In a perfect number $2^{p−1} times (2^p − 1)$, the ratio of $p$ to the digits in its perfect number...
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I was reading about Mersenne primes and perfect numbers, and how the expression $2^{p−1} times (2^p − 1)$, where $p$ is any prime number, can be used to generate perfect numbers when $2^{p−1}$ is a Mersenne prime. I also found a claim on Wikipedia that "The ratio ($p$ / digits) approaches $log(10) / log(4) = 1.6609640474ldots$" but I cannot find the proof that shows this to be true. Why is it $log(10)/log(4)$?
prime-numbers perfect-numbers
edited Dec 7 '18 at 3:48
zipirovich
11.2k11631
asked Dec 7 '18 at 3:39
Sara Sara
132
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add a comment |
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I was reading about Mersenne primes and perfect numbers, and how the expression $2^{p−1} times (2^p − 1)$, where $p$ is any prime number, can be used to generate perfect numbers when $2^{p−1}$ is a Mersenne prime. I also found a claim on Wikipedia that "The ratio ($p$ / digits) approaches $log(10) / log(4) = 1.6609640474ldots$" but I cannot find the proof that shows this to be true. Why is it $log(10)/log(4)$?
prime-numbers perfect-numbers
edited Dec 7 '18 at 3:48
zipirovich
11.2k11631
asked Dec 7 '18 at 3:39
Sara Sara
132
$endgroup$
add a comment |
$begingroup$
I was reading about Mersenne primes and perfect numbers, and how the expression $2^{p−1} times (2^p − 1)$, where $p$ is any prime number, can be used to generate perfect numbers when $2^{p−1}$ is a Mersenne prime. I also found a claim on Wikipedia that "The ratio ($p$ / digits) approaches $log(10) / log(4) = 1.6609640474ldots$" but I cannot find the proof that shows this to be true. Why is it $log(10)/log(4)$?
prime-numbers perfect-numbers
edited Dec 7 '18 at 3:48
zipirovich
11.2k11631
asked Dec 7 '18 at 3:39
Sara Sara
132
$endgroup$
I was reading about Mersenne primes and perfect numbers, and how the expression $2^{p−1} times (2^p − 1)$, where $p$ is any prime number, can be used to generate perfect numbers when $2^{p−1}$ is a Mersenne prime. I also found a claim on Wikipedia that "The ratio ($p$ / digits) approaches $log(10) / log(4) = 1.6609640474ldots$" but I cannot find the proof that shows this to be true. Why is it $log(10)/log(4)$?
prime-numbers perfect-numbers
prime-numbers perfect-numbers
edited Dec 7 '18 at 3:48
zipirovich
11.2k11631
asked Dec 7 '18 at 3:39
Sara Sara
132
edited Dec 7 '18 at 3:48
zipirovich
11.2k11631
asked Dec 7 '18 at 3:39
Sara Sara
132
edited Dec 7 '18 at 3:48
zipirovich
11.2k11631
edited Dec 7 '18 at 3:48
zipirovich
11.2k11631
edited Dec 7 '18 at 3:48
zipirovich
11.2k11631
11.2k11631
asked Dec 7 '18 at 3:39
Sara Sara
132
asked Dec 7 '18 at 3:39
Sara Sara
132
asked Dec 7 '18 at 3:39
Sara Sara
132
132
add a comment |
add a comment |
1 Answer
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$begingroup$
Because the number of digits in an integer $N$ is approximately $log_{10}N$. The precise formula is $lfloorlog_{10}Nrfloor+1$, but when $N$ is large this rounding is insignificant.
For $N=2^{p−1}times(2^p−1)$, and assuming $p$ (and therefore, $N$) is large, this gives:
$$begin{aligned}
log_{10}N&=log_{10}left(2^{p−1}times(2^p−1)right)=\
&=log_{10}left(2^{p−1}right)+log_{10}left(2^p−1right)approx\
&approxlog_{10}left(2^{p−1}right)+log_{10}left(2^pright)=\
&=(p-1)log_{10}2+plog_{10}2=\
&=(2p-1)log_{10}2approx\
&approx2plog_{10}2=\
&=plog_{10}4,
end{aligned}$$
and then the change of base formula finishes the proof.
answered Dec 7 '18 at 3:57
zipirovichzipirovich
11.2k11631
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
Because the number of digits in an integer $N$ is approximately $log_{10}N$. The precise formula is $lfloorlog_{10}Nrfloor+1$, but when $N$ is large this rounding is insignificant.
