A318439 -id:A318439 - cp's OEIS frontend

A318438 For any n >= 0 with binary expansion Sum_{k=0..w} b_k * 2^k, let h(n) = Sum_{k=0..w} b_k * (i-1)^k (where i denotes the imaginary unit); a(n) is the real part of h(n).

0, 1, -1, 0, 0, 1, -1, 0, 2, 3, 1, 2, 2, 3, 1, 2, -4, -3, -5, -4, -4, -3, -5, -4, -2, -1, -3, -2, -2, -1, -3, -2, 4, 5, 3, 4, 4, 5, 3, 4, 6, 7, 5, 6, 6, 7, 5, 6, 0, 1, -1, 0, 0, 1, -1, 0, 2, 3, 1, 2, 2, 3, 1, 2, 0, 1, -1, 0, 0, 1, -1, 0, 2, 3, 1, 2, 2, 3, 1, 2

Offset: 0

Rémy Sigrist, Aug 26 2018

See A318439 for the imaginary part of h.
See A318479 for the square of the modulus of h.
The function h corresponds to the interpretation of the binary representation of a number in base -1+i and defines a bijection from the nonnegative integers to the Gaussian integers.
The function h has nice fractal features (see scatterplot in Links section).
This sequence has similarities with A316657.

Rémy Sigrist, Table of n, a(n) for n = 0..10000
Rémy Sigrist, Colored scatterplot of (a(n), A318439(n)) for n = 0..2^20-1 (where the hue is function of n)
Wikipedia, Base -1+/-i
Index entries for sequences related to binary expansion of n

Cf. A009116, A318439 (imaginary part), A318479 (norm), A340669 (negation).

Cf. A316657 (base 2+i).

a(n) = my (d=Vecrev(digits(n,2))); real(sum(i=1, #d, d[i]*(I-1)^(i-1)))

a(2^k) = A009116(k) for any k >= 0.

A318479 For any n >= 0 with binary expansion Sum_{k=0..w} b_k * 2^k, let h(n) = Sum_{k=0..w} b_k * (i-1)^k (where i denotes the imaginary unit); a(n) is the square of the modulus of h(n).

Original entry on oeis.org

0, 1, 2, 1, 4, 5, 2, 1, 8, 13, 10, 13, 4, 9, 2, 5, 16, 9, 26, 17, 20, 13, 26, 17, 8, 5, 18, 13, 4, 1, 10, 5, 32, 41, 18, 25, 52, 61, 34, 41, 40, 53, 26, 37, 52, 65, 34, 45, 16, 17, 10, 9, 36, 37, 26, 25, 8, 13, 2, 5, 20, 25, 10, 13, 64, 65, 82, 81, 36, 37, 50

Offset: 0

Rémy Sigrist, Aug 27 2018

See A318438 for the real part of h and additional comments.

Rémy Sigrist, Table of n, a(n) for n = 0..10000
Index entries for sequences related to binary expansion of n

Cf. A318438, A318439.

a(n) = my (d=Vecrev(digits(n, 2))); norm(sum(i=1, #d, d[i]*(I-1)^(i-1)))

a(n) = A318438(n)^2 + A318439(n)^2.
a(2^k) = 2^k for any k >= 0.
a(3 * 2^k) = 2^k for any l >= 0.

A340669 Permutation of the nonnegative integers formed by negation in complex base i-1.

Original entry on oeis.org

0, 29, 58, 7, 116, 25, 14, 3, 232, 21, 50, 239, 28, 17, 6, 235, 464, 13, 42, 471, 100, 9, 478, 467, 56, 5, 34, 63, 12, 1, 470, 59, 928, 957, 26, 935, 84, 953, 942, 931, 200, 949, 18, 207, 956, 945, 934, 203, 112, 941, 10, 119, 68, 937, 126, 115, 24, 933, 2, 31

Offset: 0

Kevin Ryde, Jan 15 2021

Complex base i-1 of Khmelnik and Penney uses an integer n>=0 to represent a complex integer z(n) = A318438(n) + A318439(n)*i. a(n) is the negation of z in this representation, so that z(a(n)) = -z(n). Every z is uniquely represented, so this is a self-inverse permutation.

