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%I #9 Aug 13 2018 03:57:31
%S 1,3,12,127,2445,66939,2324026,96491718,4631150520,251413638241,
%T 15206137508067,1013223645173301,73729926406815893,
%U 5817609547850902791,494790115210979151063,45129281235546080750387,4394695321061357601501585,455127430187799524613334185,49952816657399856543050669882,5792366218971732073257841216098,707622192835283858272032714820854
%N G.f. A(x) satisfies: Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(3*n) )^n = 1.
%H Paul D. Hanna, <a href="/A317802/b317802.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)^(3*n) )^n.
%F (2) A(x) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(3*n+3) )^n.
%F (3) 1 = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(3*n+3) )^n / (1+x)^(3*n+3).
%F (4) Let B(x) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(3*n+1) )^n , then
%F B(x) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(3*n+3) )^n / (1+x)^(2*n+2).
%F (5) Let C(x) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(3*n+2) )^n , then
%F C(x) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(3*n+3) )^n / (1+x)^(n+1).
%F a(n) ~ 2^(-log(2)/6 - 5/2) * 3^n * n^n / (sqrt(1-log(2)) * exp(n) * (log(2))^(2*n+1)). - _Vaclav Kotesovec_, Aug 13 2018
%e G.f.: A(x) = 1 + 3*x + 12*x^2 + 127*x^3 + 2445*x^4 + 66939*x^5 + 2324026*x^6 + 96491718*x^7 + 4631150520*x^8 + 251413638241*x^9 + 15206137508067*x^10 + ...
%e such that
%e 1 = 1 + (1/A(x) - 1/(1+x)^3) + (1/A(x) - 1/(1+x)^6)^2 + (1/A(x) - 1/(1+x)^9)^3 + (1/A(x) - 1/(1+x)^12)^4 + (1/A(x) - 1/(1+x)^15)^5 + (1/A(x) - 1/(1+x)^18)^6 + (1/A(x) - 1/(1+x)^21)^7 + (1/A(x) - 1/(1+x)^24)^8 + ...
%e Also,
%e A(x) = 1 + (1/A(x) - 1/(1+x)^6) + (1/A(x) - 1/(1+x)^9)^2 + (1/A(x) - 1/(1+x)^12)^3 + (1/A(x) - 1/(1+x)^15)^4 + (1/A(x) - 1/(1+x)^18)^5 + (1/A(x) - 1/(1+x)^21)^6 + (1/A(x) - 1/(1+x)^24)^7 + (1/A(x) - 1/(1+x)^27)^8 + ...
%e RELATED SERIES.
%e (1) The series B(x) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(3*n+1) )^n begins
%e B(x) = 1 + x + 3*x^2 + 33*x^3 + 634*x^4 + 17326*x^5 + 601161*x^6 + 24961740*x^7 + 1198455358*x^8 + 65087157334*x^9 + 3938132342935*x^10 + ...
%e restated,
%e B(x) = 1 + (1/A(x) - 1/(1+x)^4) + (1/A(x) - 1/(1+x)^7)^2 + (1/A(x) - 1/(1+x)^10)^3 + (1/A(x) - 1/(1+x)^13)^4 + (1/A(x) - 1/(1+x)^16)^5 + (1/A(x) - 1/(1+x)^19)^6 + (1/A(x) - 1/(1+x)^22)^7 + (1/A(x) - 1/(1+x)^25)^8 + ...
%e which can also be written
%e B(x) = 1/(1+x)^2 + (1/A(x) - 1/(1+x)^6)/(1+x)^4 + (1/A(x) - 1/(1+x)^9)^2/(1+x)^6 + (1/A(x) - 1/(1+x)^12)^3/(1+x)^8 + (1/A(x) - 1/(1+x)^15)^4/(1+x)^10 + (1/A(x) - 1/(1+x)^18)^5/(1+x)^12 + (1/A(x) - 1/(1+x)^21)^6/(1+x)^14 + (1/A(x) - 1/(1+x)^24)^7/(1+x)^16 + (1/A(x) - 1/(1+x)^27)^8/(1+x)^18 + ...
%e ...
%e (2) The series C(x) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(3*n+2) )^n begins
%e C(x) = 1 + 2*x + 7*x^2 + 75*x^3 + 1442*x^4 + 39413*x^5 + 1367095*x^6 + 56736076*x^7 + 2722528369*x^8 + 147785496105*x^9 + 8937999326808*x^10 + ...
%e restated,
%e C(x) = 1 + (1/A(x) - 1/(1+x)^5) + (1/A(x) - 1/(1+x)^8)^2 + (1/A(x) - 1/(1+x)^11)^3 + (1/A(x) - 1/(1+x)^14)^4 + (1/A(x) - 1/(1+x)^17)^5 + (1/A(x) - 1/(1+x)^20)^6 + (1/A(x) - 1/(1+x)^23)^7 + (1/A(x) - 1/(1+x)^26)^8 + ...
%e which can also be written
%e C(x) = 1/(1+x) + (1/A(x) - 1/(1+x)^6)/(1+x)^2 + (1/A(x) - 1/(1+x)^9)^2/(1+x)^3 + (1/A(x) - 1/(1+x)^12)^3/(1+x)^4 + (1/A(x) - 1/(1+x)^15)^4/(1+x)^5 + (1/A(x) - 1/(1+x)^18)^5/(1+x)^6 + (1/A(x) - 1/(1+x)^21)^6/(1+x)^7 + (1/A(x) - 1/(1+x)^24)^7/(1+x)^8 + (1/A(x) - 1/(1+x)^27)^8/(1+x)^9 + ...
%e ...
%e Compare the above series to
%e 1 = 1/(1+x)^3 + (1/A(x) - 1/(1+x)^6)/(1+x)^6 + (1/A(x) - 1/(1+x)^9)^2/(1+x)^9 + (1/A(x) - 1/(1+x)^12)^3/(1+x)^12 + (1/A(x) - 1/(1+x)^15)^4/(1+x)^15 + (1/A(x) - 1/(1+x)^18)^5/(1+x)^18 + (1/A(x) - 1/(1+x)^21)^6/(1+x)^21 + (1/A(x) - 1/(1+x)^24)^7/(1+x)^24 + (1/A(x) - 1/(1+x)^27)^8/(1+x)^27 + ...
%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))^(3*m+3) )^m ) )[#A]/2 ); A[n+1]}
%o for(n=0, 25, print1(a(n), ", "))
%Y Cf. A317339, A317801, A317803, A317667.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Aug 12 2018