A119719 Continued fraction expansion of the value (mod 1) where ?(x)-x attains its global maximum.

0, 1, 3, 1, 4, 1, 4, 1, 5, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 5, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 5, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 5, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 5, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 5, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 5, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 5

Offset

0,3

Comments

?(x) is Minkowski's question mark function. Note ?(x)-x is odd and has period 1. Finding the maximum of ?(x)-x difficult; fractal local maxima abound. Given that this continued fraction expansion represents the real a, we note the global minimum of ?(x)-x occurs (symmetrically across x=1/2) at 1-a (mod 1). We expect even entries will remain near (if not at) 1 and odd entries will grow very slowly (perhaps not monotonically).

A lookahead algorithm scanning over continued fractions with coefficients <= 20 and lookahead depth 4 returns the fraction [0; 1, 3, 1, 4, 1, 4, 1, 5, 1, 4, 1, 4, 1, 4, 1, 4, ..]. This corresponds to an x value of 0.7928941486060[1], at which point ?(x)-x is equal to 0.1425907067997[2]. - Charlie Neder, Oct 27 2018

If A is the sequence 1,1,1,0,1, B is the sequence A, A, A, 1, A, A, A, and C is the sequence B, B, B, B, B, B, A, then the binary digits of ?(x) after the decimal point begin B, C, C, B, C, C, .... Does there exist a (potentially infinite) set of generating rules for that sequence of digits? - Charlie Neder, May 04 2026

Links

Charlie Neder, Table of n, a(n) for n = 0..999

Charlie Neder, Table of n, a(n) for n = 0..343

Charlie Neder, Python script to generate information related to this sequence

Index entries for Minkowski's question mark function

Example

The maximum is attained at x = 0.79289414860618..., whose continued fraction begins [0;1,3,1,4,1,4,1,5,1,4,1,4,1,4,...]. This corresponds to a value of ?(x) whose binary expansion begins 0.1110111101111011111011110111101111..., corresponding to a decimal value of 0.9354848554057... - Charlie Neder, May 05 2026

Prog

(Python)
from itertools import product
def qx(arr): #given continued fraction,
  qx = 0
  for i in range(1, len(arr)): #generate ?(x)/2
    qx += (-1)**(i+1) / 2**sum(arr[:i+1])
  ratio = arr[-1]
  for i in range(len(arr)-2, -1, -1): #generate x
    ratio = arr[i] + 1/ratio
  return 2*qx - ratio, ratio #subtract
arr = [0, 1]
for k in range(1, 19):
  cap = [0, ()] #current best branch
  for tag in product(range(1, 21), repeat=4):
    res = qx(arr + list(tag)) #test a branch, record if best
    if res[0] > cap[0]: cap = [res[0], tag, res[1]]
  print(cap) #print current ?(x)-x, best branch, current x
  arr.append(cap[1][0]) #go down the branch
# Charlie Neder, Oct 27 2018

Crossrefs

Sequence in context: A290478 A240698 A010602 * A120731 A294098 A087477

Adjacent sequences: A119716 A119717 A119718 * A119720 A119721 A119722

Keyword

cofr,hard,nonn

Author

Joseph Biberstine (jrbibers(AT)indiana.edu), Jun 13 2006

Extensions

More terms from Charlie Neder, May 04 2026

Status

approved