A339413 a(0) = 0; for n > 0, a(n) = a(n-1) if c0 == c1; a(n) = a(n-1) - c0 if c0 > c1; a(n) = a(n - 1) + c1 if c1 > c0, where c0 and c1 are respectively the number of 0's and 1's in the binary expansion of n.
0, 1, 1, 3, 1, 3, 5, 8, 5, 5, 5, 8, 8, 11, 14, 18, 14, 11, 8, 11, 8, 11, 14, 18, 15, 18, 21, 25, 28, 32, 36, 41, 36, 32, 28, 28, 24, 24, 24, 28, 24, 24, 24, 28, 28, 32, 36, 41, 37, 37, 37, 41, 41, 45, 49, 54, 54, 58, 62, 67, 71, 76, 81, 87, 81, 76, 71, 67, 62
Offset: 0
Links
- Rémy Sigrist, Table of n, a(n) for n = 0..8192
Programs
-
Mathematica
Block[{a = {0}}, Do[AppendTo[a, a[[-1]] + Which[#1 > #2, #1, #1 < #2, -#2, True, 0] & @@ DigitCount[i, 2]], {i, 68}]; a] (* Michael De Vlieger, Dec 07 2020 *) -
PARI
{ for (n=0, 68, if (n==0, v=0, b=if (n, binary(n), [0]); c0=#b-c1=vecsum(b);if (c0>c1, v-=c0, c1>c0, v+=c1)); print1 (v", ")) } \\ Rémy Sigrist, Dec 25 2020 -
Python
from collections import Counter a = [0] for i in range(1, 10000): counts = Counter(str(bin(i))[2:]) if counts['0'] > counts['1']: a.append(a[-1] - counts['0']) elif counts['1'] > counts['0']: a.append(a[-1] + counts['1']) else: a.append(a[-1]) print(a)
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