A138608 List first F(1) numbers from A016777, then first F(2) numbers from A016789, then the first F(3) numbers from A008585 (starting from 3), then the next F(4) numbers from A016777, then the next F(5) numbers from A016789, then the next F(6) numbers from A008585, etc, where F(n) = A000045(n), the n-th Fibonacci number.
1, 2, 3, 6, 4, 7, 10, 5, 8, 11, 14, 17, 9, 12, 15, 18, 21, 24, 27, 30, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 56, 59, 62, 65, 68, 71, 74, 77, 80, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84
Offset: 1
Examples
Let us separate natural numbers into three disjoint sets (A016777, A016789 and A008585):
1,4,7,10,13,16,19,22,25,28,31,...
2,5,8,11,14,17,20,23,26,29,32,...
3,6,9,12,15,18,21,24,27,30,33,...
then
S0={1}
S1={2}
S2={3,6}
S3={4,7,10}
S4={5,8,11,14,17}
S5={9,12,15,18,21,24,27,30}
...
and concatenating S0/S1/S2/S3/S4/S5/... gives this sequence.
Links
- Douglas E. Iannucci, Donna Mills-Taylor, On Generalizing the Connell Sequence, Journal of Integer Sequences, Vol. 2 (1999), Article 99.1.7
- Index entries for sequences that are permutations of the natural numbers
Formula
If n < 4, a(n) = n. If n = A000045(A072649(n)+1), then a(n) = a(n-1-A000045(A072649(n)))+3, otherwise a(n) = a(n-1)+3. - Antti Karttunen, Oct 05 2009
1. The sequence is formed by concatenating subsequences S0,S1, S2, ..., each of finite length. 2. The subsequence S0 consists of the element 1. 3. The n-th subsequence has F(n) elements, F(n) denotes n-th Fibonacci number. 4. Each subsequence is nondecreasing and the difference between two consecutive elements in the same subsequence is 3.
Extensions
Edited, extended, starting offset changed from 0 to 1, and Scheme-code added by Antti Karttunen, Oct 05 2009
Comments