# Finding 2013 in Pi

My youngest daughter asked me this morning whether you can find the number 2013 in the digits of pi. I said it must be possible, then wrote the following Python code to find where 2013 first appears.

from mpmath import mp mp.dps = 100000 digits = str(mp.pi)[2:] digits.find('2013')

I use the multi-precision math package `mpmpath`

to get pi to 100,000 significant figures. I save this as a string and cut off the “3.” at the beginning to have just the string of digits after the decimal point.

The code returns 6275, so “2013″ appears in the 6275th position in the string of digits. However, we usually count decimal places starting from 1 but count positions in a string starting from 0, so 2013 starts in the 6276th decimal place of pi.

So π = 3.14159…2013… where the first “…” represents 6,270 digits not shown.

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Now we jump off into deeper mathematical water.

For some purposes, the digits of pi are random. The digits are obviously not random — there are algorithms for calculating them — and yet they behave randomly, and random is as random does.

If the digits of pi were random, then we could almost certainly find any sequence we want if we look long enough. Can we find *any* finite sequence of digits in the decimal expansion of pi? I would assume so, but I don’t know whether that has been proven.

You might expect that not only can you find 2013 in pi, but that if you split the digits of pi into blocks of 4, then 2013 and every other particular block would occur with the same frequency in the limit. That is, one would expect that the expansion of pi is uniform in base 10,000.

More generally, you might conjecture that pi is a *normal number*, i.e. that the digits of pi are uniformly distributed in *every* base. This has not been proven. In fact, no one has proved that any *particular* number is normal [reference]. However, we do know that almost all numbers are normal. That is, the set of non-normal numbers has Lebesgue measure zero.

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