A335214 Divide the biggest term of the pair [a(n), a(n+1)] by the smallest one and keep the remainder; the successive remainders of the successive pairs rebuild the starting sequence, digit after digit. This is the lexicographically earliest sequence of distinct positive terms with this property.

10, 3, 6, 9, 15, 24, 23, 18, 8, 12, 5, 13, 14, 22, 30, 29, 27, 11, 34, 31, 17, 19, 21, 45, 90, 44, 35, 33, 26, 25, 4, 7, 32, 63, 46, 47, 38, 36, 37, 41, 87, 39, 78, 74, 70, 67, 62, 59, 28, 54, 52, 57, 53, 60, 92, 43, 20, 16, 86, 82, 75, 72, 64, 61, 55, 58, 51, 106, 88, 81, 42, 49, 155, 148, 48, 103, 206, 40, 127, 65

Offset: 1

Author

Eric Angelini and Jean-Marc Falcoz, May 27 2020

Keywords

• base
• nonn

Comments

This is conjectured to be a permutation of the positive integers.

One might enter the successive remainders as the sequence T, which would start with 1, 0, 3, 6, 9, 1, 5, 2, 4, 2, 3, 1, 8, 8, 1, 2, 5, 1, 3, 14, 2, 2, 3, 0, 2, 9, 2, 7, 1, 1, 3, 4, 31, 17, 1, 9,... We see that some remainders are > 9.

Examples

			a(1)/a(2) = 10/3 = 3 with remainder 1;
a(3)/a(2) = 6/3 = 2 with remainder 0;
a(4)/a(3) = 9/6 = 1 with remainder 3;
a(5)/a(4) = 15/9 = 1 with remainder 6;
a(6)/a(5) = 24/15 = 1 with remainder 9;
a(6)/a(7) = 24/23 = 1 with remainder 1;
a(7)/a(8) = 23/18 = 1 with remainder 5; etc.
We see that the successive remainders 1,0,3,6,9,1,5,... are the successive digits of the sequence itself 10,3,6,9,15,24,23,...
		

Links

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

Crossrefs

Cf. A334336.