%I #14 Oct 21 2024 20:24:44
%S 0,1,3,1,4
%N Continued fraction expansion of the value (mod 1) where ?(x)-x attains its global maximum.
%C ?(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).
%C 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
%H <a href="/index/Me#MinkowskiQ">Index entries for Minkowski's question mark function</a>
%H <a href="/index/Me#MinkowskiQ">Index entries for sequences related to Minkowski's question mark function</a>
%e a = [0;1,3,1,4,..?..]
%o (Python)
%o from itertools import product
%o def qx(arr): #given continued fraction,
%o qx = 0
%o for i in range(1, len(arr)): #generate ?(x)/2
%o qx += (-1)**(i+1) / 2**sum(arr[:i+1])
%o ratio = arr[-1]
%o for i in range(len(arr)-2, -1, -1): #generate x
%o ratio = arr[i] + 1/ratio
%o return 2*qx - ratio, ratio #subtract
%o arr = [0, 1]
%o for k in range(1, 19):
%o cap = [0, ()] #current best branch
%o for tag in product(range(1, 21), repeat=4):
%o res = qx(arr + list(tag)) #test a branch, record if best
%o if res[0] > cap[0]: cap = [res[0], tag, res[1]]
%o print(cap) #print current ?(x)-x, best branch, current x
%o arr.append(cap[1][0]) #go down the branch
%o # _Charlie Neder_, Oct 27 2018
%K cofr,hard,more,nonn,changed
%O 0,3
%A Joseph Biberstine (jrbibers(AT)indiana.edu), Jun 13 2006