

A233525


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.


2



1, 1, 2, 1, 5, 4, 11, 1, 32, 49, 47, 100, 41, 259, 110, 667, 323, 1678, 1229, 3805, 7256, 4159, 17609, 19822, 33005, 26461, 72554, 6829, 210833, 342316, 290183, 736765, 133784, 2076511, 1535657
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OFFSET

1,3


COMMENTS

Define 3free Fibonacci numbers as sequences where b(n) = (b(n1) + b(n2))/3^i such that 3^i is the greatest power of 2 that divides b(n1) + b(n2). Read backwards from the nth term, this sequence produces a subsequence of 3free Fibonacci numbers where we must divide by a power of 3 every time we add.


LINKS



PROG

(Python)
def minDivisionRich(n, a=1, b=1):
....yield a
....yield b
....for i in range(2, n):
........a *= 3
........while a <= b:
............a *= 3
........a, b = b, a  b
........yield b


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



