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

Author

Wesley Ivan Hurt, May 10 2019

Keywords

• nonn

Links

• Wikipedia, Integer Triangle

Crossrefs

Cf. A308065, A001840, A130518, A062781.

Programs

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}]

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

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)

Conjectures from Marc Bofill Janer, May 15 2019: (Start)

a(4*n) = a(4*n+1).

a(4*n) < a(4*n-1).

a(4*n) = A001840(n-1) = A130518(n+1) = A062781(n+2).

a(4*n-1) = a(4*n+4) = a(4*n+5).

a(4*n-1) = A001840(n) = A130518(n+2) = A062781(n+3).

a(4*n+2) = a(4*n-4) = a(4*n-3).

a(4*n+2) = A001840(n-2) for n>=2.

a(4*n+2) = A130518(n) = A062781(n+1).

(End)