) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(4*n+1) )^n begins
%e B(x,1) = 1 + x + 4*x^2 + 58*x^3 + 1482*x^4 + 53953*x^5 + 2496149*x^6 + 138245508*x^7 + 8853719964*x^8 + 641386920943*x^9 + 51762649442019*x^10 + ...
%e where B(x,1) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(4*n+4) )^n / (1+x)^(3*n+3).
%e (2) The series B(x,2) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(4*n+2) )^n begins
%e B(x,2) = 1 + 2*x + 9*x^2 + 128*x^3 + 3270*x^4 + 119002*x^5 + 5502295*x^6 + 304531768*x^7 + 19491119849*x^8 + 1411222743454*x^9 + 113839065423087*x^10 + ...
%e where B(x,2) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(4*n+4) )^n / (1+x)^(2*n+2).
%e (3) The series B(x,3) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(4*n+3) )^n begins
%e B(x,3) = 1 + 3*x + 15*x^2 + 211*x^3 + 5392*x^4 + 196341*x^5 + 9079538*x^6 + 502467023*x^7 + 32153605481*x^8 + 2327561975059*x^9 + 187722580703289*x^10 + ...
%e where B(x,3) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(4*n+4) )^n / (1+x)^(n+1).
%o (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))^(4*m+4) )^m ) )[#A]/2 ); A[n+1]}
%o for(n=0, 25, print1(a(n), ", "))
%Y Cf. A317339, A317801, A317802, A317995, A317668.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Aug 12 2018
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