

A303952


a(n) is the number of monic polynomials P(z) of degree n over the complex numbers such that P(z) divides P(z^2).


0



1, 2, 5, 17, 69, 302, 1367, 6302, 29401, 138356, 655425, 3121439, 14930541, 71675840, 345148893, 1666432817, 8064278289, 39103576700, 189949958333, 924163714217, 4502711570989, 21966152501240, 107284324830303
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OFFSET

0,2


COMMENTS

Note that if z_0 is a root of P(z), so is (z_0)^2, so z_0 must equal to 0 or 1. As a result, all such polynomials must have the form P(z) = z^d_0 * Product_{j=1..k} (z  exp(2*Pi*i*q_j))^d_j, where Sum_{j=0..k} d_j = n and {q_1, q_2, ..., q_k} is a set of k rational numbers on [0,1) such that if x belongs to it, the fractional part of 2x also belongs to it. That explains the formula a(n) = Sum_{k=1..n} binomial(n,k)*A014300(k) + 1 in the formula section, the "+1" represents the case d_0 = n and k = 0 corresponding to the polynomial P(z) = z^n.


LINKS

Table of n, a(n) for n=0..22.


FORMULA

a(n) = Sum_{k=1..n} binomial(n,k)*A014300(k) + 1. The "+1" represents the polynomial P(z) = z^n.
a(n) = A128730(n+1) + 1.
G.f.: 2x/(16x+5x^2+(1+x)sqrt(16x+5x^2)) + 1/(1x).
Dfinite with recurrence: +2*n*a(n) +(13*n+4)*a(n1) +2*(7*n+3)*a(n2) +8*(n7)*a(n3) +2*(8*n+33)*a(n4) +5*(n4)*a(n5)=0.  R. J. Mathar, Jan 27 2020


EXAMPLE

For n = 0, P(z) = 1.
For n = 1, P(z) = z or z  1.
For n = 2, P(z) = z^2, z^2  1, z^2  2z + 1, z^2 + z or z^2 + z + 1.


PROG

(PARI) x='x+O('x^50); Vec(2*x/(16*x+5*x^2+(1+x)*sqrt(16*x+5*x^2))+1/(1x))


CROSSREFS

Cf. A014300, A128728, A128730.
Sequence in context: A027361 A101971 A211387 * A162037 A319467 A326412
Adjacent sequences: A303949 A303950 A303951 * A303953 A303954 A303955


KEYWORD

nonn,easy


AUTHOR

Jianing Song, May 03 2018


STATUS

approved



