[seqfan] Re: Are these really the same?

David Wilson davidwwilson at comcast.net
Mon Jan 9 01:03:42 CET 2012

Yes, I understand that the term "normal" applies to real numbers, and 
really cannot be applied to integers, since you need the
infinite sample of digits to apply the standard definition of normality.

However, there is certainly a corresponding concept for very large 
integers, to wit, that the distribution of digits does not stray
too far from what would be expected for a number of a given size. 
Certainly you can find examples where a number n has random
looking digits while n^2 does not, but likewise you can find examples 
where non-random looking numbers n have n^2 with
random looking digits. If you need one:

1553585556595^2 = 2413628081660595947994025

Here n has a marked preponderance of the digit 5, while n^2 has a 
reasonably random digit distribution for a number of its
size. So in this case, the mapping n -> n^2 increased digital randomness 
rather than reduce it, as in your example.

The point being that on average, the mapping n -> n^2 does not affect 
randomness. As the number of digits d increases, for
a given d-digit number n, we would expect both n and n^2 to have 
randomly distributed digits. The mapping clearly has
some order-inducing digital effects, for instance, squares cannot end in 
digit 2, 3, 7 or 8; if a square ends in 0, it must end
in 00; if d-digit n is equidistributed, the first digit of n will favor 
1, etc, but I claim that these effects are negligible for very
large numbers, and that the mapping n -> n^2 can be expected to preserve 
normality, with probability 1 in the limit.
Almost all large normal integers will have normal squares.

Clearly, this is just a claim, which I readily admit I can neither 
formalize nor prove. I suspect the amount of machinery
needed to formalize my claims is significant, and proofs are beyond 
hope, given the state of mathematics with respect
to digital representations (for example, we don't know if pi is normal 
or if the digit 7 occurs in any sufficiently large power
of 2 in base 10).

Anyway, with regard to the original question, the main point is that 
almost all sufficiently large "normal" numbers n will
have a "normal" square and "normal" digitsum(x). This, along with the 
observation that n^2 and digitsum(n) tend to grow
without bound for large "normal" n, led me to expect that an n == 4 (mod 
9) could be found with

     digitsum(digitsum(digitsum(x^2)^2)) > 4

which was key to solving the problem. I was rewarded in that expectation 
by being able to find such an n.

On 1/8/2012 1:46 PM, israel at math.ubc.ca wrote:

> The actual definition of "normal" applies to real numbers, not to 
> integers. It is believed that sqrt(2), for example, is normal. This 
> would mean for example that floor(10^n*sqrt(2)) would be "normal" in 
> your sense for large n. But its square, which would be very close to 
> 2*10^(2*n), would not be "normal". So x^2 does not preserve normality.
> Robert Israel
> University of British Columbia

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