A088023 Set a(1) = 1. Then take the list of defined initial terms, reverse their order, add 1, 2, 3, ... to the reversed list in succession and append this new list to the right of the previously defined terms. Repeat this process indefinitely.

1, 2, 3, 3, 4, 5, 5, 5, 6, 7, 8, 8, 8, 9, 9, 9, 10, 11, 12, 12, 13, 14, 14, 14, 14, 15, 16, 16, 16, 17, 17, 17, 18, 19, 20, 20, 21, 22, 22, 22, 23, 24, 25, 25, 25, 26, 26, 26, 26, 27, 28, 28, 29, 30, 30, 30, 30, 31, 32, 32, 32, 33, 33, 33, 34, 35, 36, 36, 37, 38

Offset

1,2

Comments

Conjecture: a(n+1) >= a(n). Comments from Don Reble, Nov 13 2005: The conjecture is plainly true. In fact, a(n+1)-a(n) = 0 or 1. Also a(A091072(n)) = n; a(A091072(n)+1) = n+1.

Links

Andrew Howroyd, Table of n, a(n) for n = 1..10000

Formula

a(n) = 2*a(n/2)-1 if n=2^k else a(n)=a(2^k-n+1)+n-2^(k-1) if 2^(k-1)<n<2^k. (Ed.)

Example

The sequence begins 1, 2, then reverse 1, 2 = 2, 1 then add 1, 2 to the latter getting 3, 3. Then append 3, 3, to the right of 1, 2, getting 1, 2, 3, 3. Then repeating the instructions, 1, 2, 3, 3 is reversed then add 1, 2, 3, 4 to 3, 3, 2, 1, = 4, 5, 5, 5. Append the latter to 1, 2, 3, 3 getting 1, 2, 3, 3, 4, 5, 5, 5, ...; and so on.

Mathematica

Nest[Join[#, Range[Length[#]] + Reverse[#]] &, {1}, 7] (* Paolo Xausa, Sep 24 2025 *)

Prog

(PARI) a(n)=my(k=logint(n, 2)); if(n<=2, n, if(n==2^k, 2*a(n/2)-1, a(2^(k+1)-n+1)+n-2^k)) \\ Andrew Howroyd, Sep 24 2025

Crossrefs

Cf. A088742, A088743, A088744.

Sequence in context: A242453 A241151 A073092 * A324477 A287292 A260717

Adjacent sequences: A088020 A088021 A088022 * A088024 A088025 A088026

Keyword

nonn

Author

Gary W. Adamson, Sep 19 2003

Extensions

Edited by John W. Layman, Oct 10 2003

a(65) onwards from Andrew Howroyd, Sep 24 2025

Status

approved