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%I #7 Mar 11 2014 01:32:50
%S 1,1,2,2,1,1,3,2,1,2,3,2,3,1,2,1,3,3,2,1,3,1,2,1,4,2,3,1,4,3,2,1,3,2,
%T 4,1,3,4,2,1,2,3,4,1,2,4,3,2,4,1,3,2,4,3,1,2,3,1,4,2,3,4,1,2,1,3,4,2,
%U 1,4,3,3,4,1,2,3,4,2,1,3,2,1,4,3,2,4,1,3,1,2,4,3,1,4,2,4,3,1,2,4,3,2,1,4,2
%N An arrangement of permutations. Irregular table read by rows: Read A175061(n) in binary from left to right. Row n contains the lengths of the runs of 0's and 1's.
%C Let F(n) = sum{k=1 to n} k!. Then rows F(n-1)+1 to F(n) are the permutations of (1,2,3,...,n). (And each row in this range is made up of exactly n terms, obviously.)
%e A175061(10) = 536 in binary is 1000011000. This contains a run of one 1, followed by a run of four 0's, followed by a run of two 1's, followed finally by a run of three 0's. So row 10 consists of the run lengths (1,4,2,3), a permutation of (1,2,3,4).
%Y Cf. A175061
%K base,nonn,tabf
%O 1,3
%A _Leroy Quet_, Dec 12 2009
%E Extended by _Ray Chandler_, Dec 16 2009
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