A224765 -id:A224765 - cp's OEIS frontend

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

N. J. A. Sloane, Apr 02 2012

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.

			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.

Reinhard Zumkeller, Table of n, a(n) for n = 0..8191
Benjamin Chaffin, John P. Linderman, N. J. A. Sloane, and Allan R. Wilks, On Curling Numbers of Integer Sequences, Journal of Integer Sequences, Vol. 16 (2013), Article 13.4.3.
Index entries for sequences related to binary expansion of n.

Cf. A212412 (parity), A212439, A212440, A212441, A007088, A090822, A224764/A224765 (fractional curling number).

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

f[n_, e_] := Module[{d = IntegerDigits[n, 2^e]}, Length[Split[d][[-1]]] - If[SameQ @@ d && Mod[n, 2^e] < 2^(e-1), 1, 0]]; a[n_] := Max[Table[f[n, e], {e, Range[Max[1, Floor[Log2[n]]]]}]]; a[0] = 1; Array[a, 100, 0] (* Amiram Eldar, Apr 08 2025 *)

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

A224762 Define a sequence of rationals by S(1)=1; for n>=1, write S(1),...,S(n) as XY^k, Y nonempty, where the fractional exponent k is maximized, and set S(n+1)=k; sequence gives numerators of S(1), S(2), ...

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 3, 2, 6, 1, 5, 1, 3, 4, 1, 4, 3, 5, 8, 1, 6, 13, 1, 4, 5, 8, 9, 1, 6, 5, 6, 3, 16, 1, 7, 1, 3, 6, 8, 14, 1, 6, 5, 16, 1, 5, 4, 24, 1, 5, 3, 15, 1, 5, 3, 7, 1, 5, 3, 7, 2, 54, 1, 7, 31, 1, 4, 21, 1, 4, 5, 1, 4, 5, 2, 15, 25, 1, 7, 17, 1, 4, 11, 1, 4, 5, 5, 30, 1, 6, 25, 15, 17, 1, 6, 7, 1, 4, 15, 1, 4, 5, 19

Offset: 1

Conference dinner party, Workshop on Challenges in Combinatorics on Words, Fields Institute, Toronto, Apr 22 2013, entered by N. J. A. Sloane

k is the "fractional curling number" of S(1),...,S(n). The infinite sequence S(1), S(2), ... is a fractional analog of Gijswijt's sequence A090822.
For the first 1000 terms, 1 <= S(n) <= 2. Is this always true?
The fractional curling number k of S = (S(1), S(2), ..., S(n)) is defined as follows. Write S = X Y Y ... Y Y' where X may be empty, Y is nonempty, there are say i copies of Y, and Y' is a prefix of Y. There may be many ways to do this. Choose the version in which the ratio k = (i|Y|+|Y'|)/|Y| is maximized; this k is the fractional curling number of S.
For example, if S = (S(1), ..., S(6)) = (1, 1, 2, 1, 3/2, 1), the best choice is to take X = 1,1,2, Y = 1,3/2, Y' = 1, giving k = (2+1)/2 = 3/2 = S(7).

			The sequence S(1), S(2), ... begins 1, 1, 2, 1, 3/2, 1, 3/2, 2, 6/5, 1, 5/4, 1, 3/2, 4/3, 1, 4/3, 3/2, 5/4, 8/7, 1, 6/5, 13/12, 1, 4/3, 5/4, 8/7, 9/7, 1, 6/5, 5/4, 6/5, 3/2, 16/15, 1, 7/6, 1, 3/2, 6/5, 8/7, 14/13, 1, 6/5, 5/4, 16/13, 1, 5/4, 4/3, 24/23, 1, 5/4, 3/2, 15/14, 1, 5/4, 3/2, 7/4, ...

Allan Wilks, Table of n, a(n) for n = 1..10000 (terms 1..1000 from N. J. A. Sloane)
N. J. A. Sloane, Maple program for fractional curling number and S(1),S(2),...
Allan Wilks, Table of n, S(n) for n = 1..10000 [The first 1000 terms were computed by _N. J. A. Sloane_]

Cf. A224763 (denominators), A090822, A224765.

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A224764 Numerator of fractional curling number of binary expansion of n.

Original entry on oeis.org

1, 1, 1, 2, 2, 3, 1, 3, 3, 4, 2, 2, 2, 3, 1, 4, 4, 5, 5, 2, 2, 5, 5, 3, 3, 4, 2, 2, 2, 3, 1, 5, 5, 6, 3, 2, 2, 2, 3, 3, 3, 5, 3, 2, 2, 2, 3, 4, 4, 5, 5, 2, 2, 5, 2, 3, 3, 4, 2, 2, 2, 3, 1, 6, 6, 7, 3, 2, 2, 2, 7, 3, 3, 7, 5, 2, 2, 5, 7, 4

Offset: 0

N. J. A. Sloane, Apr 26 2013

See A224762 for definition and Maple program.

			1, 1, 1, 2, 2, 3/2, 1, 3, 3, 4/3, 2, 2, 2, 3/2, 1, 4, 4, 5/4, 5/3, 2, 2, 5/2, 5/3, 3, 3, 4/3, 2, 2, 2, 3/2, 1, 5, 5, 6/5, 3/2, 2, 2, 2, 3/2, 3, 3, 5/3, 3, 2, 2, 2, 3/2, 4, 4, 5/4, 5/3, 2, 2, 5/2, 2, 3, 3, 4/3, 2, 2, 2, 3/2, 1, 6, 6, 7/6, 3/2, 2, ...
For example, 18 = 10010 in binary has fractional curling number 5/4.

N. J. A. Sloane, Table of n, a(n) for n = 0..10000

Cf. A224765, A224762, A181935.

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A181935 Curling number of binary expansion of n.

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A224762 Define a sequence of rationals by S(1)=1; for n>=1, write S(1),...,S(n) as XY^k, Y nonempty, where the fractional exponent k is maximized, and set S(n+1)=k; sequence gives numerators of S(1), S(2), ...

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A224764 Numerator of fractional curling number of binary expansion of n.

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