A382632 -id:A382632 - cp's OEIS frontend

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

Daniel Mondot, Mar 31 2025

nonn

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.

			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.

Daniel Mondot, Table of n, a(n) for n = 1..10000
Allan Gottlieb, Puzzle Corner, Technology Review, December 2, 2003.

Cf. A380868, A382268, A380867, A382632.

cp's OEIS Frontend

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

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