0,2

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

A(x) = 1 + 2*x - 6*x^2 + 28*x^3 - 410*x^4 - 40548*x^5 - 25678044*x^6 -...

Illustrate 1 + 4x = Sum_{n>=0} x^n * [y^n] A(y)^{2^n} by listing

the initial coefficients in the (2^n)-th powers of A(x):

A^(2^0)=[1, 2, -6, 28, -410, -40548, -25678044, -46695237064,...];

A^(2^1)=[1, 4, -8, 32, -672, -83072, -51512576, -93492722688,...];

A^(2^2)=[1, 8, 0, 0, -1024, -172032, -103677952, -187396259840,...];

A^(2^3)=[1, 16, 64, 0, -2048, -360448, -210108416, -376451366912,...];

A^(2^4)=[1, 32, 384, 2048, 0, -786432, -432013312, -759672340480,...];

A^(2^5)=[1, 64, 1792, 28672, 278528, 0, -910163968, -1547597512704,...];

A^(2^6)=[1, 128, 7680, 286720, 7438336, 138412032, 0, -3195724103680,.];

A^(2^7)=[1, 256, 31744, 2539520, 147259392, 6585057280, 231894679552, 0,.];

the main diagonal is all zeros except for the initial [1,4].

(PARI) {a(n)=local(A=[1]); if(n==1, 2, for(i=0, n, A=Vec(Ser(concat(Vec(Ser(A)^(2^#A)), if(#A==1, 1, 0)))^(1/2^#A))); A[n+1]*4^n)}

Sequence in context: A119966 A002047 A126340 this_sequence A027109 A107375 A020126

Adjacent sequences: A136636 A136637 A136638 this_sequence A136640 A136641 A136642

sign

Paul D. Hanna (pauldhanna(AT)juno.com), Jan 19 2008