A382268 Numbers k such that a right triangle can be formed from a chain of linked rods of lengths 1, 2, 3, ..., k, with the perimeter equal to the total length.

15, 20, 24, 35, 39, 44, 48, 55, 56, 63, 75, 76, 80, 84, 91, 95, 99, 104, 111, 119, 120, 132, 135, 140, 143, 144, 152, 155, 168, 175, 176, 187, 188, 195, 203, 207, 215, 216, 219, 224, 252, 259, 260, 264, 272, 275, 279, 287, 288, 296, 299, 308, 315, 320, 324, 335, 351, 360

Offset: 1

Author

Ali Sada and Daniel Mondot, Mar 19 2025

Keywords

• nonn

Comments

The corresponding perimeters T(a(n)) must be in the intersection of T = A000217 (triangular numbers) with A010814 (perimeters of integer sided right triangles). A number k is in the sequence if there exists a solution {k1, k2, k3} with k > k1 > k2 > k3 >= 0 such that for a < b < c in { T(k1) - T(k2), T(k2) - T(k3), T(k) - T(k1) + T(k3) } one has a^2 + b^2 = c^2. - M. F. Hasler, Mar 20 2025

Examples

			The first triangle is when k = 15. The segments are [6+7+8+9+10] [11+12+13+14] [15+1+2+3+4+5]. The sums of the segments are (40, 50, 30), which is 10 times the primitive right triangle (3, 4, 5).
The second term, k = 20, corresponds to 5 distinct solutions:
  S1 = {18, 16, 9}: a = 9+...+1 + 20+19 = 84, b = 18+17 = 35, c = 16+...+10 = 91,
  S2 = {17, 11, 3}: a = 20+19+18 + 3+2+1 = 63, c = 17+...+12 = 87, b = 11+...+4 = 60,
  S3 = {17, 11, 2}: a = 20+19+18 + 2+1 = 60, c = 17+...+12 = 87, b = 11+...+3 = 63,
  S4 = {16, 9, 4}: a = 20+...+17 + 4+...+1 = 84, c = 16+...+10 = 91, b = 9+...+5 = 35,
  S5 = {15, 8, 1}: c = 20+...+16 + 1 = 91, a = 15+...+9 = 84, b = 8+...+2 = 35.
We note that S2 and S3, and S1, S4 and S5, have the same side lengths, but different decompositions.
		

Links

• Daniel Mondot, Table of n, a(n) for n = 1..3052

Crossrefs

Cf. A000217, A010814, A380867, A380868, A380875.

Programs

PARI

select( {is_A382268(n)=my(Tn=n*(n+1)\2,T1,T2,S); Tn%2==0 && is_A005279(Tn\2) && forstep(n1=n-1,sqrtint(Tn-1)+1,-1, T1=n1*(n1+1)\2; forstep(n2=n1-1,sqrtint(2*T1-Tn-1)+1,-1, T2=n2*(n2+1)\2; forstep(n3=n2-1,0,-1, #(S=Set([Tn-T1+S=n3*(n3+1)\2,T2-S,T1-T2]))>2 && S[3]^2 == S[1]^2+S[2]^2 && return(S))))}, [1..100])\\ M. F. Hasler, Mar 22 2025