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

0, 1, 3, 1, 4

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

Links

Table of n, a(n) for n=0..4.

Index entries for Minkowski's question mark function

Index entries for sequences related to Minkowski's question mark function

Example

a = [0;1,3,1,4,..?..]

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: A229755 A257634 A110790 * A125162 A174382 A123730

Adjacent sequences: A119716 A119717 A119718 * A119720 A119721 A119722

Keyword

cofr,hard,more,nonn

Author

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

Status

approved