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%I #26 Mar 24 2024 09:57:02
%S 6,12,26,104,159,741,2267,10841,16890,80904,248750,1191812,1857147,
%T 8898105,27359639,131087885,204268686,978710052,3009310946,
%U 14418474944,22467697719,107649207021,330996843827,1585901155361,2471242479810,11840434061664,36406643509430
%N Solutions a(n) of (a(n)-5)*(a(n)-6) = 21*b(n)*(b(n)-1).
%C The associated b(n) are in A181443.
%C Consider an urn with r red and b blue balls. Draw 7 balls without replacement. The probability of picking 7 red balls is (r/(r+b)) * ((r-1)/(r+b-1)) * ((r-2)/(r+b-2)) * ... * ((r-6)/(r+b-6)). The probability of picking 5 red and 2 blue balls is binomial(7,2) * r*(r-1)*(r-2)*...*(r-4)*b*(b-1)/ ((r+b)*(r+b-1)*...*(r+b-6)). For equal probability we need (r-5)*(r-6) = 21*b*(b-1). The current sequence gives the values of r, the number of red balls which allows such scenario of equal probability.
%C Diagonalizing the quadratic equation with r=(A(n)+11)/2 and b=(B(n)+1)/2, this is equivalent to the Pell equation A(n)^2 - 21*B(n)^2 = -20 with 4 fundamental solutions (1;1), (13;3), (41;9), (197;43) and the solution (55;12) for the unit form.
%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,110,-110,0,0,-1,1).
%F G.f.: (-6 - 6*x - 14*x^2 - 78*x^3 + 605*x^4 + 78*x^5 + 14*x^6 + 6*x^7 - 5*x^8) / ((x-1)*(x^8-110*x^4+1)). - _R. J. Mathar_, Feb 05 2011
%F Explicit formulas: r=sqrt(21), s=55+12*r, t=55-12*r.
%F a(4*n) = (22 + (1+r)*s^n + (1-r)*t^n))/4.
%F a(4*n+1) = (22 + (13+3*r)*s^n + (13-3*r)*t^n)/4.
%F a(4*n+2) = (22 + (41+9*r)*s^n + (41-9*r)*t^n)/4.
%F a(4*n+3) = (22 + (197+43*r)*s^n + (197-43*r)*t^n)/4.
%F a(n) = 111*a(n-4) - 111*a(n-8) + a(n-12).
%e For n=2: a(2)=26; b(2)=5; binomial(26,7)=657800.
%e binomial(26,5)*binomial(5,2)=657800.
%e The 2-tuples begin: (6, 1); (12, 2); (26, 5); (104, 22).
%p n=0: for s from 1 to 100 do r:=(sqrt(84*s^2-84*s+1)+11)/2: if (trunc(r)=r) then a(n):=r: b(n):=s: n:=n+1: end if: end do:
%Y Cf. A181443.
%K nonn,easy
%O 0,1
%A _Paul Weisenhorn_, Jan 20 2011
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