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6
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edited Nov 1 2011 at 14:26 |
Say that a number is an odd-bit number if
the count of 1-bits in its binary representation is odd.
Define an even-bit number analogously.
Thus $541 = 1000011101_2$ is an odd-bit number,
and $523 = 1000001011_2$ is an even-bit number.
Are there, asymptotically, as many odd-bit primes as even-bit primes?
For the first ten primes, we have
$$
\lbrace 10, 11, 101, 111, 1011, 1101, 10001, 10011, 10111, 11101
\rbrace
$$
with 1-bits
$$
\lbrace 1, 2, 2, 3, 3, 3, 2, 3, 4, 4 \rbrace
$$
and so ratio $5/5=1$ at the 10-th prime.
Here is a plot of this ratio up to $10^5$:
I would expect the ratio to approach $\frac{1}{2}$, except perhaps the fact that primes ($>2$) are
odd might bias the ratio. The above plot does not suggest convergence
by the 100,000-th prime (1,299,709).
Pardon the naïveness of my question.
Addendum: Extended the computation to the $10^6$-th prime (15,485,863), where it still
remains 1.5% above $\frac{1}{2}.$\frac{1}{2}$:
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5
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edited Oct 31 2011 at 19:11 |
Say that a number is an odd-bit number if
the count of 1-bits in its binary representation is odd.
Define an even-bit number analogously.
Thus $541 = 1000011101_2$ is an odd-bit number,
and $523 = 1000001011_2$ is an even-bit number.
Are there, asymptotically, as many odd-bit primes as even-bit primes?
For the first ten primes, we have
$$
\lbrace 10, 11, 101, 111, 1011, 1101, 10001, 10011, 10111, 11101
\rbrace
$$
with 1-bits
$$
\lbrace 1, 2, 2, 3, 3, 3, 2, 3, 4, 4 \rbrace
$$
and so ratio $5/5=1$ at the 10-th prime.Here is a plot of this ratio up to $10^5$:
I would expect the ratio to approach $\frac{1}{2}$, except perhaps the fact that primes ($>2$) are
odd might bias the ratio. The above plot does not suggest convergence by the 100,000-th prime (1,299,709).
Pardon the naïveness of my question.
Addendum: Extended the computation to the $10^6$-th prime (15,485,863), where it still
remains 1.5% above $\frac{1}{2}$:
\frac{1}{2}.$
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4
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edited Nov 3 2010 at 10:58 |
Say that a number is an odd-bit number if
the count of 1-bits in its binary representation is odd.
Define an even-bit number analogously.
Thus $541 = 1000011101_2$ is an odd-bit number,
and $523 = 1000001011_2$ is an even-bit number.
Are there, asymptotically, as many odd-bit primes as even-bit primes?
For the first ten primes, we have
$$
\lbrace 10, 11, 101, 111, 1011, 1101, 10001, 10011, 10111, 11101
\rbrace
$$
with 1-bits
$$
\lbrace 1, 2, 2, 3, 3, 3, 2, 3, 4, 4 \rbrace
$$
and so ratio $5/5=1$ at the 10-th prime.
Here is a plot of this ratio up to $10^5$:
I would expect the ratio to approach $\frac{1}{2}$, except perhaps the fact that primes ($>2$) are
odd might bias the ratio. The above plot does not suggest convergence
by the 100,000-th prime (1,299,709).
Pardon the naïveness of my question.
Addendum: Extended the computation to the $10^6$:
10^6$-th prime (15,485,863), where it still
remains 1.5% above $\frac{1}{2}$:
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3
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edited Nov 2 2010 at 19:50 |
Say that a number is an odd-bit number if
the count of 1-bits in its binary representation is odd.
Define an even-bit number analogously.
Thus $541 = 1000011101_2$ is an odd-bit number,
and $523 = 1000001011_2$ is an even-bit number.
Are there, asymptotically, as many odd-bit primes as even-bit primes?
For the first ten primes, we have
$$
\lbrace 10, 11, 101, 111, 1011, 1101, 10001, 10011, 10111, 11101
\rbrace
$$
with 1-bits
$$
\lbrace 1, 2, 2, 3, 3, 3, 2, 3, 4, 4 \rbrace
$$
and so ratio $5/5=1$ at the 10-th prime.
Here is a plot of this ratio up to $10^5$:
I would expect the ratio to approach $\frac{1}{2}$, except perhaps the fact that primes ($>2$) are
odd might bias the ratio. The above plot does not suggest convergence
by the 100,000-th prime (1,299,709).
Pardon the naïveness of my question.
Addendum: Extended the computation to $10^6$:
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2
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edited Nov 2 2010 at 15:46 |
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1
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asked Nov 2 2010 at 14:25 |
Odd-bit primes ratio
Say that a number is an odd-bit number if
the count of 1-bits in its binary representation is odd.
Define an even-bit number analogously.
Thus $541 = 1000011101_2$ is an odd-bit number,
and $523 = 1000001011_2$ is an even-bit number.
Are there, asymptotically, as many odd-bit primes as even-bit primes?
For the first ten primes, we have
$$
\lbrace 10, 11, 101, 111, 1011, 1101, 10001, 10011, 10111, 11101
\rbrace
$$
with 1-bits
$$
\lbrace 1, 2, 2, 3, 3, 3, 2, 3, 4, 4 \rbrace
$$
and so ratio $5/5=1$ at the 10-th prime.
Here is a plot of this ratio up to $10^5$:
I would expect the ratio to approach $\frac{1}{2}$, except perhaps the fact that primes ($>2$) are
odd might bias the ratio. The above plot does not suggest convergence
by the 100,000-th prime (1,299,709).
Pardon the naïveness of my question.
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