A385736 a(n) is the number of distinct nondegenerate triangles with perimeter n whose side lengths are triangular numbers.

0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 2, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 2, 1, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0, 2, 1, 2, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 2, 1, 0, 0, 1, 0, 1, 1, 0, 1

Offset: 0

Author

Felix Huber, Jul 16 2025

Keywords

• nonn
• easy

Comments

0, 1, 6, 10, 28, 55 are the only triangular numbers <= 10^6 that are not perimeters of triangles whose side lengths are triangular numbers. Conjecture: There are no other triangular numbers that have this property.

Examples

			The a(31) = 2 distinct nondegenerate triangles with perimeter 31 and whose side lengths are triangular numbers are [1, 15, 15] and [6, 10, 15].
		

Links

• Felix Huber, Table of n, a(n) for n = 0..10000
• Felix Huber, Maple program to compute the counted triangles
• Eric Weisstein's World of Mathematics, Triangular Number

Crossrefs

Cf. A000217, A005044, A051516, A070088, A366398, A380875, A385737, A385860, A385872.

Programs

Maple

A385736:=proc(N) # To get the first N + 1 terms.
    local p,x,y,z,i;
    p:=[];
    for z to floor((sqrt(24*N+9)-3)/6) do
        for x from z to floor((sqrt(4*N-3)-1)/2) do
            for y from max(z,floor((sqrt(1+4*(x^2+x-z^2-z))-1)/2)+1) to min(x,floor((sqrt(1+4*(2*N-x^2-x-z^2-z))-1)/2)) do
                p:=[op(p),z*(z+1)/2+y*(y+1)/2+x*(x+1)/2]
            od
        od
    od;
    return seq(numboccur(p,i),i=0..N)
end proc;
A385736(87);

Formula

Trivial upper bound: a(n) <= A005044(n).

a(A385737(n)) >= 1.