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%I #6 Jul 31 2015 01:23:24
%S 1,9,156,1696,3974,21558,82512,631294,5619414,93118405,739310894
%N a(1)=1; a(n) is the smallest integer > 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 3^n.
%e The continued fraction for S(5) = 1 + 1/9 + 1/156 + 1/1696 + 1/3974 is [1, 8, 2, 4, 2, 1, 2, 1, 5, 4, 1, 3, 2, 243, 1, 1, 3] where the largest element is 243=3^5 and 3974 is the smallest integer >1696 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[3, Max[ContinuedFraction[s + 1/k]]] != n, k++ ]; k]; Do[ Print[ a[n]], {n, 11}] (* _Robert G. Wilson v_, Aug 08 2005 *)
%o (PARI) s=1; t=1; for(n=2, 50, s=s+1/t; while(abs(3^n-vecmax(contfrac(s+1/t)))>0,t++); print1(t,","))
%K hard,nonn
%O 1,2
%A _Ryan Propper_, Aug 06 2005
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