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Start with a(1) = 1, a(2) = 1, then a(n)*3^k = a(n+1) + a(n+2), with 3^k the smallest power of 3 (k>0) such that all terms a(n) are positive integers.
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%I #25 Jun 24 2022 14:55:30

%S 1,1,2,1,5,4,11,1,32,49,47,100,41,259,110,667,323,1678,1229,3805,7256,

%T 4159,17609,19822,33005,26461,72554,6829,210833,342316,290183,736765,

%U 133784,2076511,1535657

%N Start with a(1) = 1, a(2) = 1, then a(n)*3^k = a(n+1) + a(n+2), with 3^k the smallest power of 3 (k>0) such that all terms a(n) are positive integers.

%C Define 3-free Fibonacci numbers as sequences where b(n) = (b(n-1) + b(n-2))/3^i such that 3^i is the greatest power of 2 that divides b(n-1) + b(n-2). Read backwards from the n-th term, this sequence produces a subsequence of 3-free Fibonacci numbers where we must divide by a power of 3 every time we add.

%C For other examples of n-free Fibonacci numbers, see A232666, A214684, A224382.

%H Brandon Avila, <a href="/A233525/b233525.txt">Table of n, a(n) for n = 1..1000</a>

%H Brandon Avila and Tanya Khovanova, <a href="http://arxiv.org/abs/1403.4614">Free Fibonacci Sequences</a>, arXiv preprint arXiv:1403.4614 [math.NT], 2014 and <a href="https://cs.uwaterloo.ca/journals/JIS/VOL17/Avila/avila4.html">J. Int. Seq. 17 (2014) # 14.8.5</a>.

%o (Python)

%o def minDivisionRich(n, a=1, b=1):

%o ....yield a

%o ....yield b

%o ....for i in range(2, n):

%o ........a *= 3

%o ........while a <= b:

%o ............a *= 3

%o ........a, b = b, a - b

%o ........yield b

%Y Cf. A233526.

%K nonn

%O 1,3

%A _Brandon Avila_ and _Tanya Khovanova_, Dec 11 2013