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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).
2
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).
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
KEYWORD
sign,easy,base
AUTHOR
STATUS
approved