

A080637


a(n) is smallest positive integer which is consistent with sequence being monotonically increasing and satisfying a(1)=2, a(a(n)) = 2n+1 for n>1.


5



2, 3, 5, 6, 7, 9, 11, 12, 13, 14, 15, 17, 19, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 33, 35, 37, 39, 41, 43, 45, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 96, 97, 98, 99, 100, 101, 102
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OFFSET

1,1


COMMENTS

Sequence is unique monotonic sequence satisfying a(1)=2, a(a(n)) = 2n+1 for n>1.


REFERENCES

HsienKuei 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/wpcontent/files/2016/12/aathhrr1.pdf. Also Exact and Asymptotic Solutions of a DivideandConquer Recurrence Dividing at Half: Theory and Applications, ACM Transactions on Algorithms, 13:4 (2017), #47; DOI: 10.1145/3127585


LINKS

Table of n, a(n) for n=1..70.
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(3*2^k  1 + j) = 4*2^k  1 + 3j/2 + j/2 for k >= 0, 2^k <= j < 2^k.
a(2n+1) = 2*a(n) + 1, a(2n) = a(n) + a(n1) + 1.


MAPLE

t := []; for k from 0 to 6 do for j from 2^k to 2^k1 do t := [op(t), 4*2^k  1 + 3*j/2 + abs(j)/2]; od: od: t;


CROSSREFS

Except for first term, same as A079905. Cf. A079000.
A007378, A079905, A080637, A080653 are all essentially the same sequence.
Equals A007378(n+1)1. First differences give A079882.
Sequence in context: A171886 A018559 A057196 * A124134 A007071 A242482
Adjacent sequences: A080634 A080635 A080636 * A080638 A080639 A080640


KEYWORD

nonn,easy


AUTHOR

N. J. A. Sloane and Benoit Cloitre, Feb 28 2003


STATUS

approved



