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%I #2 Mar 30 2012 18:37:16
%S 1,1,2,0,-26,0,1378,0,-141202,0,22716418,0,-5218302090,0,
%T 1619288968386,0,-653379470919714,0,333014944014777730,0,
%U -209463165121436380282,0,159492000935562428176162,0,-144654795258284936534929586,0
%N G.f. A(x) satisfies the condition that both A(x) and F(x) = A(x/F(x)^2) have zeros for every other coefficient after initial terms; g.f. of dual sequence A157307 satisfies the same condition.
%C After initial 2 terms, reversing signs yields the complementary sequence A157305, which has very similar properties.
%F For n>=1, [x^(2n)] 1/A(x)^(4n-1) = 0.
%F G.f. satisfies: A(x) = F(x*A(x)^2) where F(x) = A(x/F(x)^2) = sqrt(x/Series_Reversion(x*A(x)^2)) = g.f. of A157302.
%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 A157303.
%e G.f.: A(x) = 1 + x + 2*x^2 - 26*x^4 + 1378*x^6 - 141202*x^8 +-...
%e ...
%e Let F(x) = A(x/F(x)^2) so that A(x) = F(x*A(x)^2) then
%e F(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 COEFFICIENTS IN ODD NEGATIVE POWERS OF G.F. A(x).
%e A^1 : [(1), 1,2,0,-26,0,1378,0,-141202,0,22716418,...];
%e A^-1: [1,(-1),-1,3,25,-57,-1397,2967,143057,...];
%e A^-3: [1,-3,(0),14,57,-333,-3880,18036,415665,...];
%e A^-5: [1,-5,5,(25),50,-766,-5370,44370,637275,...];
%e A^-7: [1,-7,14,28,(0),-1246,-5334,79148,770469,...];
%e A^-9: [1,-9,27,15,-81,(-1647),-3519,117981,784998,...];
%e A^-11:[1,-11,44,-22,-165,-1859,(0),155584,662046,...];
%e A^-13:[1,-13,65,-91,-208,-1820,4836,(186576),396942,...];
%e A^-15:[1,-15,90,-200,-150,-1548,10370,206280,(0),...];
%e A^-17:[1,-17,119,-357,85,-1173,15895,211395,-504577,(-31572383),...];
%e ...
%e When scaled, the coefficients shown above in parenthesis
%e forms the coefficients of the function F(x) = A(x/F(x)^2):
%e F: [1,-1/(-1),0,25/(-5),0,-1647/(-9),0,186576/(-13),0,-31572383/(-17),...].
%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/serreverse(x*Ser(A)))[ #A], t, 0))); Vec(serreverse(x/Ser(A))/x)[n+1]}
%Y Cf. A157302, A157303, A157305 (complement), A157306, A157307 (dual).
%K sign
%O 0,3
%A _Paul D. Hanna_, Feb 28 2009
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