%I #17 Jun 18 2020 13:35:43

%S 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,

%T 12,9,18,15,15,12,22,18,18,15,26,22,22,18,30,26,26,22,35,30,30,26,40,

%U 35,35,30,45,40,40,35,51,45,45,40,57,51,51,45,63,57

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

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Integer_triangle">Integer Triangle</a>

%F 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).

%F Conjectures from _Colin Barker_, May 11 2019: (Start)

%F 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)).

%F 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.

%F (End)

%F Conjectures from _Marc Bofill Janer_, May 15 2019: (Start)

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

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

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

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

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

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

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

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

%F (End)

%t 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}]

%o (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

%Y Cf. A308065, A001840, A130518, A062781.

%K nonn

%O 1,7

%A _Wesley Ivan Hurt_, May 10 2019