»  Solutions to puzzles in my National Review Online Diary

March 2011

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In my March diary I returned to the topic of Benford's Law, which says that in almost any big list of numbers, around 30.1 percent will start with digit 1, around 17.6 percent with digit 2, and so on. For digit n the proportion is  log10(n+1) − log10n  (so the percentage of course 100 times that).

Consider an infinite sequence of numbers  u0, u1, u2, u3, u4, u5, u6, …, each one of which is bigger by 1 than the square of the previous one.

un+1 :=  un² + 1

Do the leading digits of the u's follow Benford's Law? The answer depends on your initial choice of u0.

If, for example, you use the real number

u0 = 9.94962308959395941218332124109326 …

as your starting value, then every single number in the sequence has leading digit 9.

Can you figure out why? Or find a closed form for this number?

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Solution

Just by reverse engineering, the magic number there is the limit, as n goes to infinity, of this process:

x1 = (102 − 1)½

x2 = ((104 − 1)½ − 1)½

x3 = (((108 − 1)½ − 1)½ − 1)½

x4 = ((((1016 − 1)½ − 1)½ − 1)½ − 1)½

x5 = (((((1032 − 1)½ − 1)½ − 1)½ − 1)½ − 1)½

… … …

xn = ((( … (((102n − 1)½ − 1)½ − 1)½ … − 1)½ − 1)½ − 1)½,
where the parentheses are nested n deep (i.e. there are n left-parens and n right-parens, not to mention n minus signs and n fractional exponents)

… … …

Most readers got that. Nobody could find a simple closed form for that limit, though, and I couldn't either. I therefore leave the question open.

Failing any simpler closed form, we'll just have to call it the Berger-Bunimovich-Hill number, after the authors of the paper in which I found it: "One-Dimensional Dynamical Systems and Benford's Law," by Arno Berger, Leonid A. Bunimovich, and Theodore P. Hill, Trans. Am. Math. Soc., Vol. 357, No. 1, pp. 197-219 (April 2004).