Khmelnik's table 4 is carries applied to z which become states and transitions by bits of n and certain 0<->1 bit flips in n. The result is the transformation in the formulas below. Bit flips may extend into 0-bits above the most significant bit of n causing the bit length of a(n) to be greater than the bit length of n.

			For n=1506, location z(1506) = 11-35*i.  Its negation is -(11-35*i) = z(29914) so a(1506) = 29914.  And being self-inverse conversely a(29914) = 1506.
In terms of bit flips, in the following "^^" is each 01 or 11 and F marks the bits flipped above them.
  n    =  1506 = binary  00001 0 1 11 100 01 0
                         FFF^^   F ^^ FFF ^^
  a(n) = 29914 = binary  11101 0 0 11 011 01 0

Kevin Ryde, Table of n, a(n) for n = 0..8192
Joerg Arndt, The fxt demos: bit wizardry. radix(-1+i)
Solomon I. Khmelnik, Specialized Digital Computer for Operations with Complex Numbers (in Russian), Questions of Radio Electronics, volume 12, number 2, 1964.
Kevin Ryde, Iterations of the Dragon Curve, see index MinusNeg.
Andrey Zabolotskiy, English translation of theorems from Khmelnik, Seqfan mailing list, September 2016.
Andrey Zabolotskiy, Python Code by bit flips or conversion and Python Code by Khmelnik's Pi and Beta, September 2016.

Cf. A318438, A318439, A340670.

{ a(n) = for(i=0,if(n,logint(n,2)),
  if(bittest(n,i),
    if(bittest(n,i+1), n=bitxor(n,4<

a(n) is formed by transforming n as follows. Write n in binary with four high 0-bits and consider bits from least to most significant. At a 01 pair (high 0, low 1), apply an 0<->1 flip to three bits immediately above this pair. At a 11 pair, flip one bit immediately above this pair. Repeat, each time seeking the next higher 01 or 11 pair above the bits just flipped.

A340670 Number of complex base i-1 points which can be represented within n bits and negated within those n bits.

Original entry on oeis.org

1, 1, 1, 3, 5, 15, 29, 47, 101, 199, 413, 847, 1621, 3255, 6541, 13087, 26373, 52423, 104637, 209711, 419253, 839511, 1678317, 3353919, 6710629, 13421287, 26845213, 53693007, 107366933, 214742391, 429498701, 858994271, 1718023109, 3435955975, 6871883645

Offset: 0

Kevin Ryde, Jan 15 2021

Complex base i-1 of Khmelnik and Penney uses an integer m to represent a complex integer point z(m) = A318438(m) + A318439(m)*i.  A340669(m) is the negation of z in this representation.  a(n) is how many n-bit m are negatable within those n bits, i.e., how many m in the range 0 <= m < 2^n have also 0 <= A340669(m) < 2^n.
A geometric interpretation of a(n) is to draw a unit square around each point z(0) to z(2^n-1), rotate a copy by 180 degrees about the origin, and measure its area of intersection with the original.
The bit-flip rule in A340669 gives the recurrence formula below.  A low 0-bit of m has a(n-1) negatables above it, or low 11 is one arbitrary bit then negatables above so 2*a(n-3), or low 01 is three arbitrary so 8*a(n-5).  This can be thought of the number of compositions of n (partitions with order) into parts 1,3,5 with 2 types of part 3 and 8 types of part 5.

