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a(1) = 1; a(m) = minimum numerator possible with a continued fraction [b(1);b(2),b(3),...b(m-1)], where (b(1),b(2),b(3),...b(m-1)) is a permutation of (a(1),a(2),a(3),...a(m-1)).
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%I #6 Feb 05 2014 20:18:06

%S 1,1,2,4,16,192,29984,776474136,582837534997525192,

%T 334033256143852482501323872038100184,

%U 111432026121971983026248175426087984579225579894344486903683496908882296

%N a(1) = 1; a(m) = minimum numerator possible with a continued fraction [b(1);b(2),b(3),...b(m-1)], where (b(1),b(2),b(3),...b(m-1)) is a permutation of (a(1),a(2),a(3),...a(m-1)).

%e a(6)=192 because the minimum numerator among permutations of

%e (1,1,2,4,16) happens when the continued fraction is [1:4,2,16,1]=192/157 or

%e [1:16,2,4,1]=192/181.

%t a[1] = 1; a[n_] := a[n] = Union[ Numerator /@ FromContinuedFraction /@ Permutations[ Table[ a[i], {i, n - 1}]]] [[1]]; Table[ a[n], {n, 11}]

%Y Cf. A105787.

%K nonn

%O 1,3

%A _Leroy Quet_ and _Robert G. Wilson v_, Apr 19 2005