Mixing
What's the formula for producing these series? [closed]
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
I have a function which produces a series from an integer. I currently do this iteratively.
f(0) =
f(1) = [0]
f(2) = [0, 0]
f(3) = [0, 1]
f(4) = [0, 0, 1]
f(5) = [0, 1, 1]
f(6) = [0, 0, 2]
f(7) = [0, 1, 2]
f(8) = [0, 0, 1, 2]
There is another pattern here.
= 0
[0] 2**0 = 1
[0, 0] 2**0 + 2**0 = 2
[0, 1] 2**0 + 2**1 = 3
[0, 0, 1] 2**0 + 2**0 + 2**1 = 4
[0, 1, 1] 2**0 + 2**1 + 2**1 = 5
[0, 0, 2] 2**0 + 2**0 + 2**2 = 6
[0, 1, 2] 2**0 + 2**1 + 2**2 = 7
[0, 0, 1, 2] 2**0 + 2**0 + 2**1 + 2**2 = 8
(Where double asterisk means 'raised to the power.)
Is there way to get from the argument, straight to the series, without working out all of the previous series?
Update, thanks for the interest.
Here's some Javascript code which generates the sequences, where weight is the sequence:
var numbers = ;
for (var count = 0; count < 20; count++)
for (var i = numbers.length; i > 0; i--)
if (count % (2**i) === 0)
numbers[i] = numbers[i - 1];
numbers[0] = count;
var weights = ;
for (var j = 1; j < numbers.length; j++)
// note the differences below are all exact powers of 2
weights[j] = Math.log2(numbers[j - 1] - numbers[j]);
if (numbers.length > 0)
weights[0] = 0;
console.log(`count $count numbers $numbers weight $weights`);
and the output:
count 0 numbers 0 weight
count 1 numbers 1 weight 0
count 2 numbers 2,1 weight 0,0
count 3 numbers 3,1 weight 0,1
count 4 numbers 4,3,1 weight 0,0,1
count 5 numbers 5,3,1 weight 0,1,1
count 6 numbers 6,5,1 weight 0,0,2
count 7 numbers 7,5,1 weight 0,1,2
count 8 numbers 8,7,5,1 weight 0,0,1,2
count 9 numbers 9,7,5,1 weight 0,1,1,2
count 10 numbers 10,9,5,1 weight 0,0,2,2
count 11 numbers 11,9,5,1 weight 0,1,2,2
count 12 numbers 12,11,9,1 weight 0,0,1,3
count 13 numbers 13,11,9,1 weight 0,1,1,3
count 14 numbers 14,13,9,1 weight 0,0,2,3
count 15 numbers 15,13,9,1 weight 0,1,2,3
count 16 numbers 16,15,13,9,1 weight 0,0,1,2,3
count 17 numbers 17,15,13,9,1 weight 0,1,1,2,3
count 18 numbers 18,17,13,9,1 weight 0,0,2,2,3
count 19 numbers 19,17,13,9,1 weight 0,1,2,2,3
There seems to be an emergent restriction that the values in the Nth place of the series can only take values between 0 and N.
sequences-and-series exponentiation
asked Jul 26 at 12:15
fadedbee
1385
closed as unclear what you're asking by Winther, amWhy, Lord Shark the Unknown, Jyrki Lahtonen, José Carlos Santos Jul 27 at 14:34
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |Â
up vote
2
down vote
favorite
I have a function which produces a series from an integer. I currently do this iteratively.
f(0) =
f(1) = [0]
f(2) = [0, 0]
f(3) = [0, 1]
f(4) = [0, 0, 1]
f(5) = [0, 1, 1]
f(6) = [0, 0, 2]
f(7) = [0, 1, 2]
f(8) = [0, 0, 1, 2]
There is another pattern here.
