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

%I A180483 #39 Oct 14 2024 05:53:23
%S A180483 3,6,11,38,87,354,839,3482,8283,34446,81971,340958,811407,3375114,
%T A180483 8032079,33410162,79509363,330726486,787061531,3273854678,7791105927,
%U A180483 32407820274,77123997719,320804348042,763448871243,3175635660126,7557364714691,31435552253198
%N A180483 Expansion of (3+3*x-25*x^2-3*x^3+2*x^4)/((1-x)*(1-10*x^2+x^4)).
%C A180483 Previous name was: Solutions a(n) to (a(n)-2)*(a(n)-3) = 6*b(n)*(b(n)-1).
%C A180483 The associated b(n) are in A181442.
%C A180483 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.
%C A180483 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.
%H A180483 G. C. Greubel, <a href="/A180483/b180483.txt">Table of n, a(n) for n = 0..1000</a>
%H A180483 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,10,-10,-1,1)
%F A180483 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
%F A180483 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.
%F A180483 a(n) = 11*a(n-2) - 11*a(n-4) + a(n-6).
%F A180483 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
%F A180483 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
%e A180483 For n=3: a(3) = 38; b(3) = 15; binomial(38,4) = 73815 and  binomial(38, 2)*binomial(15, 2) = 73815.
%e A180483 The 2-tuples begin (3, 1); (6, 2); (11, 4); (38, 15).
%p A180483 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:
%t A180483 LinearRecurrence[{1,10,-10,-1,1},{3,6,11,38,87},30] (* _Harvey P. Dale_, Apr 28 2018 *)
%o A180483 (Magma) R<x>:=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
%o A180483 (SageMath)
%o A180483 def b(n): return (1/2)*(1+(-1)^n)*chebyshev_U(n//2, 5)
%o A180483 def A180483(n): return (1/2)*(5 +b(n) +7*b(n-1) +7*b(n-2) +b(n-3))
%o A180483 [A180483(n) for n in (0..40)] # _G. C. Greubel_, Apr 28 2022
%Y A180483 Cf. A004189, A181442.
%K A180483 nonn,easy
%O A180483 0,1
%A A180483 _Paul Weisenhorn_, Jan 20 2011
%E A180483 New name using the g.f. by _R. J. Mathar_ from _Joerg Arndt_, Apr 27 2022