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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 (list; graph; refs; listen; history; text; internal format)
 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

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Last modified August 11 17:54 EDT 2024. Contains 375073 sequences. (Running on oeis4.)