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Define b_n(k) to be the lexicographically earliest sequence of distinct nonnegative integers with the property that two terms that contain the digit "d" are always separated by exactly "d" terms that do not contain the digit "d", in base n. a(n) is the number of terms in b_n(k).
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%I #15 Nov 16 2024 16:11:49

%S 2,4,6,10,10,18,18,22,22,30,30,34,42,42,78,78,78,78,102,102,114,114,

%T 114,114,142,142,142,142,214,214,214,214,214,214,214,222,274,274,274,

%U 274,274,354,354,354,354,354,354,642,642,642,642,642,642,642,642

%N Define b_n(k) to be the lexicographically earliest sequence of distinct nonnegative integers with the property that two terms that contain the digit "d" are always separated by exactly "d" terms that do not contain the digit "d", in base n. a(n) is the number of terms in b_n(k).

%C This process terminates only when all nonzero digits are prohibited by the restrictions in place for the next term; as b_n(2) = "10" for all n, the digit 1 is only prohibited for odd numbered terms, and as such a(n) must be even for all n. Similar logic can be applied to the digit 3 to show that for all n>3, a(n) is not divisible by 4.

%C A375232 is the sequence generated when n=10.

%H Jake Bird, <a href="/A376150/b376150.txt">Table of n, a(n) for n = 2..88</a>

%e For n = 5:

%e b_5(1) = 0; as this contains the digit 0, b_5(2), b_5(3) etc. must also contain a 0

%e b_5(2) = 10 (= 5 in decimal); must contain a 0 from b_5(1); as this contains the digit 1, b_5(4), b_5(6) etc. must also contain a 1, and all other terms must not contain a 1

%e b_5(3) = 20; must have 0 but not 1

%e b_5(4) = 100; must have 0 and 1 but not 2

%e b_5(5) = 30; must have 0 but not 1 or 2

%e b_5(6) = 102; must have 0, 1, and 2, but not 3

%e b_5(7) = 40; must have 0 but not 1, 2, or 3

%e b_5(8) = 101; must have 0 and 1 but not 2, 3, or 4

%e b_5(9) = 203; must have 0, 2, and 3, but not 1 or 4

%e b_5(10) = 110; must have 0 and 1 but not 2, 3, or 4

%e b_5(11) = ---; must have 0 but not 1, 2, 3, or 4 - the only number that fills this condition is 0, but 0 already appears in the sequence, so the sequence terminates after ten terms, and a(5) = 10

%Y Cf. A375232.

%K nonn,base

%O 2,1

%A _Jake Bird_, Sep 12 2024