A380867 Numbers k such that one can make a rectangle from a chain of linked rods of length 1, 2, 3, ..., k, with perimeter equal to the total length.

8, 15, 20, 24, 27, 32, 35, 39, 44, 48, 51, 55, 56, 63, 68, 75, 80, 84, 87, 92, 95, 99, 104, 111, 115, 116, 119, 120, 123, 124, 128, 132, 135, 140, 143, 144, 147, 152, 155, 159, 160, 164, 168, 171, 175, 176, 183, 184, 188, 195, 200, 203, 204, 207, 208, 212, 215, 216, 219, 220, 224, 231, 235, 236

Offset: 1

Author

Ali Sada and M. F. Hasler, Mar 14 2025

Keywords

• nonn
• nice

Comments

A more explicit description: Numbers n for which exist n1 > n2 > n3 > n4 >= 0 such that T(n) = 2*(A+B) with A = T(n1) - T(n2) = T(n3) - T(n4) and B = T(n2) - T(n3), where T = A000217.

If in the above solution {n1, ..., n4} we have n4 = 0, this means that the first and last rods (length 1 and n) meet in a corner. This first happens for n = 20 where we can have {0, 11, 14, 18} or {0, 5, 14, 15} with this property (and a third solution without this property).

If in such a solution we have two consecutive integers, e.g., n1 = n2 + 1, this means that one side of the rectangle is made of one single rod, here n1. (This happens in the second solution above with n1 = 15, and in the EXAMPLE n = 8, with n1 = 7.)

Examples

			The smallest such number is a(1) = 8, for which we have (n1..n4) = (2, 4, 6, 7), that is, the rectangle:
    o--+--o--o--+--+--+--+--+--+--+--o
    |  2   1             8           |
    |3                               |
    |                                |
    o                                7
    |                                |
    |4                               |
    |                                |
    |       5               6        |
    o--+--+--+--+--o--+--+--+--+--+--o
This is a special case of the 2-dimensional closed-loop self-avoiding paths on a square lattice considered in A334720.
		

Links

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

Crossrefs

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

Programs

PARI

T(n)=n*(n+1)/2 /* = A000217 */
select( {is_A380867(n)=my(Tn=T(n), T1, T2, T3, T4, n3, n4); 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 && return(n), E,)))}, [1..99])