sod(n) denotes the sum of the digits of a base ten positive integer n, and:
R_t = 11...11 (the digit 1 repeated precisely t times.)

Determine the values of t for which:

sod(Rt2) = (sod(Rt))2

(In reply to computer exploration by Charlie)

we can determine a hard upper limit as follows:
r(t)=(10^t-1)/9 and thus has t digits
r(t)^2=(10^t-1)^2/81=(10^(2t)-2*10^t-1)/81 and thus has 2t-1 digits

now since r(t) is comprised of all 1's sod(r(t))=t
thus the RHS is sod(r(t))^2=t^2
the LHS is
sod(r(t)^2)
since r(t)^2 has 2t-1 digits the upper limit on the sod is 9*(2t-1)=18t-9
thus we are certain not to find any more solutions once
18t-9<t^2
t^2-18t+9>0
t^2-18t+81>72
(t-9)^2>72
t-9>sqrt(72) or t-9<-sqrt(72)
t>9+sqrt(72) or t<9-sqrt(72)
t>17.49 or t<0.51
thus we need only check up to t=17

Posted by Daniel on 2013-10-02 10:36:11