

A072939


Define a sequence c depending on n by: 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 > infinity c(k) = infinity.


7



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

1,1


COMMENTS

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 is an integer k(n) and an integer m(n) such that for any k >= k(n) c(2k)c(2k1) = 2*m(n) and c(2k+1)c(2k)=m(n). Sometimes m(n) = (n1)/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: a(n) = A036554(n)+1.  Vladeta Jovovic, Apr 01 2003
a(n) = A036554(n)+1 = A079523(n)+2.  Ralf Stephan, Jun 09 2003
Conjecture: this sequence gives the positions of 0's in the limiting 0word of the morphism 0>11, 1>10, A285384.  Clark Kimberling, Apr 26 2017
Conjecture: This also gives the positions of the 1's in A328979.  N. J. A. Sloane, Nov 05 2019


LINKS

Table of n, a(n) for n=1..60.


FORMULA

Conjecture : lim n > infinity a(n)/n = 3.


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(2k1)= 10 c(2k+1)c(2k) =  5 for k >=2 implies that c(k) > infinity hence 41 is in the sequence.


CROSSREFS

Cf. A036554, A079523, A285384, A328979.
Sequence in context: A158938 A047529 A125667 * A171947 A287914 A291348
Adjacent sequences: A072936 A072937 A072938 * A072940 A072941 A072942


KEYWORD

nonn


AUTHOR

Benoit Cloitre, Aug 12 2002


STATUS

approved



