A181935 Curling number of binary expansion of n.

1, 1, 1, 2, 2, 1, 1, 3, 3, 1, 2, 2, 2, 1, 1, 4, 4, 1, 1, 2, 2, 2, 1, 3, 3, 1, 2, 2, 2, 1, 1, 5, 5, 1, 1, 2, 2, 2, 1, 3, 3, 1, 3, 2, 2, 2, 1, 4, 4, 1, 1, 2, 2, 2, 2, 3, 3, 1, 2, 2, 2, 1, 1, 6, 6, 1, 1, 2, 2, 2, 1, 3, 3, 2, 2, 2, 2, 1, 1, 4, 4, 1, 2, 2, 2, 3

Offset

0,4

Comments

Given a string S, write it as S = XYY...Y = XY^k, where X may be empty, and k is as large as possible; then k is the curling number of S.

A212439(n) = 2 * n + a(n) mod 2. - Reinhard Zumkeller, May 17 2012

Links

Reinhard Zumkeller, Table of n, a(n) for n = 0..8191

Benjamin Chaffin and N. J. A. Sloane, The Curling Number Conjecture, preprint.

Example

731 = 1011011011 in binary, which we could write as XY^2 with X = 10110110 and Y = 1, or as XY^3 with X = 1 and Y = 011. The latter is better, giving k = 3, so a(713) = 3.

Prog

(Haskell)
import Data.List (unfoldr, inits, tails, stripPrefix)
import Data.Maybe (fromJust)
a181935 0 = 1
a181935 n = curling $ unfoldr
   (\x -> if x == 0 then Nothing else Just $ swap $ divMod x 2) n where
   curling zs = maximum $ zipWith (\xs ys -> strip 1 xs ys)
                          (tail $ inits zs) (tail $ tails zs) where
      strip i us vs | vs' == Nothing = i
                    | otherwise      = strip (i + 1) us $ fromJust vs'
                    where vs' = stripPrefix us vs
-- Reinhard Zumkeller, May 16 2012

Crossrefs

Cf. A212412 (parity), A212440, A212441, A007088, A090822, A224764/A224765 (fractional curling number).

Sequence in context: A352911 A278566 A255559 * A027358 A332548 A352685

Adjacent sequences: A181932 A181933 A181934 * A181936 A181937 A181938

Keyword

nonn,base

Author

N. J. A. Sloane, Apr 02 2012

Status

approved