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A072939
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Let c(1)=1 and c(2)=n; c(k+2) = c(k+1)/2 + c(k)/2 if c(k+1) and c(k) have the same parity; c(k+2)=abs(c(k+1)-2*c(k)) otherwise; sequence gives values of n=c(2) such that lim k -> infinity c(k) = infinity.
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5
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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
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,1
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COMMENTS
| If c(2) is even then c(k) = 1 for k >= 2*c(2), hence there's no even value in the sequence. If n is in the sequence, there's 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). Sometime m(n) = (n-1)/2 but not always. If B(n) = a(n+1)-a(n) then B(n) = 2 or 4 only, but B(n) doesn't seem follow any pattern.
Conjecture: a(n) = A036554(n)+1. - Vladeta Jovovic (vladeta(AT)eunet.rs), Apr 01 2003
a(n) = A036554(n)+1 = A079523(n)+2. - Ralf Stephan (ralf(AT)ark.in-berlin.de), Jun 09 2003
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FORMULA
| Conjecture : lim n -> infinity a(n)/n = 3.
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EXAMPLE
| if c(2)=41 -> c(3)= 21 ->c(4)= 31 ->c(5)= 26 ->c(6)= 36 ->c(7)= 31 ->c(8)= 41 ->c(9)= 36 then c(2k)-c(2k-1)= 10 c(2k+1)-c(2k) = - 5 for k >=2 implies that c(k) -> infinity hence 41 is in the sequence.
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CROSSREFS
| Sequence in context: A047529 A125667 * A171947 A186890 A075607 A173699
Adjacent sequences: A072936 A072937 A072938 * A072940 A072941 A072942
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KEYWORD
| nonn
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AUTHOR
| Benoit Cloitre (benoit7848c(AT)orange.fr), Aug 12 2002
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