			For n=3, the a(3)=3 points of n bits are m = 0,3,7 < 2^n, which negate to A340669(0,3,7) = 0,7,3 < 2^n.  These m are located at z = 0,i,-i,
               negate        intersection
  z(0..7)    (rotate 180)   a(3) = 3 points
                * *
   * *            * *             *
     o *        * o               o
   * *            * *             *
     * *

Kevin Ryde, Table of n, a(n) for n = 0..700
Kevin Ryde, Iterations of the Dragon Curve, see index MinusNegA.
Index entries for linear recurrences with constant coefficients, signature (1,0,2,0,8).

Cf. A340669, A107920, A078069.

{ my(table=[4,-2,-2,6, -4,2,2,-6], p=Mod('x,2-'x+'x^2));
a(n) = (6<

a(n) = a(n-1) + 2*a(n-3) + 8*a(n-5).
a(n) = (2/5)*2^n + (h/15)*2^floor(n/2) + (2/3)*Im((1/2 + i*sqrt(7)/2)^(n+1)) where h = 4,-2,-2,6, -4,2,2,-6 according as n == 0 to 7 (mod 8) respectively.
a(n) = (2/5)*2^n + (1/3)*A107920(n+1) + (1/15)*A078069(n+2).
G.f.: 1/(1 - x - 2*x^3 - 8*x^5).
G.f.: (2/5)/(1-2*x) + (1/3)/(1-x+2*x^2) + (2/15)*(2+3*x)/(1+2*x+2*x^2).

A340566 Square array, read by descending antidiagonals; T(n,k) is A001057(n) + A001057(k)*i, converted to complex binary (base -1 + i), where i=sqrt(-1).

Original entry on oeis.org

0, 11, 1, 111, 1110, 11101, 1110100, 111010, 10, 1100, 100, 1110101, 110, 1111, 11100, 1110111, 101, 11001, 111011, 11111, 1101, 110011, 1010, 11101001, 1000, 11101011, 111010010, 10001, 1110000, 111110, 1110110, 111000, 11000, 111010110, 11110, 111010000

Offset: 0

Davis Smith, Jan 11 2021

Complex binary (base -1 + i) has the ability to express all positive or negative, real or complex, integers with only 2 numerical symbols ('0' and '1') as integers, without the need for a sign marking the integers as such.
Converting a real number, n, to complex binary requires one to convert it to base -4 ((n + N) xor N, N = floor(4/5*16^(ceiling(log_4(abs(n))) + 1))), then adding 10 to every digit greater than 1, then treating it as a number in base 16 and converting that to binary. (E.g., -5 => [2,3] => [12,13] => 205 => 11001101.)
Converting a complex number, n + k*i, requires one to convert X = n + k into complex binary and then convert k into the same but shift it one digit to the left. After this, one must add them together. This functions much the same way as binary addition, but the carry is '110' rather than '1' and 11 + 111 = 0.

			Square array T(n,k) begins:
  \k      0         1         2          3        4       5        6 ...
  n\
   0|     0        11       111    1110100      100 1110111   110011 ...
   1|     1      1110    111010    1110101      101    1010   111110 ...
   2| 11101        10       110      11001 11101001 1110110   110010 ...
   3|  1100      1111    111011       1000   111000    1011   111111 ...
   4| 11100     11111  11101011      11000 11101000   11011 11101111 ...
   5|  1101 111010010 111010110       1001   111001 1100110   100010 ...
   6| 10001     11110  11101010 1110100101    10101   11010 11101110 ...

T. Jamil, Complex Binary Number System, Springer, 2013.

Cf. A001057, A318438, A318439.

