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%I #2 Mar 30 2012 18:37:16
%S 1,1,1,-4,-19,134,1074,-10158,-110067,1302086,17451662,-248857456,
%T -3948994550,66104803660,1210719480268,-23304935437410,
%U -484165206834051,10541640152174406,245149591293286518,-5958110923842801192
%N G.f. A(x) satisfies the condition that both G(x) = A(x/G(x)) and H(x) = A(x*H(x)) have zeros for every other coefficient after initial terms; g.f. of dual sequence A157306 satisfies the same condition.
%F For n>=2, [x^(2n-1)] A(x)^(2n) = 0.
%F For n>=1, [x^(2n)] 1/A(x)^(2n-1) = 0.
%F G.f. satisfies: A(x) = G(x*A(x)) where G(x) = A(x/G(x)) = x/Series_Reversion(x*A(x)) = g.f. of A157302.
%F G.f. satisfies: A(x) = H(x/A(x)) where H(x) = A(x*H(x)) = Series_Reversion(x/A(x))/x = g.f. of A157304.
%e G.f.: A(x) = 1 + x + x^2 - 4*x^3 - 19*x^4 + 134*x^5 + 1074*x^6 --++...
%e ...
%e Let G(x) = A(x/G(x)) so that A(x) = G(x*A(x)) then
%e G(x) = 1 + x - 5*x^3 + 183*x^5 - 14352*x^7 + 1857199*x^9 -+...
%e has alternating zeros in the coefficients (cf. A157302):
%e [1,1,0,-5,0,183,0,-14352,0,1857199,0,-355082433,0,94134281460,0,...]
%e ...
%e Let H(x) = A(x*H(x)) so that A(x) = H(x/A(x)) then
%e H(x) = 1 + x + 2*x^2 - 26*x^4 + 1378*x^6 - 141202*x^8 +-...
%e has alternating zeros in the coefficients (cf. A157304):
%e [1,1,2,0,-26,0,1378,0,-141202,0,22716418,0,-5218302090,0,...].
%e ...
%e ZERO COEFFICIENTS IN POWERS OF G.F. A(x).
%e Even powers A(x)^(2n) yield zeros at odd positions 2n-1 for n>=2:
%e A^4: [1,4,10, 0, -105,228,5442,-24048,-535293,3588828,...];
%e A^6: [1,6,21,26,-144, 0, 8415,-17802,-849771,3596382,...];
%e A^8: [1,8,36,80,-110,-384,10608, 0, -1143105,2366472,...];
%e A^10:[1,10,55,170,65,-718,11580,26520,-1381080, 0, ...];
%e ...
%e Odd negative powers 1/A(x)^(2n-1) yield zeros at even positions 2n for n>=1:
%e A^-1: [1,-1, 0, 5,10,-168,-762,12297,87222,-1524622,...];
%e A^-3: [1,-3,3,14, 0, -549,-1173,41184,180558,-5088382,...];
%e A^-5: [1,-5,10,15,-45,-891, 0, 71760,152670,-9042640,...];
%e A^-7: [1,-7,21,0,-105,-1092,2604,99042, 0, -13000393,...]; ...
%o (PARI) {a(n)=local(A=[1, 1]); for(i=1, n, if(#A%2==1, A=concat(A, t); A[ #A]=-subst(Vec(serreverse(x/Ser(A)))[ #A], t, 0)); if(#A%2==0, A=concat(A, t); A[ #A]=-subst(Vec(x^2/serreverse(x*Ser(A)))[ #A], t, 0))); A[n+1]}
%Y Cf. A157302, A157304, A157305, A157306 (dual), A157307.
%K sign
%O 0,4
%A _Paul D. Hanna_, Feb 28 2009
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