[seqfan] Re: Positive integers m such that sum of decimal digits of (16^n - 1) equals 6*n.

[israel at math.ubc.ca]

I think it's extremely unlikely that anything close to this could be proved 
in the current state of the art. There just aren't the tools to handle such 
questions. I would be surprised if we could prove there isn't an n such 
that all digits of 16^n-1 are >= 5.

Cheers,
Robert

On Nov 28 2017, Iain Fox wrote:

> Hello all, I recently submitted a conjecture to this sequence (A165722) 
> that for n greater than 223 the sum of digits of 16^n - 1 will always be 
> less than 6*n, meaning that the sequence is finite. I have tested this up 
> to n=10^6 so it seems likely. I, however, have no clue about how to go 
> about proving this. If anyone has any ideas it would be appreciated.
>
>Full text format of sequence:
>
> %I A165722 %S A165722 
> 1,2,3,4,5,6,7,10,12,13,14,17,18,23,37,43,46,60,119,183,223 %N A165722 
> Positive integers n such that the sum of decimal digits of (16^n - 1) 
> equals 6*n. %C A165722 No other terms below 10^4. %C A165722 Conjecture: 
> For n > 223, digsum(16^n - 1) < 6*n. This would mean that no further 
> terms exist in the sequence. - _Iain Fox_, Nov 22 2017 %C A165722 No 
> other terms below 10^6. - _Iain Fox_, Nov 25 2017 %F A165722 A007953(16^n 
> - 1) = A008588(n). - _Iain Fox_, Nov 22 2017 %e A165722 For n=1, 16-1 is 
> 15 with sum of digits 6, so 1 is a term. %e A165722 For n=2, 16^2-1 is 
> 255 with sum of digits 12, so 2 is a term. %t A165722 
> Select[Range[250],6#==Total[IntegerDigits[16^#-1]]&] (* _Harvey P. Dale_, 
> Nov 13 2012 *) %o A165722 (PARI) is(n) = 6*n == sumdigits(16^n-1) \\ 
> _Iain Fox_, Nov 24 2017 %K A165722 base,more,nonn,changed %O A165722 1,2 
> %A A165722 _Max Alekseyev_, Sep 24 2009
>
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