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A066321
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Binary representation of base-(i-1) expansion of n: replace i-1 with 2 in base-(i-1) expansion of n.
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17
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0, 1, 12, 13, 464, 465, 476, 477, 448, 449, 460, 461, 272, 273, 284, 285, 256, 257, 268, 269, 3280, 3281, 3292, 3293, 3264, 3265, 3276, 3277, 3088, 3089, 3100, 3101, 3072, 3073, 3084, 3085, 3536, 3537, 3548, 3549, 3520, 3521, 3532, 3533, 3344, 3345, 3356
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OFFSET
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0,3
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COMMENTS
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Here i = sqrt(-1).
First differences follow a strange period-16 pattern: 1 11 1 XXX 1 11 1 -29 1 11 1 -189 1 11 1 -29 where XXX is given by A066322. Number of one-bits is A066323.
(Observations.)
Actually, the sequence of the first differences can be split into blocks of size of any power of 2, and there will be only one position in the block that does not repeat. In this sense, one may say that the first differences follow (almost-)period-2^s pattern for any s > 0.
Specifically, the first differences are given by the formula: a(n+1)-a(n) = A282137(A007814((n xor ...110011001100) + 1)). Here binary representation of n is bitwise-xored with the period-4 bit sequence (A021913 written right-to-left) which is infinite or simply long enough; A007814(m) does not depend on the bits of m other than the least significant 1.
A282137 gives all first differences in the order of decreasing occurrence frequency.
(End)
Penney shows that since (i-1)^4 = -4, the representation a(n) of a real integer n is found by writing n in base -4 using digits 0 to 3 (A007608), changing those digits to bit strings 0000, 0001, 1100, 1101 respectively, and interpreting as binary. - Kevin Ryde, Sep 07 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. (See also exercise 16, p. 177; answer, p. 494.)
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LINKS
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FORMULA
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In "rebase notation" a(n) = (i-1)[n]2.
G.f. g(z) satisfies g(z) = z*(1+12*z+13*z^2)/(1-z^4) + 16*z^4*(13+12*z^4+z^8)/((1-z)*(1+z^4)*(1+z^8)) + 256*(1-z^16)*g(z^16)/(z^12-z^13). - Robert Israel, Oct 21 2016
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EXAMPLE
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a(4) = 464 = 2^8 + 2^7 + 2^6 + 2^4 since (i-1)^8 + (i-1)^7 + (i-1)^6 + (i-1)^4 = 4.
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MAPLE
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f:= proc(n) option remember; local t, m;
t:= n mod 4;
procname(t) + 16*procname((t-n)/4)
end proc:
f(0):= 0: f(1):= 1: f(2):= 12: f(3):= 13:
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PROG
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(Perl) See Links section.
(Python)
from gmpy2 import c_divmod
u = ('0000', '1000', '0011', '1011')
if n == 0:
return 0
else:
s, q = '', n
while q:
q, r = c_divmod(q, -4)
s += u[r]
(PARI) a(n) = my(ret=0, p=0); while(n, ret+=[0, 1, 12, 13][n%4+1]<<p; n\=-4; p+=4); ret; \\ Kevin Ryde, Sep 07 2019
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CROSSREFS
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See A271472 for the conversion of these decimal numbers to binary.
See A009116 and A009545 for real and imaginary parts of (i-1)^n (except for signs).
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KEYWORD
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base,easy,nonn
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AUTHOR
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STATUS
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approved
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