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A194683
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Triangular array: T(n,k)=[<r^n>+<r^k>], where [ ] = floor, < > = fractional part, and r=(1+sqrt(3))/2.
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5
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0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0
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OFFSET
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1
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COMMENTS
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n-th row sum gives number of k in [0,1] for which <r^n>+<r^k> > 1; seeA194684.
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LINKS
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EXAMPLE
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First ten rows:
0
1 1
0 1 1
0 1 1 0
1 1 1 1 1
0 1 1 0 1 0
1 1 1 1 1 1 1
0 0 0 0 0 0 1 0
0 1 1 1 1 1 1 0 1
0 1 1 1 1 1 1 0 1 1
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MATHEMATICA
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r = 1/2 + Sqrt[3]/2; z = 15;
p[x_]:= FractionalPart[x]; f[x_]:= Floor[x];
w[n_, k_]:= p[r^n] + p[r^k] - p[r^n + r^k]
Flatten[Table[w[n, k], {n, 1, z}, {k, 1, n}]]
TableForm[Table[w[n, k], {n, 1, z}, {k, 1, n}]]
s[n_]:= Sum[w[n, k], {k, 1, n}]
Table[s[n], {n, 1, 100}] (*A194684*)
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}] (*A194686*)
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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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