= 0
[0] 2**0 = 1
[0, 0] 2**0 + 2**0 = 2
[0, 1] 2**0 + 2**1 = 3
[0, 0, 1] 2**0 + 2**0 + 2**1 = 4
[0, 1, 1] 2**0 + 2**1 + 2**1 = 5
[0, 0, 2] 2**0 + 2**0 + 2**2 = 6
[0, 1, 2] 2**0 + 2**1 + 2**2 = 7
[0, 0, 1, 2] 2**0 + 2**0 + 2**1 + 2**2 = 8
(Where double asterisk means 'raised to the power.)
Is there way to get from the argument, straight to the series, without working out all of the previous series?
Update, thanks for the interest.
Here's some Javascript code which generates the sequences, where weight is the sequence:
var numbers = ;
for (var count = 0; count < 20; count++)
for (var i = numbers.length; i > 0; i--)
if (count % (2**i) === 0)
numbers[i] = numbers[i - 1];
numbers[0] = count;
var weights = ;
for (var j = 1; j < numbers.length; j++)
// note the differences below are all exact powers of 2
weights[j] = Math.log2(numbers[j - 1] - numbers[j]);
if (numbers.length > 0)
weights[0] = 0;
console.log(`count $count numbers $numbers weight $weights`);
and the output:
count 0 numbers 0 weight
count 1 numbers 1 weight 0
count 2 numbers 2,1 weight 0,0
count 3 numbers 3,1 weight 0,1
count 4 numbers 4,3,1 weight 0,0,1
count 5 numbers 5,3,1 weight 0,1,1
count 6 numbers 6,5,1 weight 0,0,2
count 7 numbers 7,5,1 weight 0,1,2
count 8 numbers 8,7,5,1 weight 0,0,1,2
count 9 numbers 9,7,5,1 weight 0,1,1,2
count 10 numbers 10,9,5,1 weight 0,0,2,2
count 11 numbers 11,9,5,1 weight 0,1,2,2
count 12 numbers 12,11,9,1 weight 0,0,1,3
count 13 numbers 13,11,9,1 weight 0,1,1,3
count 14 numbers 14,13,9,1 weight 0,0,2,3
count 15 numbers 15,13,9,1 weight 0,1,2,3
count 16 numbers 16,15,13,9,1 weight 0,0,1,2,3
count 17 numbers 17,15,13,9,1 weight 0,1,1,2,3
count 18 numbers 18,17,13,9,1 weight 0,0,2,2,3
count 19 numbers 19,17,13,9,1 weight 0,1,2,2,3
There seems to be an emergent restriction that the values in the Nth place of the series can only take values between 0 and N.
sequences-and-series exponentiation
asked Jul 26 at 12:15
fadedbee
1385
closed as unclear what you're asking by Winther, amWhy, Lord Shark the Unknown, Jyrki Lahtonen, José Carlos Santos Jul 27 at 14:34
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
The pattern looks like: if there are no pairs of repeated digits add a zero to the left, if there are double digits increment the right most digit and set the left of the pair to zero. Am I right in thinking then that $f(9)=[0,1,1,2],f(10)=[0,0,2,2]$? I can't see what to then do for f(11) because there are two pairs of repeated digits, it could be $[0,0,0,3]$ or $[0,1,2,2]$, I wouldn't know how to deal with a triple digit, so I guess it would be $[0,1,2,2]$.
– Benedict W. J. Irwin
Jul 26 at 12:59
1
Following that rule through large numbers seem to end in long strings of [...,0,3,0,3,0,3,0,3] etc.
– Benedict W. J. Irwin
Jul 26 at 13:09
1
@BenedictWilliamJohnIrwin Thanks for your interest, I've posted an update.
– fadedbee
Jul 26 at 13:36
@Winther Yes, it is well defined. I've posted a javascript function which generates the series in an update.
– fadedbee
Jul 26 at 14:05
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have a function which produces a series from an integer. I currently do this iteratively.
f(0) =
f(1) = [0]
f(2) = [0, 0]
f(3) = [0, 1]
f(4) = [0, 0, 1]
f(5) = [0, 1, 1]
f(6) = [0, 0, 2]
f(7) = [0, 1, 2]
f(8) = [0, 0, 1, 2]
There is another pattern here.
