A087888 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, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 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

Offset

1,1

Comments

There are exactly three infinite sequences satisfying this relation, namely this sequence, A087889 and A087890.

Links

Table of n, a(n) for n=1..105.

Crossrefs

Cf. A000002, A087889, A087890.

Sequence in context: A255934 A358401 A085028 * A296299 A109494 A074292

Adjacent sequences: A087885 A087886 A087887 * A087889 A087890 A087891

Keyword

easy,eigen,nonn

Author

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

Extensions

The description was not quite clear to me but I hope I have edited it correctly. - N. J. A. Sloane.

Status

approved