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G.f. satisfies: A(x) = Sum_{n>=0} (A(x)^n - 1)^n * x^n / (1-2*x)^(n+1).
4

%I #5 Jan 19 2015 22:41:44

%S 1,2,6,22,102,582,3838,28006,220182,1835942,16090254,147392806,

%T 1406155462,13938397190,143329786526,1527361429606,16853799768310,

%U 192460135069030,2273147914341294,27752645471624486,350014361895293862,4556667705994310342,61182067015102861502,846481772532061942374

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

%F G.f. satisfies:

%F (1) A(x) = Sum_{n>=0} A(x)^(n^2) * x^n / (1-2*x + x*A(x)^n)^(n+1).

%F (2) A(x) = Sum_{n>=0} x^n * Sum_{k=0..n} binomial(n,k) * 2^(n-k) * (A(x)^k - 1)^k.

%e G.f.: A(x) = 1 + 2*x + 6*x^2 + 22*x^3 + 102*x^4 + 582*x^5 + 3838*x^6 +...

%e where we have the identities:

%e (0) A(x) = 1/(1-2*x) + (A(x)-1)*x/(1-2*x)^2 + (A(x)^2-1)^2*x^2/(1-2*x)^3 + (A(x)^3-1)^3*x^3/(1-2*x)^4 + (A(x)^4-1)^4*x^4/(1-2*x)^5 + (A(x)^5-1)^5*x^5/(1-2*x)^6 +...

%e (1) A(x) = 1/(1-x) + A(x)*x/(1-2*x + x*A(x))^2 + A(x)^4*x^2/(1-2*x + x*A(x)^2)^3 + A(x)^9*x^3/(1-2*x + x*A(x)^3)^4 + A(x)^16*x^4/(1-2*x + x*A(x)^4)^5 + A(x)^25*x^5/(1-2*x + x*A(x)^5)^6 + A(x)^36*x^6/(1-2*x + x*A(x)^6)^7 +...

%e (2) A(x) = 1 + x*(2 + (A(x)-1)) + x^2*(2^2 + 2*2*(A(x)-1) + (A(x)^2-1)^2) + x^3*(2^3 + 3*2^2*(A(x)-1) + 3*2*(A(x)^2-1)^2 + (A(x)^3-1)^3) + x^4*(2^4 + 4*2^3*(A(x)-1) + 6*2^2*(A(x)^2-1)^2 + 4*2*(A(x)^3-1)^3 + (A(x)^4-1)^4) + x^5*(2^5 + 5*2^4*(A(x)-1) + 10*2^3*(A(x)^2-1)^2 + 10*2^2*(A(x)^3-1)^3 + 5*2*(A(x)^4-1)^4 + (A(x)^5-1)^5) +...

%o (PARI) {a(n)=local(A=1+x); for(i=1, n, A=sum(m=0, n, (A^m - 1)^m * x^m / (1-2*x +x*O(x^n) )^(m+1) )); polcoeff(A, n)}

%o for(n=0, 25, print1(a(n), ", "))

%o (PARI) {a(n)=local(A=1+x); for(i=1, n, A=sum(m=0, n, A^(m^2) * x^m / (1-2*x + x*A^m +x*O(x^n) )^(m+1) )); polcoeff(A, n)}

%o for(n=0, 25, print1(a(n), ", "))

%o (PARI) {a(n)=local(A=1+x); for(i=1, n, A=sum(m=0, n, x^m * sum(k=0, m, binomial(m, k) * 2^(m-k) * (A^k - 1)^k +x*O(x^n)))); polcoeff(A, n)}

%o for(n=0, 25, print1(a(n), ", "))

%Y Cf. A251177, A251178, A251179, A251180, A244610.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Jan 19 2015