A340566(n,k)={my(A001057(x)=if(x%2,x\2+1,-x/2),V=vecsum(Vec(matconcat(apply(w->my(Y=if(w,A001057(k), A001057(n)+A001057(k)));if(Y, my(X=floor(4^(2*logint(abs(Y), 4)+5)/5));Vecrev(binary(shift(fromdigits(apply(z->z+(10*(z>1)), digits(bitxor(Y+X,X),4)),16),w)))),[0,1])~)~))~);
while(vecmax(V)>1,my(Z=Vec(select(x->x>1,V,1)));for(x=1,#Z,my(z=Z[x]);if(V[z]<=1,,(z+2<=#V)&&(V[z+1]>1)&&V[z+2],for(j=z,z+2,V[j]-=2^(j!=(z+2))),(z+4<=#V)&&vecmin(V[z+2..z+4]),V[z]-=2;for(j=z+2,z+4,V[j]-=1),z+1>#V,V[z]-=2;V=concat(V,[0,1,1]),V[z]-=2;for(j=z+2,z+3,if(j<=#V,V[j]+=1,V=concat(V,1))))));fromdigits(Vecrev(V))}

{ T(n,k) = my(z=n\/2*-(-1)^n + k\/2*-(-1)^k*I, ret=List([]));
  while(z, my(bit=(real(z)+imag(z))%2);
    listput(ret,bit); z=(z-bit)/(I-1));
  fromdigits(Vecrev(ret)); } \\ Kevin Ryde, Jan 12 2021

A355431 Numbers k whose binary expansion, when interpreted in base -1+i, gives a Gaussian prime.

Original entry on oeis.org

2, 5, 6, 9, 11, 13, 14, 15, 17, 19, 21, 23, 25, 27, 31, 33, 37, 39, 41, 43, 49, 51, 53, 57, 58, 59, 63, 69, 71, 73, 77, 81, 83, 89, 97, 99, 101, 111, 113, 117, 119, 123, 127, 129, 131, 133, 137, 139, 141, 147, 159, 163, 169, 177, 183, 191, 193, 197, 201, 207

Offset: 1

John-Vincent Saddic, Jul 17 2022

Complex base -1+i is a bijection between integers k and Gaussian integers z(k) = A318438(k) + A318439(k)*i.
The present sequence is those k where z(k) is a Gaussian prime.
The Gaussian primes have an 8-way symmetry in the complex plane so that this sequence is also the Gaussian primes in the conjugate complex base -1-i.
The graphs on the complex plane (see links) show the Gaussian primes mapped and connected by lines in the order in which their indices appear in {a(n)}. The numbers in base -1+i tile the complex plane in the twin dragon fractal pattern, and the Gaussian primes are numerous such that the fractal is still discernible.
The only even terms are 2, 6, 14, and 58, since even terms correspond to Gaussian integers divisible by -1+i, and the base-(-1+i) expansions of -1+i, -1-i, 1+i, and 1-i are 10, 110, 1110, and 111010 respectively. - Jianing Song, Oct 02 2022

			123 is a term since z(123) = 2+7i is a Gaussian prime.
124 is not a term because z(124) = 2+4i is not a Gaussian prime.

John-Vincent Saddic, Graphs on the complex plane
John-Vincent Saddic, Julia program
John-Vincent Saddic, Python program

Cf. A066321 (real integers in base -1+i).

Julia
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cp's OEIS Frontend

A318438 For any n >= 0 with binary expansion Sum_{k=0..w} b_k * 2^k, let h(n) = Sum_{k=0..w} b_k * (i-1)^k (where i denotes the imaginary unit); a(n) is the real part of h(n).

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A318479 For any n >= 0 with binary expansion Sum_{k=0..w} b_k * 2^k, let h(n) = Sum_{k=0..w} b_k * (i-1)^k (where i denotes the imaginary unit); a(n) is the square of the modulus of h(n).

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A340669 Permutation of the nonnegative integers formed by negation in complex base i-1.

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A340670 Number of complex base i-1 points which can be represented within n bits and negated within those n bits.

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A340566 Square array, read by descending antidiagonals; T(n,k) is A001057(n) + A001057(k)*i, converted to complex binary (base -1 + i), where i=sqrt(-1).

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A355431 Numbers k whose binary expansion, when interpreted in base -1+i, gives a Gaussian prime.

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