A260261 Concatenation of their first digits yields the difference of two subsequent terms, a(1)=1. This is the lexicographically first infinite sequence with this property.

0, 1, 12, 24, 48, 97, 188, 199, 211, 233, 255, 277, 299, 322, 355, 388, 422, 466, 511, 566, 622, 688, 755, 833, 922, 1013, 1024, 1035, 1046, 1057, 1068, 1079, 1090, 1101, 1112, 1123, 1134, 1145, 1156, 1167, 1178, 1189, 1200, 1211, 1222, 1233, 1244, 1255, 1266, 1277

Offset: 0

Author

Eric Angelini and M. F. Hasler, Jul 21 2015

Keywords

• nonn
• base

Comments

At each step, the smallest possible choice must be made for the next term.

The initial term a(0) = 0 has been added since it can be thought to make sense, but it is maybe not without controversy ("leading zero", unambiguous continuation, ...) and should therefore be considered as purely conventional or be ignored completely.

In contrast to A121805, this sequence is infinite.

Examples

			a(1)=1 cannot be followed by a(2) = 2, 3, ..., 10 or 11 because this would yield concatenated first digits equal to 12, 13, ..., 19, 11 and 11, all different from the gap a(2) - a(1). But a(2) = 12 does produce a gap of 11 equal to the first two digits concatenated.
Then must follow a term >= 12 + 11 = 23 since the concatenation of the first two digits will be at least 11, but 23 would yield 12 for the concatenation, not equal to the gap. For a(3) = 24 the same concatenation is equal to the gap.
After a(25)=1013, and also after a(282)=10044, there follow many gaps of 11, then, e.g. after a(1187)=19999, one gap of 12 followed by about 450 gaps of 22, then one gap of 23 and about 300 gaps of 33, etc. After about 100 gaps of 99 the sequence reaches a(2850) = 99911, followed by a gap of 91, and then again more than 9000 gaps of 11.
This behavior will continue for each order of magnitude, so it is easily seen that the sequence is infinite.
		

Links

• M. F. Hasler, Table of n, a(n) for n = 0..10000
• E. Angelini, A story of leftmost digit (corrected), Jul 21 2015

Crossrefs

Cf. A121805 (gap equals concatenation of the last and first digits, respectively).

Programs

PARI

a=1;for(n=1,100,print1(a",");a+=10*digits(a)[1];for(k=1,9,digits(a+k)[1]==k&&(a+=k)&&next(2));error) \\ The error never occurs.

Formula

a(10^7)=178567225.