= 0
[0] 2**0 = 1
[0, 0] 2**0 + 2**0 = 2
[0, 1] 2**0 + 2**1 = 3
[0, 0, 1] 2**0 + 2**0 + 2**1 = 4
[0, 1, 1] 2**0 + 2**1 + 2**1 = 5
[0, 0, 2] 2**0 + 2**0 + 2**2 = 6
[0, 1, 2] 2**0 + 2**1 + 2**2 = 7
[0, 0, 1, 2] 2**0 + 2**0 + 2**1 + 2**2 = 8
(Where double asterisk means 'raised to the power.)
Is there way to get from the argument, straight to the series, without working out all of the previous series?
Update, thanks for the interest.
Here's some Javascript code which generates the sequences, where weight is the sequence:
var numbers = ;
for (var count = 0; count < 20; count++)
for (var i = numbers.length; i > 0; i--)
if (count % (2**i) === 0)
numbers[i] = numbers[i - 1];
numbers[0] = count;
var weights = ;
for (var j = 1; j < numbers.length; j++)
// note the differences below are all exact powers of 2
weights[j] = Math.log2(numbers[j - 1] - numbers[j]);
if (numbers.length > 0)
weights[0] = 0;
console.log(`count $count numbers $numbers weight $weights`);
and the output:
count 0 numbers 0 weight
count 1 numbers 1 weight 0
count 2 numbers 2,1 weight 0,0
count 3 numbers 3,1 weight 0,1
count 4 numbers 4,3,1 weight 0,0,1
count 5 numbers 5,3,1 weight 0,1,1
count 6 numbers 6,5,1 weight 0,0,2
count 7 numbers 7,5,1 weight 0,1,2
count 8 numbers 8,7,5,1 weight 0,0,1,2
count 9 numbers 9,7,5,1 weight 0,1,1,2
count 10 numbers 10,9,5,1 weight 0,0,2,2
count 11 numbers 11,9,5,1 weight 0,1,2,2
count 12 numbers 12,11,9,1 weight 0,0,1,3
count 13 numbers 13,11,9,1 weight 0,1,1,3
count 14 numbers 14,13,9,1 weight 0,0,2,3
count 15 numbers 15,13,9,1 weight 0,1,2,3
count 16 numbers 16,15,13,9,1 weight 0,0,1,2,3
count 17 numbers 17,15,13,9,1 weight 0,1,1,2,3
count 18 numbers 18,17,13,9,1 weight 0,0,2,2,3
count 19 numbers 19,17,13,9,1 weight 0,1,2,2,3
There seems to be an emergent restriction that the values in the Nth place of the series can only take values between 0 and N.
sequences-and-series exponentiation
asked Jul 26 at 12:15
fadedbee
1385
I have a function which produces a series from an integer. I currently do this iteratively.
f(0) =
f(1) = [0]
f(2) = [0, 0]
f(3) = [0, 1]
f(4) = [0, 0, 1]
f(5) = [0, 1, 1]
f(6) = [0, 0, 2]
f(7) = [0, 1, 2]
f(8) = [0, 0, 1, 2]
There is another pattern here.
= 0
[0] 2**0 = 1
[0, 0] 2**0 + 2**0 = 2
[0, 1] 2**0 + 2**1 = 3
[0, 0, 1] 2**0 + 2**0 + 2**1 = 4
[0, 1, 1] 2**0 + 2**1 + 2**1 = 5
[0, 0, 2] 2**0 + 2**0 + 2**2 = 6
[0, 1, 2] 2**0 + 2**1 + 2**2 = 7
[0, 0, 1, 2] 2**0 + 2**0 + 2**1 + 2**2 = 8
(Where double asterisk means 'raised to the power.)
Is there way to get from the argument, straight to the series, without working out all of the previous series?
Update, thanks for the interest.
