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A180509
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Solutions a(n) of (a(n)-5)*(a(n)-6) = 21*b(n)*(b(n)-1).
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1
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6, 12, 26, 104, 159, 741, 2267, 10841, 16890, 80904, 248750, 1191812, 1857147, 8898105, 27359639, 131087885, 204268686, 978710052, 3009310946, 14418474944, 22467697719, 107649207021, 330996843827, 1585901155361, 2471242479810, 11840434061664, 36406643509430
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OFFSET
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0,1
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COMMENTS
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The associated b(n) are in A181443.
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.
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.
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LINKS
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FORMULA
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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
Explicit formulas: r=sqrt(21), s=55+12*r, t=55-12*r.
a(4*n) = (22 + (1+r)*s^n + (1-r)*t^n))/4.
a(4*n+1) = (22 + (13+3*r)*s^n + (13-3*r)*t^n)/4.
a(4*n+2) = (22 + (41+9*r)*s^n + (41-9*r)*t^n)/4.
a(4*n+3) = (22 + (197+43*r)*s^n + (197-43*r)*t^n)/4.
a(n) = 111*a(n-4) - 111*a(n-8) + a(n-12).
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EXAMPLE
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For n=2: a(2)=26; b(2)=5; binomial(26,7)=657800.
binomial(26,5)*binomial(5,2)=657800.
The 2-tuples begin: (6, 1); (12, 2); (26, 5); (104, 22).
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MAPLE
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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:
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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