A224382 -id:A224382 - cp's OEIS frontend

A232666 6-free Fibonacci numbers.

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 4, 93, 97, 190, 287, 477, 764, 1241, 2005, 541, 2546, 3087, 5633, 8720, 14353, 23073, 37426, 60499, 97925, 26404, 124329, 150733, 275062, 425795, 700857, 1126652, 1827509, 2954161, 796945, 3751106, 4548051, 8299157, 12847208, 21146365, 33993573

Offset: 0

Brandon Avila and Tanya Khovanova, Nov 27 2013

nonn

The sequences of n-free Fibonacci numbers were suggested by John H. Conway.
a(n) is the sum of the two previous terms divided by the largest possible power of 6.
4-free Fibonacci numbers are A224382.
The sequence coincides with the Fibonacci sequence until the first multiple of 6 in the Fibonacci sequence: 144, which in this sequence is divided by 36 to produce 4.
7-free Fibonacci numbers is A078414.

B. Avila, T. Khovanova, Free Fibonacci Sequences, J. Int. Seq. 17 (2014) # 14.8.5.

Cf. A224382, A214684.

sixPower[n_] := (a = Transpose[FactorInteger[n]]; a2 = Position[a[[1]], 2]; a3 = Position[a[[1]], 3]; If[Length[a2] == 0 || Length[a3] == 0 , res = 0, res = Min[a[[2]][[a2[[1]][[1]]]], a[[2]][[a3[[1]][[1]]]]]]; res); sixFree[n_] := n/6^sixPower[n]; appendNext6Free[list_] := Append[list, sixFree[list[[-1]] + list[[-2]]]]; Nest[appendNext6Free, {0, 1}, 50]

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.

Original entry on oeis.org

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

Brandon Avila and Tanya Khovanova, Dec 11 2013

nonn

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.

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.

Cf. A233525.

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

A219328 Number of different prime divisors >= prime(n) of sums of two consecutive terms of sequence {f_n(k)} defined in A224523.

Original entry on oeis.org

1, 1, 1, 3, 3, 3, 4, 7, 3, 3, 3, 3, 3, 3, 3, 3, 3, 9, 9, 9, 9, 9, 9, 18, 14, 6, 6, 6, 6, 6, 6, 6, 6, 6, 9, 9, 12, 19, 19, 19, 19, 19, 19, 19, 19, 5, 5, 5, 5, 5, 5, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 6

Offset: 1

Vladimir Shevelev, Apr 11 2013

nonn

a(n) shows that it is sufficient to choose a(n) primes >= prime(n) such that Fibonacci-like sequence without multiples of these primes is periodic (see comment in A078414).

			1) In case n=4, the sequence {f_4(k)} has period {1,1,2,3,5,8,1,9,10}. We see that only sums of consecutive terms 5+8=13, 9+10=19, 10+1=11 have divisors >= prime(4)=7. Thus {f_4(k)} is the Fibonacci-like sequence without multiples of 11,13,19. So a(4)=3.
2) In cases 52 <= n <= 120, prime(n) >= prime(52) = 239, every sequence {f_n(k)} has period {1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1, 988, 989, 3, 992, 995, 1, 996}. It is Fibonacci-like sequence without multiples of 659, 997, 1597, or 1987. Since 659 = prime(120), then in the considered interval every a(n)=4.

Peter J. C. Moses, Table of n, a(n) for n = 1..4000

Cf. A078414, A078412, A214684, A219255, A224382, A224523.

Corrections and terms beginning a(37) were calculated by Peter J. C. Moses, Apr 19 2013

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.

Original entry on oeis.org

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

Offset: 1

Brandon Avila and Tanya Khovanova, Dec 11 2013

nonn

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.

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

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.

Cf. A233526.

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

cp's OEIS Frontend

A232666 6-free Fibonacci numbers.

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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.

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A219328 Number of different prime divisors >= prime(n) of sums of two consecutive terms of sequence {f_n(k)} defined in A224523.

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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.

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Author

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Comments

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Python