A382605 Number of distinct solutions to the problem of folding in half a chain of linked rods of length 1, ..., n.

0, 0, 1, 1, 0, 0, 1, 2, 0, 0, 1, 1, 0, 0, 2, 1, 0, 0, 1, 3, 0, 0, 1, 2, 0, 0, 3, 1, 0, 0, 1, 2, 0, 0, 4, 1, 0, 0, 4, 1, 0, 0, 1, 4, 0, 0, 1, 2, 0, 0, 3, 1, 0, 0, 3, 3, 0, 0, 1, 1, 0, 0, 2, 1, 0, 0, 1, 3, 0, 0, 1, 1, 0, 0, 4, 1, 0, 0, 1, 4, 0, 0, 1, 5, 0, 0, 3, 1, 0, 0, 1, 3, 0, 0, 2, 1, 0, 0, 6, 1

Offset

1,8

Comments

In order to be able to fold such chain in half, the total length of the chain (A000217(n)) has to be even, which is true when n=3 (mod 4) or n=0 (mod 4).

Conjecture: Whenever the length of the chain is even, there is at least one solution. That makes A154708 the sequence that lists numbers k that have at least one solution.

A380868(n) = binomial(a(n), 2). This is because in order to make a rectangle out of a chain of linked rods, we need to be able to fold the chain in half in at least 2 ways. And the number of solutions to the rectangle problem is the triangular number of the term from this sequence minus 1. See Reddit discussion below. - Daniel Mondot, Oct 19 2025

Links

Daniel Mondot, Table of n, a(n) for n = 1..10000

Allan Gottlieb, Puzzle Corner, Technology Review, December 2, 2003.

Reddit, Can we prove that all terms of this sequence are triangular numbers?

Example

A chain of 7 rods of length 1 to 7, can be folded in half in only one way: 2+3+4+5 on one side, 6+7+1, on the other, both sides being 14 in total length. Therefore a(7) = 1.

Crossrefs

Cf. A380868, A382268, A380867, A382632, A390056.

Sequence in context: A025442 A260118 A128582 * A213185 A285716 A101606

Adjacent sequences: A382602 A382603 A382604 * A382606 A382607 A382608

Keyword

nonn

Author

Daniel Mondot, Mar 31 2025

Status

approved