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

Table of n, a(n) for n=1..72.

Wikipedia, Integer Triangle

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)

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. A308065, A001840, A130518, A062781.

Sequence in context: A336362 A336363 A308451 * A124748 A161225 A174980

Adjacent sequences: A308064 A308065 A308066 * A308068 A308069 A308070

Keyword

nonn

Author

Wesley Ivan Hurt, May 10 2019

Status

approved