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%I #11 Jul 14 2013 14:15:16
%S 3,7,7,28,45,189,799,2091,2091,8856,14329,60697,257115,673135,673135,
%T 2851444,4613733,19544085,82790071,216747219,216747219,918155952,
%U 1485607537,6293134513,26658145587,69791931223,69791931223,295643364940
%N Least positive integer k such that the fractional part of k*sqrt(5) has its n initial partial quotients all equal to 1.
%F a(n) = (Fib(12[ n/6 ] + S_(n mod 6))+1)/2 where S = (2, 5, 7, 7, 10, 11).
%F Empirical g.f.: -x*(x^12 -x^10 +3*x^9 +4*x^7 -356*x^6 +144*x^5 +17*x^4 +21*x^3 +4*x +3) / ((x -1)*(x^2 -3*x +1)*(x^2 +3*x +1)*(x^4 -3*x^3 +8*x^2 -3*x +1)*(x^4 +3*x^3 +8*x^2 +3*x +1)). - _Colin Barker_, Jul 14 2013
%K nonn
%O 1,1
%A _Clark Kimberling_
%E More terms and formula from _David W. Wilson_ May 15 1997
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