A233526 -id:A233526 - cp's OEIS frontend

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.

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

A266295 2-free tetranacci sequence beginning 1,3,5,7.

Original entry on oeis.org

1, 3, 5, 7, 1, 1, 7, 1, 5, 7, 5, 9, 13, 17, 11, 25, 33, 43, 7, 27, 55, 33, 61, 11, 5, 55, 33, 13, 53, 77, 11, 77, 109, 137, 167, 245, 329, 439, 295, 327, 695, 439, 439, 475, 1, 677, 199, 169, 523, 49, 235, 61, 217, 281, 397, 239, 567, 371, 787

Offset: 1

Jeremy F. Alm, Dec 28 2015

nonn

For n>4, a(n) = (a(n-1) + a(n-2) + a(n-3) + a(n-4)) / 2^d, where 2^d is the largest power of 2 dividing a(n-1) + a(n-2) + a(n-3) + a(n-4). In other words, sum the previous four terms, then divide by two until the result is odd.

Alm, Herald, Miller, and Sexton, 2-Free Tetranacci Sequences, unpublished.

Jeremy F. Alm, Table of n, a(n) for n = 1..10000

Cf. A000265, A232666, A233248, A233526.

nxt[{a_, b_, c_, d_}] := {b, c, d, (a + b + c + d)/2^IntegerExponent[ a + b + c + d, 2]}; NestList[nxt,{1,3,5,7},60][[All,1]] (* Harvey P. Dale, Nov 09 2020 *)

lista(nn) = {print1(x = 1, ", "); print1(y = 3, ", "); print1(z = 5, ", "); print1(t = 7, ", "); for (n=5, nn, tt = (x+y+z+t); tt /= 2^valuation(tt, 2); print1(tt, ", "); x=y; y=z; z=t; t=tt;);} \\ Michel Marcus, Dec 29 2015

### CREATES A b-FILE ###
def main():
    name = "b266295.txt"
    file = open(name, 'w')
    file.write('1' + ' ' + '1\n')
    file.write('2' + ' ' + '3\n')
    file.write('3' + ' ' + '5\n')
    file.write('4' + ' ' + '7\n')
    a, b, c, d = 1, 3, 5, 7
    for i in range(5,10001):
        x=a+b+c+d
        while x%2==0:
            x /= 2
        a, b, c, d = b, c, d, x
        file.write(str(i) + ' ' + str(int(d)) + '\n')
    file.close()
main()

a(n) = (a(n-1) + a(n-2) + a(n-3) + a(n-4)) / 2^d, where 2^d is the largest power of 2 dividing a(n-1) + a(n-2) + a(n-3) + a(n-4).

a(n) = A000265(a(n-1) + a(n-2) + a(n-3) + a(n-4)). - Michel Marcus, Dec 29 2015

cp's OEIS Frontend

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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A266295 2-free tetranacci sequence beginning 1,3,5,7.

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