A157304 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.

1, 1, 2, 0, -26, 0, 1378, 0, -141202, 0, 22716418, 0, -5218302090, 0, 1619288968386, 0, -653379470919714, 0, 333014944014777730, 0, -209463165121436380282, 0, 159492000935562428176162, 0, -144654795258284936534929586, 0

Offset

0,3

Comments

After initial 2 terms, reversing signs yields the complementary sequence A157305, which has very similar properties.

Links

Table of n, a(n) for n=0..25.

Formula

For n>=1, [x^(2n)] 1/A(x)^(4n-1) = 0.

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.

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.

Example

G.f.: A(x) = 1 + x + 2*x^2 - 26*x^4 + 1378*x^6 - 141202*x^8 +-...
...
Let F(x) = A(x/F(x)^2) so that A(x) = F(x*A(x)^2) then
F(x) = 1 + x - 5*x^3 + 183*x^5 - 14352*x^7 + 1857199*x^9 -+...
has alternating zeros in the coefficients (cf. A157302):
[1,1,0,-5,0,183,0,-14352,0,1857199,0,-355082433,0,94134281460,0,...].
...
COEFFICIENTS IN ODD NEGATIVE POWERS OF G.F. A(x).
A^1 : [(1), 1,2,0,-26,0,1378,0,-141202,0,22716418,...];
A^-1: [1,(-1),-1,3,25,-57,-1397,2967,143057,...];
A^-3: [1,-3,(0),14,57,-333,-3880,18036,415665,...];
A^-5: [1,-5,5,(25),50,-766,-5370,44370,637275,...];
A^-7: [1,-7,14,28,(0),-1246,-5334,79148,770469,...];
A^-9: [1,-9,27,15,-81,(-1647),-3519,117981,784998,...];
A^-11:[1,-11,44,-22,-165,-1859,(0),155584,662046,...];
A^-13:[1,-13,65,-91,-208,-1820,4836,(186576),396942,...];
A^-15:[1,-15,90,-200,-150,-1548,10370,206280,(0),...];
A^-17:[1,-17,119,-357,85,-1173,15895,211395,-504577,(-31572383),...];
...
When scaled, the coefficients shown above in parenthesis
forms the coefficients of the function F(x) = A(x/F(x)^2):
F: [1,-1/(-1),0,25/(-5),0,-1647/(-9),0,186576/(-13),0,-31572383/(-17),...].

Prog

(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]}

Crossrefs

Cf. A157302, A157303, A157305 (complement), A157306, A157307 (dual).

Sequence in context: A097563 A360643 A158045 * A157305 A306416 A327601

Adjacent sequences: A157301 A157302 A157303 * A157305 A157306 A157307

Keyword

sign

Author

Paul D. Hanna, Feb 28 2009

Status

approved