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A308067
Number of integer-sided triangles with perimeter n whose longest side length is odd.
0
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
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
KEYWORD
nonn
AUTHOR
Wesley Ivan Hurt, May 10 2019
STATUS
approved