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A316658
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For any n >= 0 with base-5 expansion Sum_{k=0..w} d_k * 5^k, let f(n) = Sum_{k=0..w} [d_k > 0] * (2 + i)^k * i^(d_k - 1) (where [] is an Iverson bracket and i denotes the imaginary unit); a(n) equals the imaginary part of f(n).
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6
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0, 0, 1, 0, -1, 1, 1, 2, 1, 0, 2, 2, 3, 2, 1, -1, -1, 0, -1, -2, -2, -2, -1, -2, -3, 4, 4, 5, 4, 3, 5, 5, 6, 5, 4, 6, 6, 7, 6, 5, 3, 3, 4, 3, 2, 2, 2, 3, 2, 1, 3, 3, 4, 3, 2, 4, 4, 5, 4, 3, 5, 5, 6, 5, 4, 2, 2, 3, 2, 1, 1, 1, 2, 1, 0, -4, -4, -3, -4, -5, -3
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OFFSET
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0,8
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COMMENTS
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See A316657 for the real part of f and additional comments.
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LINKS
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FORMULA
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a(5^n) = A099456(n-1) for any n > 0.
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MATHEMATICA
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a[n_] := Module[{d, z}, d = IntegerDigits[n, 5] // Reverse; z = Sum[ If[d[[i]]>0, (2+I)^(i-1)*I^(d[[i]]-1), 0], {i, 1, Length[d]}]; Im[z]];
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PROG
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(PARI) a(n) = my (d=Vecrev(digits(n, 5)), z=sum(i=1, #d, if (d[i], (2+I)^(i-1) * I^(d[i]-1), 0))); imag(z)
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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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