%I #6 Aug 18 2018 15:06:44
%S 1,3,5,7,9,122,576,1422,1876,4122,4576
%N Numbers k such that the average digit in the concatenation of the numbers from 1 through k is an integer.
%C Equivalently, numbers k such that A058183(k) divides A037123(k).
%C 4576 is the final term; 4 < A037123(k)/A058183(k) < 5 for all k > 4576.
%e 9 is a term because the average digit in 123456789 is (1+2+3+4+5+6+7+8+9)/9 = 45/9 = 5 (an integer).
%e 122 is a term because 12345789101112..119120121122 has digit sum 1032 and digit count 258, and 1032/258 = 4 (an integer).
%t Flatten@ Position[ Divide @@@ Transpose[ Accumulate /@ {Total /@ #, Length /@ #} &@ IntegerDigits@ Range@ 5000], _Integer] (* _Giovanni Resta_, Aug 12 2018 *)
%Y Cf. A033713, A034967, A037123, A058183, A114136.
%K nonn,base,fini,full
%O 1,2
%A _Jon E. Schoenfield_, Aug 11 2018