A340669 -id:A340669 - 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.

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).

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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