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.

5, 10, 15, 25, 31, 36, 38, 41, 47, 53, 58, 66, 70, 75, 85, 93, 94, 96, 99, 101, 111, 126, 134, 140, 146, 154, 165, 177, 192, 199, 206, 209, 211, 222, 225, 232, 234, 236, 239, 241, 248, 253, 259, 266, 271, 282, 287, 296, 300

Offset

1,1

Comments

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).

The lexicographically first infinite such sequence starts with 11, 13, 15, 17, 19... as it is visible here: A277268

Links

Table of n, a(n) for n=1..49.

Example

To compute a(2), add to a(1) the 5th digit of S; so a(2) = 5+5 = 10
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
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
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
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
Etc.
The sequence stops with a(49) = 300 as the next term will be also 300, which is forbidden.

Crossrefs

Cf. A277268.

Sequence in context: A344294 A151761 A008439 * A340961 A031471 A045206

Adjacent sequences: A276511 A276512 A276513 * A276515 A276516 A276517

Keyword

nonn,base,fini,full

Author

Eric Angelini and Jean-Marc Falcoz, Nov 07 2016

Status

approved