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

Max Alekseyev maxale at gmail.com
Wed Nov 29 03:41:56 CET 2017


It's worth to quote my old message regarding the phenomenon of having so
many terms in this sequence.
Below S(n) denotes the sum of decimal digits of n.

---------- Forwarded message ----------
From: Max A. <maxale at gmail.com>
Date: Sun, Oct 29, 2006 at 5:52 PM
Subject: Re: Sum of decimal digits of 16^n - 1 = 6*n

[...]

The reason is that S(n) == n (mod 9).
One can easily check that 16^n - 1 == 6*n (mod 9) holds for all n
while 16^n - 1 == 5*n (mod 9) holds only for n == 0 (mod 9).

btw, these are some other values of a and k such that the sum of
digits of a^n - 1 equals k*n for n=1,2,...,m:

m=9:
a=53941, k=21
a=539410, k=30

m=8:
a=136195 k=24
a=963055, k=27

m=7:
a=18811, k=18
a=18901, k=18
a=27244, k=18
a=40771, k=18

E.g., the sum of digits of 53941^n - 1 equals 21*n for the following n:
1, 2, 3, 4, 5, 6, 7, 8, 9, 13, 21, 31, 40, 48, 59, 67, 98, 114, 119,
130, 140, 148, 151, 156, 174, 186, 190, 242, 280, 335, 361, 367, 414,
418, 421, 425, 451, 511, 567, 569, 653, 789, 898, 1027, 1321, 1340,
2277, 2416

Max

On Tue, Nov 28, 2017 at 8:59 PM, <israel at math.ubc.ca> wrote:

> 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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>>
>>
>>
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