A180483 Expansion of (3+3*x-25*x^2-3*x^3+2*x^4)/((1-x)*(1-10*x^2+x^4)).

3, 6, 11, 38, 87, 354, 839, 3482, 8283, 34446, 81971, 340958, 811407, 3375114, 8032079, 33410162, 79509363, 330726486, 787061531, 3273854678, 7791105927, 32407820274, 77123997719, 320804348042, 763448871243, 3175635660126, 7557364714691, 31435552253198

Offset: 0

Author

Paul Weisenhorn, Jan 20 2011

Keywords

• nonn
• easy

Comments

Previous name was: Solutions a(n) to (a(n)-2)*(a(n)-3) = 6*b(n)*(b(n)-1).

The associated b(n) are in A181442.

Consider an urn with r red and b blue balls. Draw 4 balls without replacement. The probability of picking 4 red balls is r/(r+b) *(r-1)/(r+b-1) *(r-2)/(r+b-2) * (r-3)/(r+b-3). The probability of picking 2 red and 2 blue balls is binomial(2,2) * r*(r-1)*b*(b-1)/ ((r+b)*(r+b-1)..*(r+b-3)). For equal probability we need (r-2)*(r-3)=6*b*(b-1). The current sequence shows the r, the number of red balls which allow such scenario of equal probability.

The quadratic equation is diagonalized with a(n) = (A(n) + 5)/2 and b(n) = (B(n) + 1)/2, equivalent to the Pell equation A(n)^2 - 6*B(n)^2 = -5 with the 2 fundamental solutions (1; 1); (7; 3) and the solution (5; 2) for the unit form.

Examples

			For n=3: a(3) = 38; b(3) = 15; binomial(38,4) = 73815 and  binomial(38, 2)*binomial(15, 2) = 73815.
The 2-tuples begin (3, 1); (6, 2); (11, 4); (38, 15).
		

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (1,10,-10,-1,1)

Crossrefs

Cf. A004189, A181442.

Programs

Magma

R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( (3+3*x-25*x^2-3*x^3+2*x^4)/((1-x)*(1-10*x^2+x^4)) )); // G. C. Greubel, Apr 28 2022

Maple

n:=0: for s from 1 to 100 do r:=(sqrt(24*s^2-24*s+1)+5)/2: if (r=trunc(r)) then a(n):=r: b(n):=s: n:=n+1: end if: end do:

Mathematica

LinearRecurrence[{1,10,-10,-1,1},{3,6,11,38,87},30] (* Harvey P. Dale, Apr 28 2018 *)

SageMath

def b(n): return (1/2)*(1+(-1)^n)*chebyshev_U(n//2, 5)
def A180483(n): return (1/2)*(5 +b(n) +7*b(n-1) +7*b(n-2) +b(n-3))
[A180483(n) for n in (0..40)] # G. C. Greubel, Apr 28 2022

Formula

G.f.: ( 3+3*x-25*x^2-3*x^3+2*x^4 )/( (1-x)*(1-10*x^2+x^4) ). - R. J. Mathar, Feb 05 2011

Let r=sqrt(6), s=5+2*r, and t=5-2*r, then a(2*n) = (10+(1+r)*s^n+(1-r)*t^n)/4 and a(2*n+1) = (10+(7+3*r)*s^n+(7-3*r)*t^n)/4.

a(n) = 11*a(n-2) - 11*a(n-4) + a(n-6).

a(n) = +a(n-1) +10*a(n-2) -10*a(n-3) -a(n-4) +a(n-5). - R. J. Mathar, Feb 05 2011

a(n) = (1/2)*(5 +b(n) +7*b(n-1) +7*b(n-2) +b(n-3)), where b(n) = (1/2)*(1+(-1)^n)*ChebyshevU(n/2, 5). - G. C. Greubel, Apr 28 2022

Extensions

New name using the g.f. by R. J. Mathar from Joerg Arndt, Apr 27 2022