

A066323


Number of one bits in binary representation of base i1 expansion of n (where i = sqrt(1)).


2



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

0,3


COMMENTS

First differences are usually +1, occasionally 4 (because in base i1 [3]+[7]=(+i)+(i)=0) hence often a(i+j)=a(i)+a(j). Differences terms given here are period16, but for full sequence is actually period256 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 1bits each respectively.  Kevin Ryde, Sep 09 2019


REFERENCES

D. E. Knuth, The Art of Computer Programming. AddisonWesley, Reading, MA, 1969, Vol. 2, p. 172, (Also exercise 16, p. 177, answer, p. 494)


LINKS

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 247248.


FORMULA

a(n) = A000120(A066321(n)) = A007953(A007608(n)).  Kevin Ryde, Sep 09 2019


EXAMPLE

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.


PROG

(PARI) a(n) = my(ret=0); while(n, ret+=n%4; n\=4); ret; \\ Kevin Ryde, Sep 09 2019


CROSSREFS

Cf. A066321, A000120, A007608.
Sequence in context: A104415 A071074 A279648 * A245347 A278059 A329606
Adjacent sequences: A066320 A066321 A066322 * A066324 A066325 A066326


KEYWORD

base,easy,nonn


AUTHOR

Marc LeBrun, Dec 14 2001


STATUS

approved



