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A194661
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Triangular array: T(n,k)=[<n*r>+<k*r>], where [ ] = floor, < > = fractional part, and r = (1+sqrt(5))/2 (the golden ratio).
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2
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1, 0, 0, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 1, 0, 1
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OFFSET
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1
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COMMENTS
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n-th row sum gives the number of k in [0,n] for which <n*r>+<k*r> > 1; see A194662.
Triangles of this sort and their row sums are sampled by the following sequences:
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LINKS
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EXAMPLE
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First 13 rows:
1
0 0
1 1 1
1 0 1 0
0 0 0 0 0
1 0 1 1 0 1
0 0 1 0 0 1 0
1 1 1 1 1 1 1 1
1 0 1 1 0 1 0 1 1
0 0 1 0 0 0 0 1 0 0
1 1 1 1 0 1 1 1 1 0 1
1 0 1 0 0 1 0 1 0 0 1 0
0 0 0 0 0 0 0 0 0 0 0 0 0
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MATHEMATICA
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r = GoldenRatio; z = 14;
p[x_] := FractionalPart[x]; f[x_] := Floor[x];
h[n_, k_] := f[p[n*r] + p[k*r]]
Flatten[Table[h[n, k], {n, 1, z}, {k, 1, n}]]
TableForm[Table[h[n, k], {n, 1, z}, {k, 1, n}]]
t[n_] := Sum[h[n, k], {k, 1, n}]
Table[t[n], {n, 1, 100}] (* A194662 *)
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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