All the terms a(n) as well as the intermediate results will be multiples of 3:
x^3 mod 3 = x mod 3 [0^3 = 0; 1^3 = 1; (-1)^3 = -1].
Therefore (sum of cubes of digits) mod 3 = (sum of digits) mod 3.
Because the only multiple of 3 in
A046197 is 153, every number which is a multiple of 3 will end up at 153.
Some other terms (not dealt with here) may reach a cycle of length > 1:
Elizabeth Todd has shown that only numbers (1 mod 3) and (2 mod 3) may reach a cycle, and the only possible cycles are {55, 230, 130}, {136, 244}, {160, 217, 352}, {919, 1459}. That means that numbers (0 mod 3) never reach a cycle but just a single number, namely 153.
Shyam Sunder Gupta tested all the multiples of 3 less than 10^5. He found that they all reach 153, in accordance with the above statements.
The values a(n) for n>15 are really too big to be fully written out (and so are missing in the list), as Jon E. Schoenfield calculated for n=16 and n=17:
a(16) = 3.777999...999*10^61042524005486970; it has one 3, three 7's, and 61042524005486967 9's, so the sum of the cubes of its digits is 1*3^3 + 3*7^3 + 61042524005486967*9^3 = 44499999999999999999 = a(15).
a(17) consists of the digit string 45888 followed by a very, very long string of 9's. The number of 9's in that string is (a(16) - 1725)/729, which is a 61042524005486968-digit number consisting of the digit 5 followed by 753611407475147 copies of the 81-digit string 182441700960219478737997256515775034293552812071330589849108367626886145404663923 followed by a single instance of the 60-digit string 182441700960219478737997256515775034293552812071330589849106.
