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If we call "T" this sequence and consider the k-digit term a(n) of T with digits abcd...k, then a(n+1) = [a(n) + the a-th digit of T + the b-th digit of T + the c-th digit of T + ... + the k-th digit of T]. This is the lexicographically first such infinite sequence containing no duplicate term.
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%I #22 Oct 18 2019 03:58:54

%S 11,13,15,17,19,21,23,25,27,29,31,33,35,37,39,41,45,49,53,55,57,59,61,

%T 67,73,75,77,79,81,89,97,99,101,103,105,107,109,111,114,119,122,125,

%U 128,137,140,144,151,154,159,162,169,176,183,192,195,198,207,209,211,214,219,222,225,228,237,240,244,251,254,259,262,269,276,283,292,295,298,307

%N If we call "T" this sequence and consider the k-digit term a(n) of T with digits abcd...k, then a(n+1) = [a(n) + the a-th digit of T + the b-th digit of T + the c-th digit of T + ... + the k-th digit of T]. This is the lexicographically first such infinite sequence containing no duplicate term.

%C There are 3 lexicographically earlier sequences with this property, but they all stop at some point. The first one is A276514, starting with 5,10,15,25,31,... and halting with a(49) = 300. The second one starts with 9,10,19,29,31,... and stops with a(246) = 3003. The third one starts with 10,11,13,15,17,... and ends with a(8) = 22, as shown here: 10, 11, 13, 15, 17, 19, 21, 22. [To compute a hypothetical a(9), one has to add to 22 the second digit of the sequence (which is zero) and (again) the second digit of the sequence (again zero): 22+0+0 = 22.]

%C As this sequence, starting with 11, 13, 15, 17, 19, ..., shows no zero digit among its first 10 digits, it will never stop.

%H Eric Angelini, <a href="/A277268/b277268.txt">Table of n, a(n) for n = 1..1001</a>

%e To compute a(2), add to a(1) the 1st digit of T and the 1st digit of T, so a(2) = 11+1+1 = 13;

%e to compute a(3), add to a(2) the 1st digit of T and the 3rd digit of T, so a(3) = 13+1+1 = 15;

%e to compute a(4), add to a(3) the 1st digit of T and the 5th digit of T, so a(4) = 15+1+1 = 17;

%e to compute a(5), add to a(4) the 1st digit of T and the 7th digit of T, so a(5) = 17+1+1 = 19;

%e to compute a(6), add to a(5) the 1st digit of T and the 9th digit of T, so a(6) = 19+1+1 = 21;

%e to compute a(7), add to a(6) the 2nd digit of T and the 1st digit of T, so a(7) = 19+1+1 = 23;

%e etc.

%Y Cf. A276514.

%K nonn,base

%O 1,1

%A _Eric Angelini_ and _Jean-Marc Falcoz_, Nov 07 2016