## 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 log_{10}(*n*+1) − log_{10}*n* (so the percentage of course 100 times
that).

Consider an infinite sequence of numbersu_{0},u_{1},u_{2},u_{3},u_{4},u_{5},u_{6}, …, each one of which is bigger by 1 than the square of the previous one.

u_{n+1}:=u_{n}² + 1

Do the leading digits of theu's follow Benford's Law? The answer depends on your initial choice ofu_{0}.

If, for example, you use the real number

u_{0}= 9.94962308959395941218332124109326 …

as your starting value, thenevery single number in the sequencehas 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:

x_{1}= (10^{2}− 1)^{½}

x_{2}= ((10^{4}− 1)^{½}− 1)^{½}

x_{3}= (((10^{8}− 1)^{½}− 1)^{½}− 1)^{½}

x_{4}= ((((10^{16}− 1)^{½}− 1)^{½}− 1)^{½}− 1)^{½}

x_{5}= (((((10^{32}− 1)^{½}− 1)^{½}− 1)^{½}− 1)^{½}− 1)^{½}

… … …

x_{n}= ((( … (((10^{2n}− 1)^{½}− 1)^{½}− 1)^{½}… − 1)^{½}− 1)^{½}− 1)^{½},

where the parentheses are nestedndeep (i.e. there arenleft-parens andnright-parens, not to mentionnminus signs andnfractional 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).