A334841 a(0) = 0; for n > 0, a(n) = (number of 1's and 3's in base 4 representation of n) - (number of 0's and 2's in base 4 representation of n).

0, 1, -1, 1, 0, 2, 0, 2, -2, 0, -2, 0, 0, 2, 0, 2, -1, 1, -1, 1, 1, 3, 1, 3, -1, 1, -1, 1, 1, 3, 1, 3, -3, -1, -3, -1, -1, 1, -1, 1, -3, -1, -3, -1, -1, 1, -1, 1, -1, 1, -1, 1, 1, 3, 1, 3, -1, 1, -1, 1, 1, 3, 1, 3, -2, 0, -2, 0, 0, 2, 0, 2, -2, 0, -2, 0, 0, 2, 0, 2, 0, 2, 0, 2, 2, 4, 2, 4, 0

Offset

0,6

Comments

Values are even for base 4 representations of n with an even number of digits, and odd for base 4 representations of n with an odd number of digits, except for a(0).

Links

Table of n, a(n) for n=0..88.

Formula

a(n) = 2*A139351(n) - A110592(n), n>0. - R. J. Mathar, Sep 02 2020

Example

      n in    #odd    #even
  n  base 4  digits - digits = a(n)
  =  ======  =======================
  0    0        0   -            0
  1    1        1   -    0   =   1
  2    2        0   -    1   =  -1
  3    3        1   -    0   =   1
  4   10        1   -    1   =   0
  5   11        2   -    0   =   2
  6   12        1   -    1   =   0
  7   13        2   -    0   =   2

Maple

a:= n-> `if`(n=0, 0, add(`if`(i in [1, 3], 1, -1), i=convert(n, base, 4))):
seq(a(n), n=0..100);  # Alois P. Heinz, May 30 2020

Mathematica

a[0] = 0; a[n_] := Total[-(-1)^(r = Range[0, 3]) * DigitCount[n, 4, r]]; Array[a, 100, 0] (* Amiram Eldar, May 13 2020 *)
Join[{0}, Table[Total[If[EvenQ[#], -1, 1]&/@IntegerDigits[n, 4]], {n, 90}]] (* Harvey P. Dale, Sep 06 2020 *)

Prog

(R)
qnary = function(n, e, q){
  e = floor(n/4)
  q = n%%4
  if(n == 0 ){return(0)}
  if(e == 0){return(q)}
  else{return(c(qnary(e), (q)))}
}
m = 400
s = seq(2, m)
v = c(0)
for(i in s){
  x = qnary(i-1)
  x[which(x%%2!=0)] = 1
  x[which(x%%2==0)] = -1
  v[i] = sum(x)
}
(Python)
import numpy as np
def qnary(n):
    e = n//4
    q = n%4
    if n == 0 : return 0
    if e == 0 : return q
    if e != 0 : return np.append(qnary(e), q)
m = 400
v = [0]
for i in range(1, m+1) :
    t = np.array(qnary(i))
    t[t%2 != 0] = 1
    t[t%2 == 0] = -1
    v = np.append(v, np.sum(t))
(PARI) a(n) = my(ret=0); if(n, forstep(i=0, logint(n, 2), 2, if(bittest(n, i), ret++, ret--))); ret; \\ Kevin Ryde, May 24 2020
(Python)
def A334841(n):
    return 2*bin(n)[-1:1:-2].count('1')-(len(bin(n))-1)//2 if n > 0 else 0 # Chai Wah Wu, Sep 03 2020

Crossrefs

Cf. A053737, A010065, A037863, A007090, A301336, A333596.

Sequence in context: A106277 A088627 A230205 * A024713 A361166 A123530

Adjacent sequences: A334838 A334839 A334840 * A334842 A334843 A334844

Keyword

sign,easy,base

Author

Alexander Van Plantinga, May 13 2020

Status

approved