There are innumerable fascinating questions here. I’ll just offer a glimpse of one.
Think of a number, not necessarily a whole number. (And if it’s negative, that’s fine, though there is not much point.) Square it (see?), then add 1. Take your answer, square it, and add 1. Take the answer to that, square it, and add 1. Continue forever.
A bit more formally: Think of a number u0. Form the new number u1 =u0² + 1. Then u2 =u1² + 1, and so on.
Still more formally: Consider the sequence generated recursively by the rule un+1
² + 1, for some given initial u0
. In what follows I’ll refer to this rule as Q
However formally it’s posed, the process gives you 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. The question is: Do the leading digits of theu’s follow Benford’s Law?
The answer depends on your initial choice of u0. Taking the set R of all possible real numbers as our domain of interest for u0, it divides into two subsets: one — call itBQ — of all the u0 that generate Benford-compliant sequences under the rule Q, and another — call it B′Q — of all the u0 that don’t. Any given real number is in either BQ or B′Q.
It can be proved that both BQ and B′Q are uncountably infinite, but that B′Q has Lebesgue measure zero. That is to say, though both sets are uncountably infinite, BQ is way more uncountably infinite than B′Q. (If you don’t follow this, I’m afraid there is nothing for it but to go and read up on the theory of Lebesgue measure.)
The loose way mathematicians use to describe a situation like this is: Almost all real numbers are inBQ. An alternative, even looser way: Though B′Q contains an uncountable infinity of numbers, the probability of any given number being in B′Q is zero.
If you have a decent math/stats package, or are serving a long prison sentence, you might want to try that out for some particular real number.
π is an obvious candidate. Then u0 is 3.141592 . . . Square and add one: That gives u1 = 10.869604 . . . Squaring that and adding one, u2 = 119.148299 . . . Proceeding, you get u3 = 14197.317353 . . . , u4 = 201563821.046002 . . . , u5 = 40627973954664757.854260 . . . and so on. Leading digits are 3, 1, 1, 1, 2, 4 . . . Hey, it looks Benford-compliant already! (It is.)
That’s all well and good, you may say, but how do I know, without experimentation, whether some particular value of u0 will generate a Benford-compliant sequence under the rule Q? (With, or course, an infinity of corresponding questions for an infinity of other rules you might think up.)
So far as I know, there is no general method for determining this, though you can of course always take the experimental-math approach: Generate the sequence out to ubazillion or so and run the stats on its leading digits, as I just did for the primes. With u0 = 0, for example, you get a sequence that sure looks Benford-compliant, but whose compliance no one has been able to prove.
It is possible to figure out particular numbers thatdon’tgenerate Benford-compliant sequences — numbers, that is to say, that are members of the set B′Q, even though the chance of any given number being in B′Q are infinitely small. If, for example, you use the real number
u0= 9.94962308959395941218332124109326 . . .
as your starting value for the rule Q, then every single number in the sequence has leading digit 9. How noncompliant is that!
Can you figure out why? Or find a closed form for this number?