

A086844


Odd numbers m such that the sequence defined by b(1) = m; for k>1, b(k) = floor((1+sqrt(3))*b(k1)) contains only odd numbers.


4



5, 7, 13, 19, 21, 27, 29, 35, 37, 43, 49, 51, 57, 59, 65, 67, 71, 73, 79, 81, 87, 89, 95, 97, 101, 103, 109, 111, 117, 119, 125, 131, 133, 139, 141, 147, 149, 155, 161, 163, 169, 171, 177, 179, 183, 185, 191, 193, 199, 201, 207, 213, 215, 221, 223, 229, 231, 237
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OFFSET

1,1


COMMENTS

Conjecture: let r(z)= (1/2) *(z+sqrt(z^2+4*z)), for any integral z>=1. Then the sequence a(n)4n (where a(n) is the sequence of odd numbers m such that the sequence defined by b(1) = m; for k>1, b(k) = floor(r(z)*b(k1)) contains only odd numbers) becomes ultimately periodic. Benoit Cloitre, Aug 10 2003


LINKS

Table of n, a(n) for n=1..58.


FORMULA

Observation: a(n+1)a(n) = 2, 4 or 6 for every n, a(n)=4n+O(1) and more precisely it seems that abs(a(n)4n)<=9. Is the sequence a(n)4n ultimately periodic ? Benoit Cloitre, Aug 10 2003


EXAMPLE

For m = 5 we get 5, 13, 35, 95, 259, 707, 1931, 5275, 14411, 39371, ... (cf. A057960).


CROSSREFS

Cf. A086843.
Sequence in context: A336581 A248920 A314325 * A109269 A216752 A216740
Adjacent sequences: A086841 A086842 A086843 * A086845 A086846 A086847


KEYWORD

nonn


AUTHOR

Philippe Deléham, Aug 09 2003


STATUS

approved



