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%I #19 Sep 02 2023 20:07:19
%S 4,8,24,43,179,783,1504,6668,29604,56983,253079,1124043,2163724,
%T 9610208,42683904,82164403,364934699,1620864183,3120083464,
%U 13857908228,61550154924,118481007103,526235577839,2337285022803,4499158186324,19983094049528
%N Place a(n) red and b(n) blue balls in an urn; draw 5 balls without replacement; Probability(5 red balls) = Probability(3 red and 2 blue balls).
%C This is equivalent to the Pell equation A(n)^2 - 10*B(n)^2 = -9 with a(n) = (A(n)+7)/2, b(n) = (B(n)+1)/2, and the 3 fundamental solutions (1,1), (9,3), (41,13), and the solution (19,6) for the unit form.
%H G. C. Greubel, <a href="/A180002/b180002.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,38,-38,0,-1,1).
%F G.f.: x*(4 +4*x +16*x^2 -133*x^3 -16*x^4 -4*x^5 +3*x^6)/((1-x)*(1 -38*x^3 +x^6)).
%F a(n+9) = 39*a(n+6) - 39*a(n+3) + a(n).
%F Let r = sqrt(10) then:
%F a(3*n+1) = (14 + (1+r)*(19+6*r)^n + (1-r)*(19-6*r)^n)/4.
%F a(3*n+2) = (14 + 3*(3+r)*(19+6*r)^n + 3*(3-r)*(19-6*r)^n)/4.
%F a(3*n+3) = (14 + (41+13*r)*(19+6*r)^n + (41-13*r)*(19-6*r)^n)/4.
%F a(n) = a(n-1) + 38*a(n-3) - 38*a(n-4) - a(n-6) + a(n-7). - _G. C. Greubel_, Mar 20 2019
%e For n=3: a(3)=24 and b(3)=7 since binomial(24,5) = binomial(24,3)*binomial(7,2) = 42504.
%t Rest[CoefficientList[Series[x*(4+4*x+16*x^2-133*x^3-16*x^4-4*x^5 +3*x^6 )/((1-x)*(1-38*x^3+x^6)), {x,0,30}], x]] (* _G. C. Greubel_, Mar 20 2019 *)
%o (PARI) my(x='x+O('x^30)); Vec(x*(4+4*x+16*x^2-133*x^3-16*x^4-4*x^5+3*x^6) /((1-x)*(1-38*x^3+x^6))) \\ _G. C. Greubel_, Mar 20 2019
%o (Magma) R<x>:=PowerSeriesRing(Integers(), 30); Coefficients(R!( x*(4+4*x+ 16*x^2-133*x^3-16*x^4-4*x^5+3*x^6)/((1-x)*(1-38*x^3+x^6)) )); // _G. C. Greubel_, Mar 20 2019
%o (Sage) a=(x*(4+4*x+16*x^2-133*x^3-16*x^4-4*x^5+3*x^6)/((1-x)*(1-38*x^3 +x^6))).series(x, 30).coefficients(x, sparse=False); a[1:] # _G. C. Greubel_, Mar 20 2019
%Y Cf. A180003 (b(n)).
%K nonn
%O 1,1
%A _Paul Weisenhorn_, Aug 05 2010
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