For $N=2^{p−1}times(2^p−1)$, and assuming $p$ (and therefore, $N$) is large, this gives:
$$begin{aligned}
log_{10}N&=log_{10}left(2^{p−1}times(2^p−1)right)=\
&=log_{10}left(2^{p−1}right)+log_{10}left(2^p−1right)approx\
&approxlog_{10}left(2^{p−1}right)+log_{10}left(2^pright)=\
&=(p-1)log_{10}2+plog_{10}2=\
&=(2p-1)log_{10}2approx\
&approx2plog_{10}2=\
&=plog_{10}4,
end{aligned}$$
and then the change of base formula finishes the proof.
answered Dec 7 '18 at 3:57
zipirovichzipirovich
11.2k11631
$endgroup$
add a comment |
$begingroup$
Because the number of digits in an integer $N$ is approximately $log_{10}N$. The precise formula is $lfloorlog_{10}Nrfloor+1$, but when $N$ is large this rounding is insignificant.
For $N=2^{p−1}times(2^p−1)$, and assuming $p$ (and therefore, $N$) is large, this gives:
$$begin{aligned}
log_{10}N&=log_{10}left(2^{p−1}times(2^p−1)right)=\
&=log_{10}left(2^{p−1}right)+log_{10}left(2^p−1right)approx\
&approxlog_{10}left(2^{p−1}right)+log_{10}left(2^pright)=\
&=(p-1)log_{10}2+plog_{10}2=\
&=(2p-1)log_{10}2approx\
&approx2plog_{10}2=\
&=plog_{10}4,
end{aligned}$$
and then the change of base formula finishes the proof.
answered Dec 7 '18 at 3:57
zipirovichzipirovich
11.2k11631
$endgroup$
add a comment |
$begingroup$
Because the number of digits in an integer $N$ is approximately $log_{10}N$. The precise formula is $lfloorlog_{10}Nrfloor+1$, but when $N$ is large this rounding is insignificant.
For $N=2^{p−1}times(2^p−1)$, and assuming $p$ (and therefore, $N$) is large, this gives:
$$begin{aligned}
log_{10}N&=log_{10}left(2^{p−1}times(2^p−1)right)=\
&=log_{10}left(2^{p−1}right)+log_{10}left(2^p−1right)approx\
&approxlog_{10}left(2^{p−1}right)+log_{10}left(2^pright)=\
&=(p-1)log_{10}2+plog_{10}2=\
&=(2p-1)log_{10}2approx\
&approx2plog_{10}2=\
&=plog_{10}4,
end{aligned}$$
and then the change of base formula finishes the proof.
answered Dec 7 '18 at 3:57
zipirovichzipirovich
11.2k11631
$endgroup$
Because the number of digits in an integer $N$ is approximately $log_{10}N$. The precise formula is $lfloorlog_{10}Nrfloor+1$, but when $N$ is large this rounding is insignificant.
For $N=2^{p−1}times(2^p−1)$, and assuming $p$ (and therefore, $N$) is large, this gives:
$$begin{aligned}
log_{10}N&=log_{10}left(2^{p−1}times(2^p−1)right)=\
&=log_{10}left(2^{p−1}right)+log_{10}left(2^p−1right)approx\
&approxlog_{10}left(2^{p−1}right)+log_{10}left(2^pright)=\
&=(p-1)log_{10}2+plog_{10}2=\
&=(2p-1)log_{10}2approx\
&approx2plog_{10}2=\
&=plog_{10}4,
end{aligned}$$
and then the change of base formula finishes the proof.
answered Dec 7 '18 at 3:57
zipirovichzipirovich
11.2k11631
answered Dec 7 '18 at 3:57
zipirovichzipirovich
11.2k11631
answered Dec 7 '18 at 3:57
zipirovichzipirovich
11.2k11631
answered Dec 7 '18 at 3:57
zipirovichzipirovich
11.2k11631
11.2k11631
add a comment |
add a comment |
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