A326936 Consider an empty list L, and for k = 1, 2, ...: if L contains a pair of consecutive terms summing to k, then let (u, v) be the first such pair: replace the two terms u and v in L with a single term k and set a(u) = -k and a(v) = +k, otherwise append k to L.

-3, 3, -7, 7, -11, 11, -18, -17, 17, -22, 18, 22, -27, 27, -31, 31, 35, -35, -39, 39, -44, -49, 44, 68, -51, 51, 49, -57, 57, -62, -70, 62, -67, 67, -84, -73, 73, -78, 70, 78, -83, 83, -88, -68, 88, -93, 93, -98, 84, 98, -108, -105, 105, -109, 109, -114, 108

Offset: 1

Author

Rémy Sigrist, Oct 22 2019

Keywords

• sign

Comments

The sequence is well defined: sketch of proof, by contradiction:

- if the sequence were not well defined, then we would have some m > 0 such that L(1) = m for any k >= m,

- to prevent L(1) from being merged to L(2) forever, L(2) must always be merged to L(3) for some L(3) < m

- this can happen only finitely many times as the number of terms < m in L strictly decreases at each such merge,

- so at some time, L(1) < L(3) and L(1) merges with L(2) at k = L(1) + L(2), and then L(1) > m, a contradiction, QED.

The given procedure leads to a kind of infinite binary tree T:

- each node has a positive integer value,

- the node with value n has as parent the node with value abs(a(n)) and as sibling the node with value abs(a(n)) - n (see A328654 for the sibling of a node),

- each node has finitely many descendant nodes,

- each node has infinitely many ancestor nodes, so the tree is not rooted (see A328653 for the ancestors of 1),

- two nodes have always a common ancestor,

- see illustration in Links section.

Examples

			For k = 1:
- we set L = (1).
For k = 2:
- we set L = (1, 2).
For k = 3:
- the first two terms, (1, 2), sum to 3,
- so a(1) = -3 and a(2) = +3,
- we set L = (3).
For k = 4:
- we set L = (3, 4).
For k = 5:
- we set L = (3, 4, 5).
For k = 6:
- we set L = (3, 4, 5, 6).
For k = 7:
- the first two terms, (3, 4), sum to 7,
- so a(3) = -1 and a(4) = +7,
- we set L = (7, 5, 6).
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 1..10000
• Rémy Sigrist, Illustration of the first terms
• Rémy Sigrist, C++ program for A326936

Crossrefs

Cf. A328653, A328654.

Formula

If a(m) + a(n) = 0, then abs(a(m)) = abs(a(n)) = m + n.