A181935 - OEIS

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

(list; graph; refs; listen; history; text; internal format)

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