Sum of decimal digits of 16^n - 1 = 6*n

[zak seidov]

--- Ignacio Larrosa Cañestro <ilarrosa at mundo-r.com>
wrote:
>If S(n) = sum of decimal digits of 16^n - 1, we have
that...
>1.204119982*n < S(n) < 10.83707983*n

Me: Actually, at n<10^4,
S(n)/n is largest for n=8 and is 57/8=7.125,
S(n)/n is least for n=42 and is 9/2=4.5.  


> Although for large n,
> S(n) ~= 4.5*(FLOOR(n*LOG(16)/LOG(10)) + 1) ~=
>5.418539918*n
Me: This seems to be OK. 
Congratulations!
Zak

--- Ignacio Larrosa Cañestro <ilarrosa at mundo-r.com>
wrote:

> Sunday, October 29, 2006 10:14 AM [GMT+1=CET],
> Max A. <maxale at gmail.com> escribió:
> 
> > The sum of decimal digits of 16^n - 1 equals 6*n
> for the following n
> > (complete for n up to 10^4):
> >
> > 1, 2, 3, 4, 5, 6, 7, 10, 12, 13, 14, 17, 18, 23,
> 37, 43, 46, 60, 119,
> > 183, 223
> > Likely, this sequence is finite and there are no
> other terms.
> >
> > Does it make sense to add this sequence to OEIS?
> >
> 
> If S(n) = sum of decimal digits of 16^n - 1, we have
> that
> 
> FLOOR(n*LOG(16)/LOG(10)) + 1 < S(n) <
> 9*(FLOOR(n*LOG(16)/LOG(10)) + 1)
> 
> Aprox FLOOR(n*LOG(16)/LOG(10)) + 1 by
> n*LOG(16)/LOG(10)),
> 
> 1.204119982*n < S(n) < 10.83707983*n
> 
> Althought for large n,
> 
> S(n) ~= 4.5*(FLOOR(n*LOG(16)/LOG(10)) + 1) ~=
> 5.418539918*n
> 
> It is curious that S(n) = k*n, with n in [1, 1000],
> only for k = 6 and the 
> indicated values of n, and for k = 7 and n = 9, but
> not for k = 5, a priore 
> more likely.
> 
> Then, there isn`t reason by that the sequence can't
> be infinite. Of course, 
> the values would be more sparse for large n, if
> there ara any other value.
> 
> Best regards,
> 
> Ignacio Larrosa Cañestro
> A Coruña (España)
> ilarrosa at mundo-r.com 
> 



 
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