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Given a sequence u consisting just of 1's and 2's, let f(u)(n) be the length of n-th run. Then we may define a sequence u = {a(n)} by a(n)=f^(n-1)(u)(1) (starting with n=1).
2

%I #3 Mar 30 2012 16:49:49

%S 2,1,1,2,1,2,2,1,2,2,1,1,2,1,2,2,1,2,1,1,2,1,1,2,2,1,2,2,1,1,2,1,2,2,

%T 1,2,2,1,1,2,1,1,2,1,2,2,1

%N Given a sequence u consisting just of 1's and 2's, let f(u)(n) be the length of n-th run. Then we may define a sequence u = {a(n)} by a(n)=f^(n-1)(u)(1) (starting with n=1).

%C There are exactly three infinite sequences satisfying this relation, namely this sequence, A087888 and A087889.

%Y Cf. A000002, A087889, A087888.

%K easy,eigen,nonn

%O 1,1

%A Vincent Nesme (vincent.nesme(AT)ens-lyon.fr), Oct 13 2003

%E The description was not quite clear to me but I hope I have edited it correctly. - _N. J. A. Sloane_.