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A079253
a(n) is taken to be the smallest positive integer greater than a(n-1) which is consistent with the condition "n is a member of the sequence if and only if a(n) is even".
12
0, 3, 5, 6, 7, 8, 10, 12, 14, 15, 16, 17, 18, 19, 20, 22, 24, 26, 28, 30, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 94, 96
OFFSET
0,2
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; https://algo.stat.sinica.edu.tw/hk/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
Benoit Cloitre, N. J. A. Sloane and M. J. Vandermast, Numerical analogues of Aronson's sequence, J. Integer Seqs., Vol. 6 (2003), #03.2.2.
Benoit Cloitre, N. J. A. Sloane and M. J. Vandermast, Numerical analogues of Aronson's sequence, arXiv:math/0305308 [math.NT], 2003.
FORMULA
For a formula for a(n) see A079000.
a(a(n)) = 2n+4 for n >= 1.
EXAMPLE
a(1) cannot be 1 because that would imply that the first term is even; it cannot be 2 because then the first term would be even despite 1's not being in the sequence; therefore a(1)=3, which creates no contradictions and the third term is the first even term of the sequence.
MATHEMATICA
a[0] = 0; a[n_] := With[{k = 2^Floor[Log[2, (n+4)/6]]}, (Abs[n - 9k + 4] - 3k + 3n + 6)/2 - 1];
Table[a[n], {n, 0, 70}] (* Jean-François Alcover, Aug 31 2018 *)
CROSSREFS
Equals A079000 - 1.
Sequence in context: A047330 A093511 A039041 * A076054 A253201 A139636
KEYWORD
nonn
AUTHOR
STATUS
approved