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A072939
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Define a sequence c depending on n as follows: c(1)=1 and c(2)=n; c(k+2) = (c(k+1) + c(k))/2 if c(k+1) and c(k) have the same parity; otherwise c(k+2) = abs(c(k+1) - 2*c(k)); sequence gives values of n such that lim_{k->oo} c(k) = infinity.
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7
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3, 7, 9, 11, 15, 19, 23, 25, 27, 31, 33, 35, 39, 41, 43, 47, 51, 55, 57, 59, 63, 67, 71, 73, 75, 79, 83, 87, 89, 91, 95, 97, 99, 103, 105, 107, 111, 115, 119, 121, 123, 127, 129, 131, 135, 137, 139, 143, 147, 151, 153, 155, 159, 161, 163, 167, 169, 171, 175, 179
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OFFSET
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1,1
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COMMENTS
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If c(2) is even then c(k) = 1 for k >= 2*c(2), hence there is no even value in the sequence. If n is in the sequence, there exist an integer k(n) and an integer m(n) such that, for any k >= k(n), c(2k) - c(2k-1) = 2*m(n) and c(2k+1) - c(2k) = -m(n). Sometimes m(n) = (n-1)/2 but not always. If B(n) = a(n+1) - a(n) then B(n) = 2 or 4, but B(n) does not seem to follow any pattern.
Conjecture: this sequence gives the positions of 0's in the limiting 0-word of the morphism 0->11, 1->10, A285384. - Clark Kimberling, Apr 26 2017
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LINKS
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FORMULA
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Conjecture : lim_{n->oo} a(n)/n = 3.
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EXAMPLE
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41 is in the sequence: if c(2)=41, then it follows that c(3)=21, c(4)=31, c(5)=26, c(6)=36, c(7)=31, c(8)=41, c(9)=36, ...; for k >= 2, c(2k) - c(2k-1) = 10 and c(2k+1) - c(2k) = -5, which implies that c(k) -> infinity.
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PROG
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(Python)
from itertools import count, islice
def A072939_gen(startvalue=2): return filter(lambda n:(~(n-1)&(n-2)).bit_length()&1, count(max(startvalue, 2))) # generator of terms >= startvalue
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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