A284439 - OEIS

A284439

Sum the last digit of a(n) and the first digit of a(n+1); keep only the leftmost digit of the result. The succession of those "leftmost digits" is the succession of the digits of the sequence itself.

1, 91, 8, 2, 61, 11, 5, 6, 4, 62, 3, 31, 32, 41, 12, 111, 21, 92, 121, 13, 14, 7, 51, 15, 9, 16, 71, 17, 33, 611, 18, 81, 19, 22, 82, 112, 83, 113, 42, 34, 63, 72, 35, 43, 73, 36, 114, 64, 65, 211, 212, 115, 116, 44, 66, 45, 37, 117, 38, 23, 621, 131, 122, 631, 118, 24, 641, 123, 52, 119, 25, 511, 213, 141, 132, 133, 124

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,2

COMMENTS

The sequence is started with a(1) a= 1 and always extended with the smallest integer not yet present and not leading to a contradiction. There are no "0" digits in the sequence as a this 0 would be the leftmost digit of a later sum -- which is impossible.

Is this a permutation of the list (A052382) of numbers with no zero digits? - N. J. A. Sloane, Mar 27 2017

LINKS

Jean-Marc Falcoz, Table of n, a(n) for n = 1..10001

EXAMPLE

The 1st sum is 1+9, with result 10, and leftmost digit 1;

The 2nd sum is 1+8, with result 9, and leftmost digit 9;

The 3rd sum is 8+2, with result 10, and leftmost digit 1;

The 4th sum is 2+6, with result 8, and leftmost digit 8;

The 5th sum is 1+1, with result 2, and leftmost digit 2;

The 6th sum is 1+5, with result 6, and leftmost digit 6, etc.

The 6 "leftmost digits" so far are [1,9,1,8,2,6]; those are precisely the first 6 digits of the sequence.

CROSSREFS