%I #6 Jul 31 2015 01:23:11

%S 1,4,30,85,91,401,1160,2338,13392,31765,39040,442431,667330,12260875,

%T 12882668,33163533,35682489

%N a(1) = 1; a(n) is the smallest integer greater than a(n-1) such that the largest element in the simple continued fraction for S(n)=1/a(1)+1/a(2)+...+1/a(n) equals 2^n.

%e The continued fraction for S(5) = 1 + 1/4 + 1/30 + 1/85 + 1/91 is [1, 3, 3, 1, 2, 1, 11, 32, 5] where the largest element is 32 = 2^5 and 91 is the smallest integer > 85 with this property.

%t a[1] = 1; a[n_] := a[n] = Block[{k = a[n - 1] + 1, s = Plus @@ (1/Table[a[i], {i, n - 1}])}, While[ Log[2, Max[ContinuedFraction[s + 1/k]]] != n, k++ ]; k]; Do[ Print[ a[n]], {n, 17}] (* _Robert G. Wilson v_, Aug 08 2005 *)

%o (PARI) s=1; t=1; for(n=2, 50, s=s+1/t; while(abs(2^n-vecmax(contfrac(s+1/t)))>0,t++); print1(t,","))

%K nonn

%O 1,2

%A _Ryan Propper_, Aug 06 2005