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 A179934 a(n) red balls and b(n) blue balls in an urn; draw 2 balls without replacement; Probability(2 red balls)=6*Probability(2 blue balls); b(n)=A181442(n); 0
 4, 9, 36, 85, 352, 837, 3480, 8281, 34444, 81969, 340956, 811405, 3375112, 8032077, 33410160, 79509361, 330726484, 787061529, 3273854676, 7791105925, 32407820272, 77123997717, 320804348040, 763448871241, 3175635660124 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS the last digit has the period (4,9,6,5,2,7,0,1); LINKS FORMULA a(n)=(1+sqrt(1+24*b(n)+(b(n)-1))/2; this is equivalent to the Pell equation A(n)^2-6*B(n)^2=-5 with the two fundamental solutions (7;3) and (17;7) and the solution (5;2) for the unit form; a(n)=(A(n)+1)/2; b(n)=(B(n)+1)/2; a(n+4)=10*a(n+2)-a(n)-4; a(n+6)=11*(a(n+4)-a(n+2))+a(n); a(2*n+1)=(2+(7+3*r)*(5+2*r)^n+(7-3*r)*(5-2*r)^n)/4; a(2*n+2)=(2+(17+7*r)*(5+2*r)^n+(17-7*r)*(5-2*r)^n)/4; a(n)= +a(n-1) +10*a(n-2) -10*a(n-3) -a(n-4) +a(n-5). G.f.: -x*(4+5*x-13*x^2-x^3+x^4) / ( (x-1)*(x^4-10*x^2+1) ). [From R. J. Mathar, Aug 03 2010] EXAMPLE For n=6: a(6)=837; b(6)=342; binomial(837,2)=349866; 6*binomial(342,2)=349866; MAPLE for n from 0 to 20 do a(2*n+1):=round((2+(7+3*r)*(5+2*r)^n)/4): a(2*n+2):=round((2+(17+7*r)*(5+2*r)^n)/4): end do: r:=sqrt(6); MATHEMATICA LinearRecurrence[{1, 10, -10, -1, 1}, {4, 9, 36, 85, 352}, 30] (* Harvey P. Dale, Dec 23 2012 *) CROSSREFS Sequence in context: A029806 A133125 A126161 * A018224 A149137 A149138 Adjacent sequences:  A179931 A179932 A179933 * A179935 A179936 A179937 KEYWORD nonn,uned AUTHOR Paul Weisenhorn, Aug 02 2010 STATUS approved

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