%I
%S 5,10,15,25,31,36,38,41,47,53,58,66,70,75,85,93,94,96,99,101,111,126,
%T 134,140,146,154,165,177,192,199,206,209,211,222,225,232,234,236,239,
%U 241,248,253,259,266,271,282,287,296,300
%N If we call "S" this sequence and consider the kdigit term a(n) of S with digits abcd...k, then a(n+1) = [a(n) + the ath digit of S + the bth digit of S + the cth digit of S + ... + the kth digit of S]. This is the lexicographically first such finite sequence with no duplicate term.
%C No such sequence is possible starting with a(1)<5. The sequence is finite and stops with 300, the 49th term (as the third digit of the sequence is a zero, leading to 300+0+0+0 = 300).
%C The lexicographically first infinite such sequence starts with 11, 13, 15, 17, 19... as it is visible here: A277268
%e To compute a(2), add to a(1) the 5th digit of S; so a(2) = 5+5 = 10
%e To compute a(3), add to a(2) the 1st digit of S and the 0th digit of S; so a(3) = 10+5+0 = 15
%e To compute a(4), add to a(3) the 1st digit of S and the 5th digit of S; so a(4) = 15+5+5 = 25
%e To compute a(5), add to a(4) the 2nd digit of S and the 5th digit of S; so a(5) = 25+1+5 = 31
%e To compute a(6), add to a(5) the 3rd digit of S and the 1st digit of S; so a(6) = 31+0+5 = 36
%e Etc.
%e The sequence stops with a(49) = 300 as the next term will be also 300, which is forbidden.
%Y Cf. A277268.
%K nonn,base,fini,full
%O 1,1
%A _Eric Angelini_ and _JeanMarc Falcoz_, Nov 07 2016
