A380868 Number of distinct solutions {n1, n2, n3, n4} to the problem of forming a rectangle with sides made of linked rods of length 1, ..., n.

0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 3, 0, 0, 0, 1, 0, 0, 3, 0, 0, 0, 0, 1, 0, 0, 6, 0, 0, 0, 6, 0, 0, 0, 0, 6, 0, 0, 0, 1, 0, 0, 3, 0, 0, 0, 3, 3, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 6, 0, 0, 0, 0, 6, 0, 0, 0, 10, 0, 0, 3, 0, 0, 0, 0, 3, 0, 0, 1, 0, 0, 0, 15

Offset: 1

Author

M. F. Hasler, Mar 14 2025

Keywords

• nonn

Comments

A solution to the problem is given by n1 > n2 > n3 > n4 >= 0 such that T(n) = 2*(A+B) with side lengths A = T(n1) - T(n2) = T(n3) - T(n4) and B = T(n2) - T(n3), where T = A000217.

When n4 = 0, this means that the rods of length 1 and n meet in a corner. When there are two consecutive n_k's (e.g. n1 = 7, n2 = 6 for n = 8) this means that one side is made of one single rod, of length equal to the larger of the two n_k's

Nonzero terms are at indices listed in A380867.

From Daniel Mondot, Mar 17 2025: (Start)

Conjecture 1: All terms are triangular numbers.

Conjecture 2: All triangular numbers will eventually appear in this sequence.(End)

Examples

			The only solution for n = 8 is depicted in A380867.
a(n = 20) = 3 is the first case where we have several distinct solutions,
 {5, 11, 15, 18}, {0, 11, 14, 18} and {0, 5, 14, 15}, The corresponding triangular numbers, whose differences give the side lengths, are {15, 66, 120, 171}, {0, 66, 105, 171} and {0, 15, 105, 120}.
		

Links

• Daniel Mondot, Table of n, a(n) for n = 1..10000
• Daniel Mondot, in reply to Ali Sada, Re: A chain problem, SeqFan list, Mar 13, 2025
• Daniel Mondot, First instances of a new number of solutions

Crossrefs

Cf. A000217 (triangular numbers), A334720 (2D cycles on square lattice).

Cf. A380867 (indices of nonzero terms).

Programs

PARI

T(n)=n*(n+1)/2 \\ = A000217
apply( {A380868(n)=my(Tn=T(n), T1, T2, T3, T4, n3, n4, cnt=0); Tn%2==0 && forstep(n1=n-1, 3, -1, T1=T(n1); forstep(n2=n1-1, 2, -1, (B = Tn/2 - A = T1 - T2 = T(n2)) < 3 && break; iferr((1+n3=sqrtint(2*T3 = T2-B))*n3==2*T3 && (1+n4=sqrtint(2*T4 = T3-A))*n4==2*T4 && cnt++, E, ))); cnt}, [1..99])