A333596 a(0) = 0; for n > 0, a(n) = a(n-1) + (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, 0, 1, 1, 3, 3, 5, 3, 3, 1, 1, 1, 3, 3, 5, 4, 5, 4, 5, 6, 9, 10, 13, 12, 13, 12, 13, 14, 17, 18, 21, 18, 17, 14, 13, 12, 13, 12, 13, 10, 9, 6, 5, 4, 5, 4, 5, 4, 5, 4, 5, 6, 9, 10, 13, 12, 13, 12, 13, 14, 17, 18, 21, 19, 19, 17, 17, 17, 19, 19, 21, 19, 19

Offset: 0

Author

Alexander Van Plantinga, Mar 27 2020

Keywords

• nonn
• base
• easy

Comments

Local maxima values minus 1 are divisible by 4.

For a digit-wise recurrence, it's convenient to sum n terms so b(n) = a(n-1) = Sum_{i=0..n-1} A334841(i). Then b(4n+r) = 4*b(n) + r*A334841(n) + (1 if r even), for 0 <= r <= 3 and 4n+r >= 1. This is 4 copies of terms 0..n-1 and r copies of the following n. The new lowest digits cancel when r is odd, or net +1 when r is even. Repeatedly expanding gives the PARI code below. - Kevin Ryde, Jun 02 2020

Examples

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

Crossrefs

Cf. A053737, A010065, A037863, A268289, A007090, A231664, A301336, A334841.

Programs

Maple

a:= proc(n) option remember; `if`(n=0, 0, a(n-1) +add(
     `if`(i in [1, 3], 1, -1), i=convert(n, base, 4)))
    end:
seq(a(n), n=0..80);  # Alois P. Heinz, May 30 2020

Mathematica

f[n_] := Total[(-1)^(r = Range[0, 3]) * DigitCount[n, 4, r]]; a[0] = 0; a[n_] := a[n] = a[n - 1] - f[n]; Array[a, 100, 0] (* Amiram Eldar, Apr 24 2020 *)

PARI

a(n) = my(v=digits(n+1,4),s=0); for(i=1,#v, my(t=v[i]); v[i]=t*s+!(t%2); s-=(-1)^t); fromdigits(v,4); \\ Kevin Ryde, May 30 2020

PARI

b(n)=my(d=digits(n,4)); -sum(i=1,#d,(-1)^d[i])
first(n)=my(s); concat(0,vector(n,k,s+=b(k))) \\ Charles R Greathouse IV, Jul 04 2020

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([np.sum(t), v[i-1]]))

Python

from itertools import accumulate
def A334841(n):
    return 2*bin(n)[-1:1:-2].count('1')-(len(bin(n))-1)//2 if n > 0 else 0
A333596_list = list(accumulate(A334841(n) for n in range(10000))) # Chai Wah Wu, Sep 03 2020

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,v[i-1])
}