A080645 a(1) = 1; for n>1, a(n) is taken to be the smallest integer greater than a(n-1) which is consistent with the condition "for n>1, if n is a member of the sequence then a(n) is even".
1, 2, 4, 6, 7, 8, 10, 12, 13, 14, 15, 16, 18, 20, 22, 24, 25, 26, 27, 28, 29, 30, 31, 32, 34, 36, 38, 40, 42, 44, 46, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107
Offset: 1
References
- Hsien-Kuei Hwang, S Janson, TH Tsai, Exact and asymptotic solutions of the recurrence f(n) = f(floor(n/2)) + f(ceiling(n/2)) + g(n): theory and applications, Preprint, 2016; http://140.109.74.92/hk/wp-content/files/2016/12/aat-hhrr-1.pdf. Also Exact and Asymptotic Solutions of a Divide-and-Conquer Recurrence Dividing at Half: Theory and Applications, ACM Transactions on Algorithms, 13:4 (2017), #47; DOI: 10.1145/3127585
Links
- B. Cloitre, N. J. A. Sloane and M. J. Vandermast, Numerical analogues of Aronson's sequence, J. Integer Seqs., Vol. 6 (2003), #03.2.2.
- B. Cloitre, N. J. A. Sloane and M. J. Vandermast, Numerical analogues of Aronson's sequence (math.NT/0305308)
- Index entries for sequences of the a(a(n)) = 2n family
Formula
a(1)=1, a(2)=2, a(3)=4; then for k>=1, abs(j)<=2^k: a(3*2^k+j)=4*2^k+3/2*j+abs(j)/2.
{a(a(n))} = {1, 2, 2i, i >= 3}.