[seqfan] Re: Are these really the same?

[David Wilson]

On 1/7/2012 6:20 PM, Veikko Pohjola wrote:
> I produced a sequence identical to A156638 as follows: Numbers n such that the digit sum of the digit sum of the square of the digit sum of n^2 = 4. Can it be proved or disproved that these are really the same?
>
> Veikko Pohjola
>
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>
A156638 consists of all numbers == 4 or 5 (mod 9). A fairly 
straightforward argument modulo 9 shows
that all the numbers you describe are in A156638. However, there are 
numbers in A156638 that are
not in your sequence. For example, let

n =
96019407833469901546749865786386112278884519841349077509706112489402136224573901
82990862369255038178669819071254543866242079874381931885715723232715356395266377
02691450018096418683273171301219560970962323567670685363467717396610425927598342
139

n == 4 (mod 9), so n is in A156638.

Letting d(x) be the digit sum of x, you can also show that

            d(n^2) = 2167
     ==> d(n^2)^2 = 4695889
     ==> d(d(n^2)^2) = 49
     ==> d(d(d(n^2)^2)) = 13

since 13 is not 4, n is not one of your numbers. Hence your sequence is 
not A156638.

The counterexample n has 243 digits. n is certainly not the smallest 
counterexample, but my gut tells me
that the smallest counterexample likely has > 100 digits, leastwise, 
your sequence and A156638 coincide for
a VERY VERY long time before a difference is encountered.

If your sequence description had yet another "the digit sum of" in front 
of it, then the n above would also be
in your sequence. The sequence would still not be the same as A156638, 
however, if we could write digits on
subatomic particles, the known universe would not contain enough 
particles to write down the first
counterexample.