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

[Max A.]

On 10/29/06, 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

[...]

> 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.

No, it is not. 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 == 6*n (mod 9) holds only for n == 0 (mod 9).

> 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.

The distribution of S(n)/n tends to normal with the mean m =
log(16)/log(10)*4.5 ~= 5.41854 and the variance sigma^2 = 8.25/n.
Therefore, the expectation of the number of n such that S(n)/n=6 in
the interval (10^4,oo) can be estimates as
SUM[n=10^4..oo] 1/sqrt(2*Pi) sqrt(n/8.25) *exp(-(6-m)^2 * n / (2*8.25))
~= INT[n=10^4..oo] 1/sqrt(2*Pi) sqrt(n/8.25) *exp(-(6-m)^2 * n / (2*8.25))
~= 7*10^(-87)  (according to Maple)

Max