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 A276514 If we call "S" this sequence and consider the k-digit term a(n) of S with digits abcd...k, then a(n+1) = [a(n) + the a-th digit of S + the b-th digit of S + the c-th digit of S + ... + the k-th digit of S]. This is the lexicographically first such finite sequence with no duplicate term. 1

%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 k-digit term a(n) of S with digits abcd...k, then a(n+1) = [a(n) + the a-th digit of S + the b-th digit of S + the c-th digit of S + ... + the k-th 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 _Jean-Marc Falcoz_, Nov 07 2016

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Last modified May 16 17:10 EDT 2021. Contains 343949 sequences. (Running on oeis4.)