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%I #29 Dec 13 2022 02:04:14
%S 1,1,2,2,4,4,5,3,3,5,6,6,7,10,10,7,9,12,11,9,12,8,8,14,14,11,15,13,13,
%T 17,23,20,16,15,17,23,24,16,18,18,19,26,21,28,25,19,22,22,29,24,20,30,
%U 27,21,32,29,30,35,26,34,36,25,31,32,34,37,39,36,28,27
%N "Forest Fire" sequence with the additional condition that no progression of the form ABA is allowed for any terms A and B
%C It is easy to see that a number can occur no more than twice: 1) If a number occurs twice, one term with that value must be at an odd n and the other at an even n. This is because otherwise you could always find a progression of the form ABA. 2) Once two terms of the same value are in the sequence on an even and odd n, no third term with that value can be added without creating a progression of form ABA.
%H Michael S. Branicky, <a href="/A357256/b357256.txt">Table of n, a(n) for n = 1..10000</a>
%e a(4)=2 because if a(4) were 1 the 2-4th terms would be the ABA-form progression 1,2,1. 2 here is the smallest number which forms neither an arithmetic nor ABA progression.
%o (Python)
%o from itertools import count, islice
%o def agen(): # generator of terms
%o alst, mink, aba = [0], [1, 1], [set(), set()] # even, odd appearances
%o for n in count(1):
%o k = mink[n&1]
%o ff = set(2*alst[n-i] - alst[n-2*i] for i in range(1, (n+1)//2))
%o while k in ff or k in aba[n&1]: k += 1
%o alst.append(k); aba[n&1].add(k); yield k
%o while mink[n&1] in aba[n&1]: mink[n&1] += 1
%o print(list(islice(agen(), 70))) # _Michael S. Branicky_, Dec 12 2022
%Y Cf. A229037.
%K nonn
%O 1,3
%A _Neal Gersh Tolunsky_, Dec 11 2022
%E More terms from _Michael S. Branicky_, Dec 12 2022
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