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 A119719 Continued fraction expansion of the value (mod 1) where ?(x)-x attains its global maximum. 0
 0, 1, 3, 1, 4 (list; graph; refs; listen; history; text; internal format)
 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 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

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Last modified October 1 17:45 EDT 2020. Contains 337444 sequences. (Running on oeis4.)