A277558 A variation on Recamán's sequence (A005132): to get a(n), we first try to subtract n from a(n-1): a(n) = a(n-1)-n if positive and not already in the sequence; if not then a(n) = a(n-1)+n-i, where i >= 0 is the smallest number such that a(n-1)+n-i has not already appeared.

0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9, 24, 8, 25, 43, 62, 42, 63, 41, 18, 40, 15, 39, 66, 38, 67, 37, 68, 36, 69, 35, 70, 34, 71, 33, 72, 32, 73, 31, 74, 30, 75, 29, 76, 28, 77, 27, 78, 26, 79, 133, 188, 132, 189, 131, 190, 130, 191, 129, 192

Offset: 0

Author

Benjamin Chaffin, Oct 19 2016

Keywords

• nonn

Comments

Is it ever impossible to extend the sequence -- meaning there is no number less than a(n-1)+n which has not appeared?

After 10^11 terms, the smallest number which has not appeared is 609790506.

Examples

			a(23) = 18. To get a(24) we try 18-24, but that is negative; so we try 18+24 = 42, but 42 has already appeared; so we try 18+24-1, but 41 has also already appeared; so we try 18+24-2. 40 is positive and has not yet appeared, and so a(24) = 40.
		

Links

• Benjamin Chaffin, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A005132, A064387 (chooses a(n-1)+n+i instead of a(n-1)+n-i).