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%I #26 May 15 2016 16:50:31
%S 4,3,1,16,9,7,256,81,175,65536,6561,58975,4294967296,43046721,
%T 4251920575,18446744073709551616,1853020188851841,
%U 18444891053520699775,340282366920938463463374607431768211456,3433683820292512484657849089281
%N a(3*n+1) = 4^(2^n), a(3*n+2) = 3^(2^n), a(3*n+3) = 4^(2^n) - 3^(2^n).
%C Previous name was: Consider pairs of fractions (x,y) starting (4,3) and updated via z:=1/(1/x+1/y), x->x-z, y->y-z. The sequence shows the triples (numerator(x), numerator(y), numerator(x)-numerator(y)) after each update.
%F a(3*n+1) = 4^(2^n), a(3*n+2) = 3^(2^n), a(3*n+3) = 4^(2^n) - 3^(2^n). - _Philippe Deléham_ , Oct 29 2013
%e (x=4,y=3) is shown as the first triple (4,3,1) in the sequence. This generates z=12/7 which generates the new pair (x,y) = (16/7,9/7) shown as (16,9,7). - _R. J. Mathar_, Feb 09 2011
%p x := 4 ; y := 3 ;
%p for loo from 1 to 7 do printf("%d, %d, %d, ", numer(x), numer(y), numer(x)-numer(y)) ; z := 1/(1/x+1/y) ; x := x-z ; y := y-z ; end do: # _R. J. Mathar_, Feb 09 2011
%K nonn,easy,less
%O 1,1
%A _Jamel Ghanouchi_, Jan 27 2011
%E Corrected by _Philippe Deléham_, Oct 29 2013
%E New name using _Philippe Deléham_'s formula, _Joerg Arndt_, Nov 14 2014