A152467 a(n) = floor(n/6).

0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14

Offset: 0

Author

Vladimir Joseph Stephan Orlovsky, Dec 05 2008

Keywords

• nonn
• easy

Comments

Apart from initial terms, same as A097992. - Philippe Deléham, Dec 06 2008

From Michel Lagneau, Apr 11 2025: (Start)

Consider the polynomial P(n,z)=(z+1)^n = z*Q(n,z)+1. The sequence 2*a(n) lists the numbers of zeros of Q(n,z) strictly inside the unit circle.

Proof:

The roots of the equation (z+1)^n=1 are:

z_k = exp(2*Pi*i*k/n)-1 for k=0,1,...,n-1. These are n complex roots regularly distributed around the unit circle, translated by -1.

We see how many of these values have a modulus >1.

e^2*Pi*i*k/n are the n-th roots of unity, all of which lie on the unit circle (with modulus 1).

By subtracting 1: z_k = e^2*Pi*i*k/n - 1, we move the unit circle to the center -1.

The z_k are distributed around the circle centered at -1, and we find how many are at a distance >1 from the center (0,0). Let z_k=e^2*Pi*i*k/n, then let Z_k = z_k - 1.

So, abs(Z_k) = abs(z_k - 1) =abs(e^2*i*Pi*k/n) = 2*abs(sin(Pi*k)/n).

So we are looking for the integers k in {1,2,...,n-1} such that:

2*abs(sin(Pi*k/n))>1 =>abs(sin(Pi*k/n))<1/2.

We know the values of sin: abs(sin(teta))<1/2 => teta in {0,Pi/6} union {5*Pi/6,Pi}. So Pi*k/n < Pi/6 => k 5*Pi/6 => k >5*n/6. Conclusion for each n, the number of roots strictly in the unit circle is 2*E[n/6]. (End)

Links

• Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,1,-1).

Crossrefs

Cf. A097992.

Programs

Maple

A152467:=n->floor(n/6); seq(A152467(n), n=0..100); # Wesley Ivan Hurt, Dec 10 2013

Mathematica

Table[Floor[n/6], {n, 0, 100}] (* Wesley Ivan Hurt, Dec 10 2013 *)

PARI

a(n)=n\6 \\ Charles R Greathouse IV, Jun 04 2011

Sage

[floor(n/6) for n in range(0,90)] # Zerinvary Lajos, Dec 02 2009

Formula

From R. J. Mathar and Philippe Deléham, Dec 06 2008: (Start)

a(n) = floor(n/6) = a(n-6) + 1.

G.f.: x^6/((1-x)^2*(1+x)*(1+x+x^2)*(x^2-x+1)). (End)

a(n) = (6*n - 15 + 3*(-1)^n + 12*sin( (2*n+1)*Pi/6 ) + 4*sqrt(3)*sin( (2n+1)*Pi/3) )/36.

a(n) = floor( (3*n-2)/2 - (4*n-3)/3 ). - Robert G. Wilson v, Jun 04 2011

E.g.f.: (6*cos(sqrt(3)*x/2)*cosh(x/2) + 3*(x - 2)*cosh(x) + 2*sqrt(3)*sin(sqrt(3)*x/2)*(2*cosh(x/2) + sinh(x/2)) + 3*(x - 3)*sinh(x))/18. - Stefano Spezia, Nov 13 2022