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

[Max A.]

On 10/29/06, Max A. <maxale at gmail.com> wrote:

> > 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).

Oh, that should be read as:
"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