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 A066321 Binary representation of base-(i-1) expansion of n: replace i-1 with 2 in base-(i-1) expansion of n. 10
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 COMMENTS 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. From Andrey Zabolotskiy, Feb 06 2017: (Start) (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 REFERENCES 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.) LINKS Paul Tek, Table of n, a(n) for n = 0..10000 Joerg Arndt, The fxt demos: bit wizardry, radix(-1+i), C++ radix-m1pi.h function bin_real_to_radm1pi(). Solomon I. Khmelnik, Specialized Digital Computer for Operations with Complex Numbers, Questions of Radio Electronics, 12 (1964), 60-82 [in Russian]. W. J. Penney, A "binary" system for complex numbers, JACM 12 (1965), 247-248. N. J. A. Sloane, Table of n, (I-1)^n for n = 0..100 Paul Tek, Perl program for this sequence Andrey Zabolotskiy, negation & addition of Gaussian integers written in base i-1 [Python script]. FORMULA 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 EXAMPLE 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. MAPLE 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: seq(f(i), i=0..100); # Robert Israel, Oct 21 2016 PROG (Perl) See Links section. (Python) from gmpy2 import c_divmod u = ('0000', '1000', '0011', '1011') def A066321(n):     if n == 0:         return 0     else:         s, q = '', n         while q:             q, r = c_divmod(q, -4)             s += u[r]         return int(s[::-1], 2) # Chai Wah Wu, Apr 09 2016 (PARI) a(n) = my(ret=0, p=0); while(n, ret+=[0, 1, 12, 13][n%4+1]<

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Last modified January 21 15:16 EST 2021. Contains 340352 sequences. (Running on oeis4.)