Mixing
More about odd numbers in Pascal's triangle
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Do you have any references to this interesting result? I could not find any...
The total number of odd numbers in the first $2^n$ rows of Pascal's triangle is $3^n$, $n>=0$.
It's easy to prove by induction based on the formula for the number of odds in row $n$ ($2^m$, where $m$ is the number of ones in the binary expansion of $n$).
combinatorics binomial-coefficients
asked Jan 29 at 19:50
Mircea DraghicescuMircea Draghicescu
184
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add a comment |
$begingroup$
Do you have any references to this interesting result? I could not find any...
The total number of odd numbers in the first $2^n$ rows of Pascal's triangle is $3^n$, $n>=0$.
It's easy to prove by induction based on the formula for the number of odds in row $n$ ($2^m$, where $m$ is the number of ones in the binary expansion of $n$).
combinatorics binomial-coefficients
asked Jan 29 at 19:50
Mircea DraghicescuMircea Draghicescu
184
$endgroup$
$begingroup$
Just saw it: the answer by @Micah to the "Odds in Pascal's Triangle" post includes a direct proof of this fact.
$endgroup$
– Mircea Draghicescu
Jan 29 at 22:10
add a comment |
$begingroup$
Do you have any references to this interesting result? I could not find any...
The total number of odd numbers in the first $2^n$ rows of Pascal's triangle is $3^n$, $n>=0$.
It's easy to prove by induction based on the formula for the number of odds in row $n$ ($2^m$, where $m$ is the number of ones in the binary expansion of $n$).
combinatorics binomial-coefficients
asked Jan 29 at 19:50
Mircea DraghicescuMircea Draghicescu
184
$endgroup$
Do you have any references to this interesting result? I could not find any...
The total number of odd numbers in the first $2^n$ rows of Pascal's triangle is $3^n$, $n>=0$.
It's easy to prove by induction based on the formula for the number of odds in row $n$ ($2^m$, where $m$ is the number of ones in the binary expansion of $n$).
combinatorics binomial-coefficients
combinatorics binomial-coefficients
asked Jan 29 at 19:50
Mircea DraghicescuMircea Draghicescu
184
asked Jan 29 at 19:50
Mircea DraghicescuMircea Draghicescu
184
asked Jan 29 at 19:50
Mircea DraghicescuMircea Draghicescu
184
asked Jan 29 at 19:50
Mircea DraghicescuMircea Draghicescu
184
asked Jan 29 at 19:50
Mircea DraghicescuMircea Draghicescu
184
184
$begingroup$
Just saw it: the answer by @Micah to the "Odds in Pascal's Triangle" post includes a direct proof of this fact.
$endgroup$
– Mircea Draghicescu
Jan 29 at 22:10
add a comment |
$begingroup$
Just saw it: the answer by @Micah to the "Odds in Pascal's Triangle" post includes a direct proof of this fact.
$endgroup$
– Mircea Draghicescu
Jan 29 at 22:10
$begingroup$
Just saw it: the answer by @Micah to the "Odds in Pascal's Triangle" post includes a direct proof of this fact.
$endgroup$
– Mircea Draghicescu
Jan 29 at 22:10
$begingroup$
Just saw it: the answer by @Micah to the "Odds in Pascal's Triangle" post includes a direct proof of this fact.
$endgroup$
– Mircea Draghicescu
Jan 29 at 22:10
add a comment |
1 Answer
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The Online Encyclopedia of Integer Sequences lists as entry A006046
Total number of odd entries in first n rows of Pascal's triangle: a(0) = 0, a(1) = 1, a(2k) = 3*a(k), a(2k+1) = 2*a(k) + a(k+1)
which doesn't give the claim explicitly but makes it very clear implicitly. Further down, in the formula section, you can find it explicitly:
a(2^n) = 3^n. - Henry Bottomley, Apr 05 2001
answered Jan 29 at 22:11
Peter TaylorPeter Taylor
9,15712343
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1 Answer
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active
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1 Answer
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active
oldest
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$begingroup$
The Online Encyclopedia of Integer Sequences lists as entry A006046
Total number of odd entries in first n rows of Pascal's triangle: a(0) = 0, a(1) = 1, a(2k) = 3*a(k), a(2k+1) = 2*a(k) + a(k+1)
which doesn't give the claim explicitly but makes it very clear implicitly. Further down, in the formula section, you can find it explicitly:
a(2^n) = 3^n. - Henry Bottomley, Apr 05 2001
answered Jan 29 at 22:11
Peter TaylorPeter Taylor
9,15712343
$endgroup$
add a comment |
$begingroup$
The Online Encyclopedia of Integer Sequences lists as entry A006046
Total number of odd entries in first n rows of Pascal's triangle: a(0) = 0, a(1) = 1, a(2k) = 3*a(k), a(2k+1) = 2*a(k) + a(k+1)
which doesn't give the claim explicitly but makes it very clear implicitly. Further down, in the formula section, you can find it explicitly:
a(2^n) = 3^n. - Henry Bottomley, Apr 05 2001
answered Jan 29 at 22:11
Peter TaylorPeter Taylor
9,15712343
$endgroup$
add a comment |
$begingroup$
The Online Encyclopedia of Integer Sequences lists as entry A006046
Total number of odd entries in first n rows of Pascal's triangle: a(0) = 0, a(1) = 1, a(2k) = 3*a(k), a(2k+1) = 2*a(k) + a(k+1)
which doesn't give the claim explicitly but makes it very clear implicitly. Further down, in the formula section, you can find it explicitly:
a(2^n) = 3^n. - Henry Bottomley, Apr 05 2001
answered Jan 29 at 22:11
Peter TaylorPeter Taylor
9,15712343
$endgroup$
The Online Encyclopedia of Integer Sequences lists as entry A006046
Total number of odd entries in first n rows of Pascal's triangle: a(0) = 0, a(1) = 1, a(2k) = 3*a(k), a(2k+1) = 2*a(k) + a(k+1)
which doesn't give the claim explicitly but makes it very clear implicitly. Further down, in the formula section, you can find it explicitly:
a(2^n) = 3^n. - Henry Bottomley, Apr 05 2001
answered Jan 29 at 22:11
Peter TaylorPeter Taylor
9,15712343
answered Jan 29 at 22:11
Peter TaylorPeter Taylor
9,15712343
answered Jan 29 at 22:11
Peter TaylorPeter Taylor
9,15712343
answered Jan 29 at 22:11
Peter TaylorPeter Taylor
9,15712343
9,15712343
add a comment |
add a comment |
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$begingroup$
Just saw it: the answer by @Micah to the "Odds in Pascal's Triangle" post includes a direct proof of this fact.
$endgroup$
– Mircea Draghicescu
Jan 29 at 22:10