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%I #11 Nov 05 2019 07:22:45
%S 1,1,0,0,1,-3,7,-10,-5,84,-248,90,2160,-7541,-5846,122824,-186259,
%T -2036532,8665409,36714511,-345711246,-517802065,14415153844,
%U -9423161197,-653074772917,1896978939457,32374651932128,-184814895196023,-1733326272860598,16839263882542991,96742403684106435
%N G.f. A(x) satisfies A(A(x)) = B(x) such that the coefficients of B(x) consist of all 1's and 2's, with A(0) = 0.
%H Kenny Lau, <a href="/A112105/b112105.txt">Table of n, a(n) for n = 1..553</a>
%e A(x) = x + x^2 + x^5 - 3*x^6 + 7*x^7 - 10*x^8 - 5*x^9 +...
%e where A(A(x)) = x + 2*x^2 + 2*x^3 + x^4 + 2*x^5 + x^6 +... is the g.f. of A112104.
%t kmax = 40;
%t A[x_] := Sum[a[k] x^k, {k, kmax}];
%t B[x_] := Sum[b[k] x^k, {k, kmax}];
%t sol = {a[1] -> 1, b[1] -> 1};
%t Do[sc = SeriesCoefficient[A[(A[x] /. sol) + O[x]^(k+1)] - B[x], {x, 0, k}] /. sol; r = Reduce[(b[k] == 1 || b[k] == 2) && sc == 0, {a[k], b[k]}, Integers]; sol = Join[r // ToRules, sol], {k, 2, kmax}];
%t a /@ Range[kmax] /. sol (* _Jean-François Alcover_, Nov 05 2019 *)
%o (PARI) {a(n,m=2)=local(F=x+x^2+x*O(x^n),G);if(n<1,0, for(k=3,n, G=F+x*O(x^k);for(i=1,m-1,G=subst(F,x,G)); F=F-((polcoeff(G,k)-1)\m)*x^k); return(polcoeff(F,n,x)))}
%Y Cf. A112104, A112106-A112127.
%K sign
%O 1,6
%A _Paul D. Hanna_, Aug 27 2005
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