A238158 Bicycle lock numbers: triangle T(n,k) with 1<=k<=n, where T(n,k) is the maximum value of min{xy, (n-x)(k-y)} over 0 <= x <= n, 0 <= y <= k for integers x, y.
0, 0, 1, 0, 1, 2, 0, 2, 2, 4, 0, 2, 3, 4, 6, 0, 3, 4, 6, 6, 9, 0, 3, 4, 6, 8, 9, 12, 0, 4, 5, 8, 9, 12, 12, 16, 0, 4, 6, 8, 10, 12, 15, 16, 20, 0, 5, 6, 10, 12, 15, 16, 20, 20, 25, 0, 5, 7, 10, 12, 15, 18, 20, 24, 25, 30, 0, 6, 8, 12, 14, 18, 20, 24, 25, 30
Offset: 1
Examples
For n=5, k=4, the maximum value is attained at x=2, y=2, so T(5, 4) = 2*2 = 4. The first few rows of the triangle are: 0 0 1 0 1 2 0 2 2 4 0 2 3 4 6 0 3 4 6 6 9 0 3 4 6 8 9 12 0 4 5 8 9 12 12 16 0 4 6 8 10 12 15 16 20 0 5 6 10 12 15 16 20 20 25 0 5 7 10 12 15 18 20 24 25 30 0 6 8 12 14 18 20 24 25 30 30 36 0 6 8 12 15 18 21 24 28 30 35 36 42
Links
- Robin Houston, Symmetry of bicycle lock numbers.
Programs
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Mathematica
t[a_, b_] := Max[Table[Min[x*y, (a - x)*(b - y)], {x, 0, a}, {y, 0, b}]]
Formula
T(n,k) = max { min{xy, (n-x)(k-y)} | 0<=x<=n, 0<=y<=k; x, y integers }.
Comments