Here's some Javascript code which generates the sequences, where weight is the sequence:
var numbers = ;
for (var count = 0; count < 20; count++)
for (var i = numbers.length; i > 0; i--)
if (count % (2**i) === 0)
numbers[i] = numbers[i - 1];
numbers[0] = count;
var weights = ;
for (var j = 1; j < numbers.length; j++)
// note the differences below are all exact powers of 2
weights[j] = Math.log2(numbers[j - 1] - numbers[j]);
if (numbers.length > 0)
weights[0] = 0;
console.log(`count $count numbers $numbers weight $weights`);
and the output:
count 0 numbers 0 weight
count 1 numbers 1 weight 0
count 2 numbers 2,1 weight 0,0
count 3 numbers 3,1 weight 0,1
count 4 numbers 4,3,1 weight 0,0,1
count 5 numbers 5,3,1 weight 0,1,1
count 6 numbers 6,5,1 weight 0,0,2
count 7 numbers 7,5,1 weight 0,1,2
count 8 numbers 8,7,5,1 weight 0,0,1,2
count 9 numbers 9,7,5,1 weight 0,1,1,2
count 10 numbers 10,9,5,1 weight 0,0,2,2
count 11 numbers 11,9,5,1 weight 0,1,2,2
count 12 numbers 12,11,9,1 weight 0,0,1,3
count 13 numbers 13,11,9,1 weight 0,1,1,3
count 14 numbers 14,13,9,1 weight 0,0,2,3
count 15 numbers 15,13,9,1 weight 0,1,2,3
count 16 numbers 16,15,13,9,1 weight 0,0,1,2,3
count 17 numbers 17,15,13,9,1 weight 0,1,1,2,3
count 18 numbers 18,17,13,9,1 weight 0,0,2,2,3
count 19 numbers 19,17,13,9,1 weight 0,1,2,2,3
There seems to be an emergent restriction that the values in the Nth place of the series can only take values between 0 and N.
sequences-and-series exponentiation
asked Jul 26 at 12:15
fadedbee
1385
edited Jul 26 at 14:12
asked Jul 26 at 12:15
fadedbee
1385
asked Jul 26 at 12:15
fadedbee
1385
asked Jul 26 at 12:15
fadedbee
1385
1385
closed as unclear what you're asking by Winther, amWhy, Lord Shark the Unknown, Jyrki Lahtonen, José Carlos Santos Jul 27 at 14:34
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Winther, amWhy, Lord Shark the Unknown, Jyrki Lahtonen, José Carlos Santos Jul 27 at 14:34
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
The pattern looks like: if there are no pairs of repeated digits add a zero to the left, if there are double digits increment the right most digit and set the left of the pair to zero. Am I right in thinking then that $f(9)=[0,1,1,2],f(10)=[0,0,2,2]$? I can't see what to then do for f(11) because there are two pairs of repeated digits, it could be $[0,0,0,3]$ or $[0,1,2,2]$, I wouldn't know how to deal with a triple digit, so I guess it would be $[0,1,2,2]$.
– Benedict W. J. Irwin
Jul 26 at 12:59
1
Following that rule through large numbers seem to end in long strings of [...,0,3,0,3,0,3,0,3] etc.
– Benedict W. J. Irwin
Jul 26 at 13:09
1
@BenedictWilliamJohnIrwin Thanks for your interest, I've posted an update.
– fadedbee
Jul 26 at 13:36
@Winther Yes, it is well defined. I've posted a javascript function which generates the series in an update.
– fadedbee
Jul 26 at 14:05
add a comment |Â
1
The pattern looks like: if there are no pairs of repeated digits add a zero to the left, if there are double digits increment the right most digit and set the left of the pair to zero. Am I right in thinking then that $f(9)=[0,1,1,2],f(10)=[0,0,2,2]$? I can't see what to then do for f(11) because there are two pairs of repeated digits, it could be $[0,0,0,3]$ or $[0,1,2,2]$, I wouldn't know how to deal with a triple digit, so I guess it would be $[0,1,2,2]$.
– Benedict W. J. Irwin
Jul 26 at 12:59
1
Following that rule through large numbers seem to end in long strings of [...,0,3,0,3,0,3,0,3] etc.
