Math Games: Evil Numbers

From the page: “Evil numbers are easy to find, as one might expect with an arbitrarily defined property. Interestingly, π is doubly evil. Many numbers aren’t evil. For example, e isn’t evil. The important sums are 665 and 668 after 141 decimal digits — e goes over 666 without hitting it. Chris Lomont notes that you can take every 15th digit of e, and show that e is somewhat evil. The farthest he had to reach for any constant was with Catalan’s constant, which wasn’t evil until he took every 28th digit.

It appears that most numbers are at least somewhat evil. With enough fiddling, almost anything is likely to be findable.

How common are evil numbers? (number of non-zero digits / sum of digits) gives a good estimate. For a base 10 number, the estimate gives 9/(1+2+3+4+5+6+7+8+9) = 20% = 1/5. Thus, any random number has a 1 in 5 chance of having this property.

A fifth is an estimate. Chris Lomont found a recursion that could give an exact value:

p[1] = 1/9;
p[k_ /; k <10] := p[k] = 1/9 + 1/9 Sum[p[j], {j, 1, k – 1}];
p[k_ /; k > 9] := p[k] = 1/9 Sum[p[k – d], {d, 1, 9}];
Timing[Table[p[100 n], {n, 1, 10}]][[1]]

The exact probability of 666 being hit is the following:

(large number edited out as it was breaking my blog)

Which is approximately .2000000000000000000000000000000000000000000000000000000000000002166222683713523944720537405934866672. That’s very close to 1/5. The third term in the continued fraction expansion of p[666] is 184653222869167741981875869102352405779668736930185085305398884.

The probability of evilness converges quickly to 1/5, but it does so in a highly oscillatory way. Note that each term is being multiplied by 1.243n.

Figure 1. ListPlot[Table[(1/5 – p[n]) (1243/1000)^n, {n, 30, 900}]]

Another method Chris Lomont discovered for calculating evilness involved the series expansion of (1-t9)/(t10-10t+9). In this series expansion, coefficient n is identical to p[n].

It seemed like there should be a closed form for all of this, but it was beyond Chris’s ability to calculate. Also, I wasn’t able to calculate the evilness probability for the continued fraction of a random irrational number. This calls for wisdom. If anyone has insight, let them calculate the equations.”

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