A367646 a(0) = 0, a(1) = 1; for n > 1, a(n) = a(n-1) - GCT(a(n-2),a(n-1)) if nonnegative and not already in the sequence, else a(n) = a(n-1) + CT(a(n-2),a(n-1)), where CT(a,b) is the Comma transform (cf. A367360) of a and b, while GCT(a,b) is the largest possible generalized Command transform (cf. A367635) where at least one digit of both a and b can be chosen.

0, 1, 2, 14, 35, 78, 21, 103, 92, 53, 28, 60, 146, 132, 71, 44, 30, 73, 66, 102, 41, 17, 6, 82, 150, 129, 117, 26, 98, 29, 111, 20, 8, 16, 97, 166, 95, 164, 113, 72, 109, 88, 186, 105, 166, 115, 54, 109, 68, 164, 83, 131, 100, 89, 81, 179, 62, 158, 137, 56, 131, 70, 87, 79, 156, 65, 131, 80

Offset: 0

Author

Scott R. Shannon, Nov 25 2023

Keywords

• nonn
• base

Comments

This is a variation of A367578, where one can choose more than one digit from both a(n-2) and a(n-1) to create the largest possible step to a nonnegative number which has not previously appeared to form a(n). If all such numbers have already appeared the smallest possible forward step is chosen, which is just the standard Comma transform of a(n-2) and a(n-1).

It is conjectured that all nonnegative numbers appear in the sequence. After the first 10 million terms the only fixed points are 0, 1, 2, 29, 65, 84, 222, 377, 491, 499, and it is likely no more exist. The first number to appear twice is a(35) = a(44) = 166.

Examples

			a(3) = 14 as CT(a(1),a(2)) = CT(1,2) = 12, so a(3) = a(2) + 12 = 14.
a(6) = 21 as GCT(a(4),a(5)) = GCT(35,78) = 57, so a(6) = a(5) - 57 = 21, as 21 is nonnegative and not already in the sequence.
a(13) = 132 as GCT(a(11),a(12)) = GCT(60,146) = 14, so a(13) = a(12) - 14 = 132, as 132 is nonnegative and not already in the sequence. This is the first term to differ from A367578.
		

Links

• Scott R. Shannon, Table of n, a(n) for n = 0..10000
• Scott R. Shannon, Image of the first 100000 terms. The green line is a(n) = n.
• Scott R. Shannon, Image of the first 10 million terms.

Crossrefs

Cf. A367360, A367635, A367578, A005132, A121805.