A308067 Number of integer-sided triangles with perimeter n whose longest side length is odd.

0, 0, 1, 0, 0, 0, 2, 1, 1, 0, 3, 2, 2, 1, 5, 3, 3, 2, 7, 5, 5, 3, 9, 7, 7, 5, 12, 9, 9, 7, 15, 12, 12, 9, 18, 15, 15, 12, 22, 18, 18, 15, 26, 22, 22, 18, 30, 26, 26, 22, 35, 30, 30, 26, 40, 35, 35, 30, 45, 40, 40, 35, 51, 45, 45, 40, 57, 51, 51, 45, 63, 57

Offset

1,7

Links

Andrew Howroyd, Table of n, a(n) for n = 1..1000

Wikipedia, Integer triangle.

Index entries for linear recurrences with constant coefficients, signature (1,-1,1,1,-1,2,-2,1,-1,-1,1,-1,1).

Formula

a(n) = Sum_{k=1..floor(n/3)} Sum_{i=k..floor((n-k)/2)} sign(floor((i+k)/(n-i-k+1))) * ((n-i-k) mod 2).

Conjectures from Colin Barker, May 11 2019: (Start)

G.f.: x^3*(1 - x + x^2 - x^3 + x^4) / ((1 - x)^3*(1 + x)^2*(1 - x + x^2)*(1 + x^2)^2*(1 + x + x^2)).

a(n) = a(n-1) - a(n-2) + a(n-3) + a(n-4) - a(n-5) + 2*a(n-6) - 2*a(n-7) + a(n-8) - a(n-9) - a(n-10) + a(n-11) - a(n-12) + a(n-13) for n>13. (End)

Conjecture: a(4*n) = a(4*n+1) = a(4*n+6) = a(4*n-5) = A001840(n-1). - Marc Bofill Janer, May 15 2019

The above conjectures are all correct. - Andrew Howroyd, Nov 06 2025

Mathematica

Table[Sum[Sum[Mod[n - i - k, 2]*Sign[Floor[(i + k)/(n - i - k + 1)]], {i, k, Floor[(n - k)/2]}], {k, Floor[n/3]}], {n, 100}]

Prog

(PARI) a(n) = sum(k=1, n\3, sum(i=k, (n-k)\2, sign((i+k)\(n-i-k+1))*((n-i-k) % 2))); \\ Michel Marcus, May 15 2019

Crossrefs

Cf. A005044, A308066, A001840, A130518, A062781.

Sequence in context: A336362 A336363 A308451 * A124748 A161225 A174980

Adjacent sequences: A308064 A308065 A308066 * A308068 A308069 A308070

Keyword

nonn,easy

Author

Wesley Ivan Hurt, May 10 2019

Status

approved