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A066323
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Number of one bits in binary representation of base i-1 expansion of n (where i = sqrt(-1)).
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2
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0, 1, 2, 3, 4, 5, 6, 7, 3, 4, 5, 6, 2, 3, 4, 5, 1, 2, 3, 4, 5, 6, 7, 8, 4, 5, 6, 7, 3, 4, 5, 6, 2, 3, 4, 5, 6, 7, 8, 9, 5, 6, 7, 8, 4, 5, 6, 7, 3, 4, 5, 6, 7, 8, 9, 10, 6, 7, 8, 9, 5, 6, 7, 8, 4, 5, 6, 7, 8, 9, 10, 11, 7, 8, 9, 10, 6, 7, 8, 9, 5, 6, 7, 8, 9, 10, 11, 12, 8, 9, 10, 11, 7, 8, 9, 10, 6, 7, 8, 9
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OFFSET
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0,3
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COMMENTS
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First differences are usually +1, occasionally -4 (because in base i-1 [3]+[7]=(+i)+(-i)=0) hence often a(i+j)=a(i)+a(j). Differences terms given here are period-16, but for full sequence is actually period-256 at least.
a(n) is the sum of the digits of n when written in base -4 using digits 0 to 3 (A007608). This is since in Penney's digit substitution for A066321, the base -4 digits 0 to 3 become bit strings of exactly 0 to 3 many 1-bits each respectively. - Kevin Ryde, Sep 09 2019
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REFERENCES
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D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, 1969, Vol. 2, p. 172, (Also exercise 16, p. 177, answer, p. 494)
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LINKS
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Table of n, a(n) for n=0..99.
Walter Penney, A Binary System for Complex Numbers, NSA Technical Journal, 1965.
Walter Penney, A Binary System for Complex Numbers, Journal of the Association for Computing Machinery (JACM), volume 12, number 2, April 1965, pages 247-248.
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FORMULA
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a(n) = A000120(A066321(n)) = A007953(A007608(n)). - Kevin Ryde, Sep 09 2019
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EXAMPLE
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A066321(4) = 464 = 111010000 (binary) so a(4) = 4. Or A007608(4) == 130 in base -4 and sum of digits is a(4) = 1+3+0 = 4.
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PROG
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(PARI) a(n) = my(ret=0); while(n, ret+=n%4; n\=-4); ret; \\ Kevin Ryde, Sep 09 2019
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CROSSREFS
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Cf. A066321, A000120, A007608.
Sequence in context: A104415 A071074 A279648 * A245347 A278059 A329606
Adjacent sequences: A066320 A066321 A066322 * A066324 A066325 A066326
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KEYWORD
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base,easy,nonn
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AUTHOR
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Marc LeBrun, Dec 14 2001
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STATUS
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approved
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