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

1,2

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.

Prog

(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

Crossrefs

Cf. A233525.

Sequence in context: A060819 A318661 A089654 * A344674 A097062 A350948

Adjacent sequences: A233523 A233524 A233525 * A233527 A233528 A233529

Keyword

nonn

Author

Brandon Avila and Tanya Khovanova, Dec 11 2013

Status

approved