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First term is 1; thereafter, if u and v are consecutive terms, with decimal expansions u = bc...ef, v = hi...jk, then v-u has decimal expansion efhi, formed by concatenating the last two digits of u and the first two digits of v. If there is a choice for v, pick the smallest.
2

%I #9 Nov 23 2023 13:03:27

%S 1,112,1325,3863,10173,17490,26516,28144,32576,40216,41857,47604,

%T 48052,53305,53858,59717,61478,69347,74121,76297,86083,94477,102187,

%U 110898,120710,121722,123934,127346,131959,137872,145086,153701,153816,155431,158546,163162

%N First term is 1; thereafter, if u and v are consecutive terms, with decimal expansions u = bc...ef, v = hi...jk, then v-u has decimal expansion efhi, formed by concatenating the last two digits of u and the first two digits of v. If there is a choice for v, pick the smallest.

%C A generalization of A121805.

%H Michael S. Branicky, <a href="/A367363/b367363.txt">Table of n, a(n) for n = 1..10000</a>

%e a(1) = 1, so ef = "01" = 1. So v-u will be a four-digit number (with a leading zero in this case), say v-u = 0xyz, with v = 1 + xyz. This suggests that we try x=1 and y=1, v = 1 + xyz = 1 + 11*, where * = 1+z. The smallest choice is z = 0, giving "efhi" = "0111" = 111, and a(2) = 1 + 111 = 112 works.

%o (Python)

%o from itertools import islice

%o def agen(): # generator of terms

%o an, y = 1, 1

%o while y:

%o yield an

%o an, y = an + 100*(an%100), 1

%o y = next((y for y in range(1, 100) if str(an+y)[:2] == str(y)), 0)

%o an += y

%o print(list(islice(agen(), 36))) # _Michael S. Branicky_, Nov 23 2023

%Y Cf. A121805.

%K nonn,base

%O 1,2

%A _N. J. A. Sloane_, Nov 23 2023

%E More terms from _Michael S. Branicky_, Nov 23 2023