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

[zak seidov]

See attached graph of S(n)/n,
Zak

--- zak seidov <zakseidov at yahoo.com> wrote:

> --- 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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