OFFSET
1,1
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
LINKS
Daniel Mondot, Table of n, a(n) for n = 1..3052
EXAMPLE
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.
PROG
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Ali Sada and Daniel Mondot, Mar 19 2025
STATUS
approved