– Benedict W. J. Irwin
Jul 26 at 13:09
1
@BenedictWilliamJohnIrwin Thanks for your interest, I've posted an update.
– fadedbee
Jul 26 at 13:36
@Winther Yes, it is well defined. I've posted a javascript function which generates the series in an update.
– fadedbee
Jul 26 at 14:05
1
1
The pattern looks like: if there are no pairs of repeated digits add a zero to the left, if there are double digits increment the right most digit and set the left of the pair to zero. Am I right in thinking then that $f(9)=[0,1,1,2],f(10)=[0,0,2,2]$? I can't see what to then do for f(11) because there are two pairs of repeated digits, it could be $[0,0,0,3]$ or $[0,1,2,2]$, I wouldn't know how to deal with a triple digit, so I guess it would be $[0,1,2,2]$.
– Benedict W. J. Irwin
Jul 26 at 12:59
The pattern looks like: if there are no pairs of repeated digits add a zero to the left, if there are double digits increment the right most digit and set the left of the pair to zero. Am I right in thinking then that $f(9)=[0,1,1,2],f(10)=[0,0,2,2]$? I can't see what to then do for f(11) because there are two pairs of repeated digits, it could be $[0,0,0,3]$ or $[0,1,2,2]$, I wouldn't know how to deal with a triple digit, so I guess it would be $[0,1,2,2]$.
– Benedict W. J. Irwin
Jul 26 at 12:59
1
1
Following that rule through large numbers seem to end in long strings of [...,0,3,0,3,0,3,0,3] etc.
– Benedict W. J. Irwin
Jul 26 at 13:09
Following that rule through large numbers seem to end in long strings of [...,0,3,0,3,0,3,0,3] etc.
– Benedict W. J. Irwin
Jul 26 at 13:09
1
1
@BenedictWilliamJohnIrwin Thanks for your interest, I've posted an update.
– fadedbee
Jul 26 at 13:36
@BenedictWilliamJohnIrwin Thanks for your interest, I've posted an update.
– fadedbee
Jul 26 at 13:36
@Winther Yes, it is well defined. I've posted a javascript function which generates the series in an update.
– fadedbee
Jul 26 at 14:05
@Winther Yes, it is well defined. I've posted a javascript function which generates the series in an update.
– fadedbee
Jul 26 at 14:05
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
I have code in PARI/GP which is easy to translate:
f(n) = if( n<=0, , my(m = exponent(n));
vector(m+1, j, my(i = j-2); if(i<0, 0, n(2^i)%2 + i)));
Here $ textttexponent(n) = floor(Log(2,n)). $ Equvalent Mathematica code:
f[n_] := If[ n<=0, , With[m = IntegerLength[n, 2] - 1,
Table[ If[ i<0, 0, Mod[Quotient[n, 2^i], 2] + i], i, -1, m-1]]];
answered Jul 26 at 20:46
Somos
11k1831
Thanks for a great answer. How did you work it out?
– fadedbee
Jul 27 at 9:34
1
I observed the pattern of the first digit (always 0), the 2nd digit (alternating 0,1), and so on. Then I wrote the code to implement those patterns.
– Somos
Jul 27 at 10:33
add a comment |Â
up vote
1
down vote
This is just a translation of @Somos' answer.
