A274606 - OEIS

A274606

Pick any two successive integers in the sequence; there is a larger one (L) and a smaller one (S); the last digit of L is the remainder of L/S.

1, 10, 2, 11, 5, 12, 60, 3, 30, 6, 31, 15, 32, 160, 4, 20, 21, 210, 7, 70, 14, 71, 35, 72, 360, 8, 40, 41, 410, 82, 16, 80, 81, 810, 9, 90, 18, 91, 45, 92, 460, 23, 230, 46, 231, 115, 22, 110, 55, 25, 50, 51, 510, 17, 170, 34, 171, 85, 172, 860, 43, 430, 86, 431, 215, 42, 211, 105, 100, 101, 1010, 202, 200, 201, 2010, 67

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OFFSET

1,2

COMMENTS

The sequence starts with a(1)=1 and is always extended with the smallest integer not yet in the sequence and not leading to a contradiction.

This sequence is probably a permutation of the positive integers.

After 10000 terms, the smallest integer not yet present in the sequence is 913 and the largest one present is 1144810.

LINKS

Eric Angelini, Table of n, a(n) for n = 1..10589

EXAMPLE

The sequence starts with 1,10,2,11,5,12,60,3,30,6,31,15,32,160,... and indeed, 10/1 has 0 as remainder [the last digit of "10", which is the biggest term of the pair (1,10)]; 10/2 has 0 as remainder (again, the last digit of "10"); 11/2 has 1 as remainder; 11/5 has 1 as remainder; 12/5 has 2 as remainder; 60/12 has 0 as remainder; etc.

CROSSREFS

Sequence in context: A303783 A341816 A339206 * A293869 A365625 A323821

Adjacent sequences: A274603 A274604 A274605 * A274607 A274608 A274609

KEYWORD

nonn,base

AUTHOR

Eric Angelini and Jean-Marc Falcoz, Jun 30 2016

STATUS

approved