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A181997
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G.f. satisfies: x = Sum_{n>=1} 1/A(x)^(3*n) * Product_{k=1..n} (1 - 1/A(x)^k).
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6
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1, 1, 2, 9, 46, 259, 1539, 9484, 59961, 386319, 2524940, 16687599, 111264335, 747080253, 5044629212, 34218868880, 232964088130, 1590660486297, 10885758313976, 74627209920879, 512254418843196, 3519150502675731, 24187028454513735, 166249089897708930
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OFFSET
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0,3
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COMMENTS
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Compare the g.f. to the identity:
G(x) = Sum_{n>=0} 1/G(x)^n * Product_{k=1..n} (1 - 1/G(x)^k)
which holds for all power series G(x) such that G(0)=1.
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LINKS
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FORMULA
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G.f. satisfies: 1+x = A(y) where y = x - 2*x^2 - x^3 + 4*x^4 + 4*x^5 + x^6.
G.f. satisfies: x = Sum_{n>=1} 1/A(x)^(n*(n+7)/2) * Product_{k=1..n} (A(x)^k - 1).
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EXAMPLE
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G.f.: A(x) = 1 + x + 2*x^2 + 9*x^3 + 46*x^4 + 259*x^5 + 1539*x^6 +...
The g.f. satisfies:
x = (A(x)-1)/A(x)^4 + (A(x)-1)*(A(x)^2-1)/A(x)^9 + (A(x)-1)*(A(x)^2-1)*(A(x)^3-1)/A(x)^15 + (A(x)-1)*(A(x)^2-1)*(A(x)^3-1)*(A(x)^4-1)/A(x)^22 + (A(x)-1)*(A(x)^2-1)*(A(x)^3-1)*(A(x)^4-1)*(A(x)^5-1)/A(x)^30 +...
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MATHEMATICA
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nmax = 20; aa = ConstantArray[0, nmax]; aa[[1]] = 1; Do[AGF = 1+Sum[aa[[n]]*x^n, {n, 1, j-1}]+koef*x^j; sol=Solve[SeriesCoefficient[Sum[Product[(1-1/AGF^m)/AGF^3, {m, 1, k}], {k, 1, j}], {x, 0, j}]==0, koef][[1]]; aa[[j]]=koef/.sol[[1]], {j, 2, nmax}]; Flatten[{1, aa}] (* Vaclav Kotesovec, Dec 01 2014 *)
CoefficientList[1+InverseSeries[Series[x - 2*x^2 - x^3 + 4*x^4 + 4*x^5 + x^6, {x, 0, 20}], x], x] (* Vaclav Kotesovec, Dec 01 2014 *)
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PROG
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(PARI) {a(n)=if(n<0, 0, polcoeff(1 + serreverse(x - 2*x^2 - x^3 + 4*x^4 + 4*x^5 + x^6 +x^2*O(x^n)), n))}
(PARI) {a(n)=local(A=[1, 1]); for(i=1, n, A=concat(A, 0); A[#A]=-polcoeff(sum(m=1, #A, 1/Ser(A)^(3*m)*prod(k=1, m, 1-1/Ser(A)^k)), #A-1)); A[n+1]}
for(n=0, 25, print1(a(n), ", "))
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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