A086844 Odd numbers m such that the sequence defined by b(1) = m; for k>1, b(k) = floor((1+sqrt(3))*b(k-1)) contains only odd numbers.

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

Offset: 1

Author

Philippe Deléham, Aug 09 2003

Keywords

• nonn

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(k-1)) contains only odd numbers) becomes ultimately periodic. Benoit Cloitre, Aug 10 2003

Examples

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

Crossrefs

Cf. A086843.

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