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G.f. A(x) satisfies A(x/A(x)) = 1/(1-x)^6.
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%I #12 Nov 01 2019 11:48:54

%S 1,6,57,866,18444,492924,15424611,542166480,20861518935,864061112296,

%T 38081996557383,1771322835258594,86425203984341130,

%U 4402953230795279532,233372023965531945057,12832558973488295874402,730347857708249147767893

%N G.f. A(x) satisfies A(x/A(x)) = 1/(1-x)^6.

%H Robert Israel, <a href="/A145170/b145170.txt">Table of n, a(n) for n = 0..317</a>

%F Self-convolution 6th power of A145167.

%F Self-convolution cube of A145168.

%F Self-convolution square of A145169.

%p A[0]:= x -> 1+c*x:

%p for n from 1 to 20 do

%p cc:= coeff(series(A[n-1](x/A[n-1](x))-1/(1-x)^6, x, n+1),x,n);

%p A[n]:= unapply(eval(A[n-1](x),c=solve(cc,c))+c*x^(n+1),x);

%p od:

%p seq(coeff(A[20](x),x,j),j=0..20); # _Robert Israel_, Aug 19 2018

%t nmax = 16; sol = {a[0] -> 1};

%t Do[A[x_] = Sum[a[k] x^k, {k, 0, n}] /. sol; eq = CoefficientList[A[x/A[x]] - 1/(1 - x)^6 + O[x]^(n + 1), x] == 0 /. sol; sol = sol ~Join~ Solve[eq] [[1]], {n, 1, nmax}]; sol /. Rule -> Set;

%t a /@ Range[0, nmax] (* _Jean-François Alcover_, Nov 01 2019 *)

%o (PARI) {a(n)=local(A=1+x+x*O(x^n),B);for(n=0,n,B=serreverse(x/A);A=1/(1-B)^6);polcoeff(A,n)}

%Y Cf. A145167, A145168, A145169.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Oct 03 2008