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%I #24 Jun 12 2015 08:38:56
%S 2,3,5,8,13,21,34,56,92,152,251,414,684,1130,1867,3084,5095,8418,
%T 13908,22979,37966,62727,103638,171232,282911,467429,772292,1275990,
%U 2108206,3483204,5754993,9508472,15710018
%N a(1) = 2; for n>1, a(n) = largest integer such that the sequence b(n) = a(n)^(1/n) is decreasing.
%C (n -> oo) lim (log a(n+1)/log a(n))^n = e. - _Thomas Ordowski_, Oct 06 2014
%F a(1) = 2, a(2) = 3; a(n+1) = floor( a(n)^(1+1/n) ) for n > 1.
%F a(n) = 1 + floor(q^n), where q = lim a(n+1)/a(n) = 1.652213...
%F a(2) = 3; a(n) = floor( exp( (n/(n-1)) * log a(n-1) ) ) for n > 2.
%p A207976 := proc(n)
%p if n = 1 then
%p 2;
%p elif n = 2 then
%p 3;
%p else
%p floor( exp(n/(n-1)*log(procname(n-1))) );
%p end if;
%p end proc: # _R. J. Mathar_, Mar 22 2012
%o (PARI) {a(n) = if( n<3, max(1, n+1), floor( exp( log(a(n-1)) * n/(n-1) )))}; /* _Michael Somos_, Oct 06 2014 */
%K nonn
%O 1,1
%A Thomas Ordowski, Feb 20 2012
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