%I #21 Sep 12 2017 02:46:27
%S 2,8,15,28,25,33,59,77,90,79,109,111,134,137,129,148,184,189,218,197,
%T 204,226,250,231,284,284,297,313,325,309,341,398,354,418,418,414,488,
%U 440,438,478,529,465,545,554,531,628,658,561,620,677,624,697,697,657,785,818,735,865,835,762,851,851
%N The n-th iteration of A062028 starting with n.
%C a(n) == n*2^n (mod 9). This has a period of 18. - _Robert Israel_, Sep 11 2017
%H Robert Israel, <a href="/A292202/b292202.txt">Table of n, a(n) for n = 1..10000</a>
%H Robert Israel, <a href="/A292202/a292202.png">Plot of a(n)/n for n = 1 .. 4000</a>
%e n=5: 5+5=10, 10+1+0=11, 11+1+1=13, 13+1+3=17, 17+1+7=25. After 5 iterations you get 25, so a(5)=25.
%p A062028:= proc(t) option remember; t + convert(convert(t,base,10),`+`):end proc:
%p seq((A062028@@n)(n), n=1..100); # _Robert Israel_, Sep 11 2017
%t Table[Nest[# + Total@ IntegerDigits@ # &, n, n], {n, 62}] (* _Michael De Vlieger_, Sep 11 2017 *)
%o (PARI) a(n) = my(x=n); for (k=1, n, x += sumdigits(x)); x; \\ _Michel Marcus_, Sep 12 2017
%Y Cf. A007618, A006507, A062028, A016096.
%K nonn,base,look
%O 1,1
%A _Peter Weiss_, Sep 11 2017