is there a special name for a function that generates a binary tree
0
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If you graph t vs i for n = 7*11^i+1 (= 2^k*j + 2^t - 1) with t being the number of trailing 1s in the binary representation, you will see an inverted binary tree. Is there any special significance to this?
elementary-number-theory elementary-functions
asked Jan 11 at 23:45
smichrsmichr
1636
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add a comment |
0
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If you graph t vs i for n = 7*11^i+1 (= 2^k*j + 2^t - 1) with t being the number of trailing 1s in the binary representation, you will see an inverted binary tree. Is there any special significance to this?
elementary-number-theory elementary-functions
asked Jan 11 at 23:45
smichrsmichr
1636
$endgroup$
$begingroup$
Please show the graph, to clarify your question.
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– David G. Stork
Jan 11 at 23:58
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It's not exactly a binary tree and the upper left point should not have the left leg connecting to anything, but it gives the gist of what I mean.
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– smichr
Jan 12 at 2:40
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This is just the consequence of parity considerations when multiplying. For the powers (repeated multiplication), every other will end in the same way, every 4 will end the same way, etc... Multiplying by something else changes where these endings start (e.g. the ending with the least 0s before 1 might occur for odd or even exponent) but the overall pattern is the same.
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– smichr
Jan 15 at 23:35
add a comment |
0
0
0
$begingroup$
If you graph t vs i for n = 7*11^i+1 (= 2^k*j + 2^t - 1) with t being the number of trailing 1s in the binary representation, you will see an inverted binary tree. Is there any special significance to this?
elementary-number-theory elementary-functions
asked Jan 11 at 23:45
smichrsmichr
1636
$endgroup$
If you graph t vs i for n = 7*11^i+1 (= 2^k*j + 2^t - 1) with t being the number of trailing 1s in the binary representation, you will see an inverted binary tree. Is there any special significance to this?
elementary-number-theory elementary-functions
elementary-number-theory elementary-functions
asked Jan 11 at 23:45
smichrsmichr
1636
asked Jan 11 at 23:45
smichrsmichr
1636
edited Jan 12 at 0:29
smichr
asked Jan 11 at 23:45
smichrsmichr
1636
asked Jan 11 at 23:45
smichrsmichr
1636
asked Jan 11 at 23:45
smichrsmichr
1636
1636
$begingroup$
Please show the graph, to clarify your question.
$endgroup$
– David G. Stork
Jan 11 at 23:58
$begingroup$
It's not exactly a binary tree and the upper left point should not have the left leg connecting to anything, but it gives the gist of what I mean.
$endgroup$
– smichr
Jan 12 at 2:40
$begingroup$
This is just the consequence of parity considerations when multiplying. For the powers (repeated multiplication), every other will end in the same way, every 4 will end the same way, etc... Multiplying by something else changes where these endings start (e.g. the ending with the least 0s before 1 might occur for odd or even exponent) but the overall pattern is the same.
$endgroup$
– smichr
Jan 15 at 23:35
add a comment |
$begingroup$
Please show the graph, to clarify your question.
$endgroup$
– David G. Stork
Jan 11 at 23:58
$begingroup$
It's not exactly a binary tree and the upper left point should not have the left leg connecting to anything, but it gives the gist of what I mean.
$endgroup$
– smichr
Jan 12 at 2:40
$begingroup$
This is just the consequence of parity considerations when multiplying. For the powers (repeated multiplication), every other will end in the same way, every 4 will end the same way, etc... Multiplying by something else changes where these endings start (e.g. the ending with the least 0s before 1 might occur for odd or even exponent) but the overall pattern is the same.
$endgroup$
– smichr
Jan 15 at 23:35
$begingroup$
Please show the graph, to clarify your question.
$endgroup$
– David G. Stork
Jan 11 at 23:58
$begingroup$
Please show the graph, to clarify your question.
$endgroup$
– David G. Stork
Jan 11 at 23:58
$begingroup$
It's not exactly a binary tree and the upper left point should not have the left leg connecting to anything, but it gives the gist of what I mean.
$endgroup$
– smichr
Jan 12 at 2:40
$begingroup$
It's not exactly a binary tree and the upper left point should not have the left leg connecting to anything, but it gives the gist of what I mean.
$endgroup$
– smichr
Jan 12 at 2:40
$begingroup$
This is just the consequence of parity considerations when multiplying. For the powers (repeated multiplication), every other will end in the same way, every 4 will end the same way, etc... Multiplying by something else changes where these endings start (e.g. the ending with the least 0s before 1 might occur for odd or even exponent) but the overall pattern is the same.
$endgroup$
– smichr
Jan 15 at 23:35
$begingroup$
This is just the consequence of parity considerations when multiplying. For the powers (repeated multiplication), every other will end in the same way, every 4 will end the same way, etc... Multiplying by something else changes where these endings start (e.g. the ending with the least 0s before 1 might occur for odd or even exponent) but the overall pattern is the same.
$endgroup$
– smichr
Jan 15 at 23:35
add a comment |
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$begingroup$
Please show the graph, to clarify your question.
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– David G. Stork
Jan 11 at 23:58
$begingroup$
It's not exactly a binary tree and the upper left point should not have the left leg connecting to anything, but it gives the gist of what I mean.
$endgroup$
– smichr
Jan 12 at 2:40
$begingroup$
This is just the consequence of parity considerations when multiplying. For the powers (repeated multiplication), every other will end in the same way, every 4 will end the same way, etc... Multiplying by something else changes where these endings start (e.g. the ending with the least 0s before 1 might occur for odd or even exponent) but the overall pattern is the same.
$endgroup$
– smichr
Jan 15 at 23:35