A233526 Start with a(1) = 1, a(2) = 3, then a(n)*2^k = a(n+1) + a(n+2), with 2^k the smallest power of 2 (k>0) such that all terms a(n) are positive integers.

1, 3, 1, 5, 3, 7, 5, 9, 1, 17, 15, 19, 11, 27, 17, 37, 31, 43, 19, 67, 9, 125, 19, 231, 73, 389, 195, 583, 197, 969, 607, 1331, 1097, 1565, 629, 2501

Offset: 1

Author

Brandon Avila and Tanya Khovanova, Dec 11 2013

Keywords

• nonn

Comments

Define 2-free Fibonacci numbers as sequences where b(n) = (b(n-1) + b(n-2))/2^i such that 2^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 2-free Fibonacci numbers where we must divide by a power of 2 every time we add.

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

Links

• Brandon Avila, Table of n, a(n) for n = 1..1000
• Brandon Avila and Tanya Khovanova, Free Fibonacci Sequences, arXiv preprint arXiv:1403.4614 [math.NT], 2014 and J. Int. Seq. 17 (2014) # 14.8.5.

Crossrefs

Cf. A233525.

Programs

Python

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