function f(n)
if (n <= 0) return ;
const m = Math.log2(n);
for (var j = 0, ret = ; j <= m; j++)
var i = j - 1;
if (i < 0)
ret[j] = 0;
else
ret[j] = Math.floor((n / 2**i) % 2 + i);
return ret;
for (var n = 0; n < 20; n++)
console.log(`n $n, series $f(n)`);
It outputs:
n 0, series
n 1, series 0
n 2, series 0,0
n 3, series 0,1
n 4, series 0,0,1
n 5, series 0,1,1
n 6, series 0,0,2
n 7, series 0,1,2
n 8, series 0,0,1,2
n 9, series 0,1,1,2
n 10, series 0,0,2,2
n 11, series 0,1,2,2
n 12, series 0,0,1,3
n 13, series 0,1,1,3
n 14, series 0,0,2,3
n 15, series 0,1,2,3
n 16, series 0,0,1,2,3
n 17, series 0,1,1,2,3
n 18, series 0,0,2,2,3
n 19, series 0,1,2,2,3
answered Jul 27 at 9:18
fadedbee
1385
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
I have code in PARI/GP which is easy to translate:
f(n) = if( n<=0, , my(m = exponent(n));
vector(m+1, j, my(i = j-2); if(i<0, 0, n(2^i)%2 + i)));
Here $ textttexponent(n) = floor(Log(2,n)). $ Equvalent Mathematica code:
f[n_] := If[ n<=0, , With[m = IntegerLength[n, 2] - 1,
Table[ If[ i<0, 0, Mod[Quotient[n, 2^i], 2] + i], i, -1, m-1]]];
answered Jul 26 at 20:46
Somos
11k1831
Thanks for a great answer. How did you work it out?
– fadedbee
Jul 27 at 9:34
1
I observed the pattern of the first digit (always 0), the 2nd digit (alternating 0,1), and so on. Then I wrote the code to implement those patterns.
– Somos
Jul 27 at 10:33
add a comment |Â
up vote
1
down vote
accepted
I have code in PARI/GP which is easy to translate:
f(n) = if( n<=0, , my(m = exponent(n));
vector(m+1, j, my(i = j-2); if(i<0, 0, n(2^i)%2 + i)));
Here $ textttexponent(n) = floor(Log(2,n)). $ Equvalent Mathematica code:
f[n_] := If[ n<=0, , With[m = IntegerLength[n, 2] - 1,
Table[ If[ i<0, 0, Mod[Quotient[n, 2^i], 2] + i], i, -1, m-1]]];
answered Jul 26 at 20:46
Somos
11k1831
Thanks for a great answer. How did you work it out?
– fadedbee
Jul 27 at 9:34
1
I observed the pattern of the first digit (always 0), the 2nd digit (alternating 0,1), and so on. Then I wrote the code to implement those patterns.
– Somos
Jul 27 at 10:33
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
I have code in PARI/GP which is easy to translate:
f(n) = if( n<=0, , my(m = exponent(n));
vector(m+1, j, my(i = j-2); if(i<0, 0, n(2^i)%2 + i)));
Here $ textttexponent(n) = floor(Log(2,n)). $ Equvalent Mathematica code:
f[n_] := If[ n<=0, , With[m = IntegerLength[n, 2] - 1,
Table[ If[ i<0, 0, Mod[Quotient[n, 2^i], 2] + i], i, -1, m-1]]];
answered Jul 26 at 20:46
Somos
11k1831
I have code in PARI/GP which is easy to translate:
f(n) = if( n<=0, , my(m = exponent(n));
vector(m+1, j, my(i = j-2); if(i<0, 0, n(2^i)%2 + i)));
Here $ textttexponent(n) = floor(Log(2,n)). $ Equvalent Mathematica code:
f[n_] := If[ n<=0, , With[m = IntegerLength[n, 2] - 1,
Table[ If[ i<0, 0, Mod[Quotient[n, 2^i], 2] + i], i, -1, m-1]]];
answered Jul 26 at 20:46
Somos
11k1831
edited Jul 26 at 21:18
answered Jul 26 at 20:46
Somos
11k1831
answered Jul 26 at 20:46
Somos
11k1831
answered Jul 26 at 20:46
Somos
11k1831
11k1831
Thanks for a great answer. How did you work it out?
– fadedbee
Jul 27 at 9:34
1
I observed the pattern of the first digit (always 0), the 2nd digit (alternating 0,1), and so on. Then I wrote the code to implement those patterns.
– Somos
Jul 27 at 10:33
add a comment |Â
Thanks for a great answer. How did you work it out?
– fadedbee
Jul 27 at 9:34
1
I observed the pattern of the first digit (always 0), the 2nd digit (alternating 0,1), and so on. Then I wrote the code to implement those patterns.
– Somos
Jul 27 at 10:33
Thanks for a great answer. How did you work it out?
