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

[Ignacio Larrosa Cañestro]

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