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A221151
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The generalized Fibonacci word f^[4].
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6
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0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1
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OFFSET
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0
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LINKS
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FORMULA
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Set S_0=0, S_1=0001; thereafter S_n = S_{n-1}S_{n-2}; sequence is S_{oo}.
a(n) = floor((n + 2)/(phi + 3)) - floor((n + 1)/(phi + 3)) where phi = 1/2*(1 + sqrt(5)) denotes the golden ratio.
If we read the present sequence as the digits of a decimal constant c = 0.00010 00010 00100 00100 00100 .... then we have the series representation c = sum {n >= 1} 1/10^floor(n*(phi + 3)). An alternative representation is c = 9*sum {n >= 1} floor(n/(phi + 3)) /10^n.
The constant 9*c has the simple continued fraction representation [0; 1111, 10, 10^4, 10^5, 10^9, ..., 10^A000285(n), ...] (see Adams and Davison).
Using this result we can find the alternating series representation c = 9*sum {n >= 1} (-1)^(n+1)*(1 + 10^A000285(3*n-1))/( (10^A000285(3*n-3) - 1)*(10^A000285(3*n) - 1) ).
The series converges very rapidly: for example, the first 10 terms of the series give a value for c accurate to more than 10 million decimal places. Cf. A005614 and A221150. (End)
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MAPLE
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# fibi and fibonni implemented in A221150.
fibonni(n, 4) ;
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MATHEMATICA
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a[n_] := Floor[(n+2)/(GoldenRatio+3)] - Floor[(n+1)/(GoldenRatio+3)];
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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