A079905 a(1)=1; then a(n) is smallest positive integer which is consistent with sequence being monotonically increasing and satisfying a(a(n)) = 2n+1 for n>1.

1, 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

Offset: 1

Author

N. J. A. Sloane, Feb 21 2003

Keywords

• nonn
• easy

Comments

Alternate definition: a(n) is taken to be smallest positive integer greater than a(n-1) such that the condition "a(a(n)) is always odd" can be satisfied. - Matthew Vandermast, Mar 03 2003

Also: a(n)=smallest positive integer > a(n-1) such that the condition "n is in the sequence if and only if a(n) is even" is false; that is, the condition "either n is not in the sequence and a(n) is odd or n is in the sequence and a(n) is even" is satisfied. - Matthew Vandermast, Mar 05 2003

Links

• Yifan Xie, Table of n, a(n) for n = 1..10000
• 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, arXiv:math/0305308 [math.NT], 2003.
• Hsien-Kuei Hwang, S. Janson, and T.-H. 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.
• Hsien-Kuei Hwang, S. Janson, and T.-H. Tsai, 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.
• Index entries for sequences of the a(a(n)) = 2n family

Crossrefs

See A080637 for a nicer version. Cf. A079000.

Equals A007378(n+1)-1, n>1.

A007378, A079905, A080637, A080653 are all essentially the same sequence.

Cf. A169956, A169957.

Union of A079946 and A005408 (the odd numbers).

Programs

Mathematica

b[n_] := b[n] = If[n<4, n+1, If[OddQ[n], b[(n-1)/2+1]+b[(n-1)/2], 2b[n/2]]];
a[n_] := If[n == 1, 1, b[n+1] - 1];
a /@ Range[70] (* Jean-François Alcover, Oct 18 2019, after Vladeta Jovovic in A007378 *)

Formula

a(1)=1, a(2)=3, then a(3*2^k - 1 + j) = 4*2^k - 1 + 3j/2 + |j|/2 for k >= 1, -2^k <= j < 2^k.

a(n) = 1+A079945(n-1)-A079944(n-1) for n>1, a(1)=1. - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Feb 23 2003

Extensions

More terms from Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Feb 23 2003