A088720 Unique monotone sequence satisfying a(a(a(n))) = 2n.

4, 5, 6, 8, 9, 10, 11, 12, 14, 16, 17, 18, 19, 20, 21, 22, 23, 24, 26, 28, 30, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 50, 52, 54, 56, 58, 60, 62, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89

Offset: 3

Author

Colin Mallows, Oct 16 2003

Keywords

• easy
• nonn

Comments

For k >= 1 and m >= 2, a monotone a(n) such that a^(k+1)(n) = m*n is unique only when m = 2 or (k,m) = (1,3).

Numbers k > 2 whose binary representation starts with 10 or ends with 0. - Yifan Xie, Jan 21 2025

Examples

			a(a(a(3))) = a(a(4)) = a(5) = 6.
		

Links

• Robert Israel, Table of n, a(n) for n = 3..10000
• 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.

Crossrefs

Cf. A007378, A003605, A088721.

Programs

Maple

seq(op([seq(n+2^m,n=3*2^m .. 5*2^m-1), seq(2*n-4*2^m, n=5*2^m..6*2^m-1)]), m=0..10); # Robert Israel, Apr 05 2017

PARI

a(n)={my(m=logint(n/3, 2)); if(n<5*2^m, n+2^m, 2*(n-2^(m+1)))}; \\ Yifan Xie, Jan 31 2024

Formula

For a^(k+1)(n) = 2n, we have for (k+1)2^m <= n <= (2k+1)2^m, a(n) = n+2^m; for (2k+1)2^m <= n <= (2k+2)2^m, a(n) = 2n-2k*2^m.

From Robert Israel, Apr 05 2017: (Start)

a(2n) = 2*a(n).

a(4n+1) = a(2n+1) + 2*a(n).

a(4n+3) = 3*a(2n+1) - 2*a(n).

G.f. g(z) satisfies g(z) = 4*z^3 + 5*z^4 + 2*z^5 - 3*z^7 + 5*z^9 - 4*z^11 + (2+1/(2*z)+3*z/2)*g(z^2) - (1/(2*z)+3*z/2)*g(-z^2) + (2*z-2*z^3)*g(z^4).

(End)

Extensions

More terms from John W. Layman, Oct 18 2003