A332251 a(n) is the real part of f(n) defined by f(0) = 0 and f(n+1) = f(n) + i^A000120(n) (where i denotes the imaginary unit). Sequence A332252 gives imaginary parts.
0, 1, 1, 1, 0, 0, -1, -2, -2, -2, -3, -4, -4, -5, -5, -5, -4, -4, -5, -6, -6, -7, -7, -7, -6, -7, -7, -7, -6, -6, -5, -4, -4, -4, -5, -6, -6, -7, -7, -7, -6, -7, -7, -7, -6, -6, -5, -4, -4, -5, -5, -5, -4, -4, -3, -2, -2, -2, -1, 0, 0, 1, 1, 1, 0, 0, -1, -2
Offset: 0
Examples
The first terms, alongside f(n) and A000120(n), are: n a(n) f(n) A000120(n) -- ---- ------ ---------- 0 0 0 0 1 1 1 1 2 1 1+i 1 3 1 1+2*i 2 4 0 2*i 1 5 0 3*i 2 6 -1 -1+3*i 2 7 -2 -2+3*i 3 8 -2 -2+2*i 1 9 -2 -2+3*i 2 10 -3 -3+3*i 2 11 -4 -4+3*i 3 12 -4 -4+2*i 2 13 -5 -5+2*i 3 14 -5 -5+i 3 15 -5 -5 4 16 -4 -4 1 From _Kevin Ryde_, Sep 24 2020: (Start) n = 2^9 + 2^8 + 2^5 + 2^2 + 2^1 = 806 f(n) = 1*b^9 + i*b^8 + i^2*b^5 + i^3*b^2 + i^4*b^1 = 23 + 37*i so a(806) = 23 and A332252(806) = 37. (End)
Links
- Rémy Sigrist, Table of n, a(n) for n = 0..8192
- Michael Beeler, R. William Gosper, and Richard Schroeppel, HAKMEM, MIT Artificial Intelligence Laboratory report AIM-239, February 1972. Item 135, The "C" Curve, by Gosper, page 65. Also HTML transcription.
- Robert Ferréol (MathCurve), Courbe du C (ou courbe de Lévy) [in French]
- Paul Lévy, Les courbes planes ou gauches et les surfaces composées de parties semblables au tout, Journal de l'École Polytechnique, July 1938 pages 227-247, and continued October 1938 pages 249-292.
- Kevin Ryde, Iterations of the Lévy C Curve, section Coordinates.
- Rémy Sigrist, Colored representation of f(n) in the complex plane for n = 0..2^20 (where the hue is function of n)
- Rémy Sigrist, Representation of f(n) for n=0..32 in the complex plan
- Wikipedia, Lévy C Curve
- Index entries for sequences related to coordinates of 2D curves
Crossrefs
Programs
-
PARI
{ z=0; for (n=0, 67, print1 (real(z) ", "); z += I^hammingweight(n)) } -
PARI
a(n) = my(v=binary(n),s=1); for(i=2,#v, if(v[i],v[i]=(s*=I))); real(subst(Pol(v),'x,1+I)); \\ Kevin Ryde, Sep 24 2020
Formula
For any k >= 0:
- a(2^(4*k)) = (-4)^k,
- a(2^(4*k+1)) = (-4)^k,
- a(2^(4*k+2)) = 0,
- a(2^(4*k+3)) = -2*(-4)^k.
From Kevin Ryde, Sep 24 2020: (Start)
With complex b = 1+i,
f(2*n) = b*f(n) and f(2*n+1) = f(2*n) + i^A000120(2*n), expand and step.
f(2^k + r) = b^k + i*f(r), for 0 <= r < 2^k, replication.
f(n) = Sum_{j=0..t} i^j*b^k[j] where binary n = 2^k[0] + ... + 2^k[t] with descending powers k[0] > ... > k[t] >= 0, so change binary to base b with rotating coefficient i^0, i^1, i^2, ... at each 1-bit.
(End)
Comments