– fadedbee
Jul 27 at 9:34
Thanks for a great answer. How did you work it out?
– fadedbee
Jul 27 at 9:34
1
1
I observed the pattern of the first digit (always 0), the 2nd digit (alternating 0,1), and so on. Then I wrote the code to implement those patterns.
– Somos
Jul 27 at 10:33
I observed the pattern of the first digit (always 0), the 2nd digit (alternating 0,1), and so on. Then I wrote the code to implement those patterns.
– Somos
Jul 27 at 10:33
add a comment |Â
up vote
1
down vote
This is just a translation of @Somos' answer.
function f(n)
if (n <= 0) return ;
const m = Math.log2(n);
for (var j = 0, ret = ; j <= m; j++)
var i = j - 1;
if (i < 0)
ret[j] = 0;
else
ret[j] = Math.floor((n / 2**i) % 2 + i);
return ret;
for (var n = 0; n < 20; n++)
console.log(`n $n, series $f(n)`);
It outputs:
n 0, series
n 1, series 0
n 2, series 0,0
n 3, series 0,1
n 4, series 0,0,1
n 5, series 0,1,1
n 6, series 0,0,2
n 7, series 0,1,2
n 8, series 0,0,1,2
n 9, series 0,1,1,2
n 10, series 0,0,2,2
n 11, series 0,1,2,2
n 12, series 0,0,1,3
n 13, series 0,1,1,3
n 14, series 0,0,2,3
n 15, series 0,1,2,3
n 16, series 0,0,1,2,3
n 17, series 0,1,1,2,3
n 18, series 0,0,2,2,3
n 19, series 0,1,2,2,3
answered Jul 27 at 9:18
fadedbee
1385
add a comment |Â
up vote
1
down vote
This is just a translation of @Somos' answer.
function f(n)
if (n <= 0) return ;
const m = Math.log2(n);
for (var j = 0, ret = ; j <= m; j++)
var i = j - 1;
if (i < 0)
ret[j] = 0;
else
ret[j] = Math.floor((n / 2**i) % 2 + i);
return ret;
for (var n = 0; n < 20; n++)
console.log(`n $n, series $f(n)`);
It outputs:
n 0, series
n 1, series 0
n 2, series 0,0
n 3, series 0,1
n 4, series 0,0,1
n 5, series 0,1,1
n 6, series 0,0,2
n 7, series 0,1,2
n 8, series 0,0,1,2
n 9, series 0,1,1,2
n 10, series 0,0,2,2
n 11, series 0,1,2,2
n 12, series 0,0,1,3
n 13, series 0,1,1,3
n 14, series 0,0,2,3
n 15, series 0,1,2,3
n 16, series 0,0,1,2,3
n 17, series 0,1,1,2,3
n 18, series 0,0,2,2,3
n 19, series 0,1,2,2,3
answered Jul 27 at 9:18
fadedbee
1385
add a comment |Â
up vote
1
down vote
up vote
1
down vote
This is just a translation of @Somos' answer.
function f(n)
if (n <= 0) return ;
const m = Math.log2(n);
for (var j = 0, ret = ; j <= m; j++)
var i = j - 1;
if (i < 0)
ret[j] = 0;
else
ret[j] = Math.floor((n / 2**i) % 2 + i);
return ret;
for (var n = 0; n < 20; n++)
console.log(`n $n, series $f(n)`);
It outputs:
n 0, series
n 1, series 0
n 2, series 0,0
n 3, series 0,1
n 4, series 0,0,1
n 5, series 0,1,1
n 6, series 0,0,2
n 7, series 0,1,2
n 8, series 0,0,1,2
n 9, series 0,1,1,2
n 10, series 0,0,2,2
n 11, series 0,1,2,2
n 12, series 0,0,1,3
n 13, series 0,1,1,3
n 14, series 0,0,2,3
n 15, series 0,1,2,3
n 16, series 0,0,1,2,3
n 17, series 0,1,1,2,3
n 18, series 0,0,2,2,3
n 19, series 0,1,2,2,3
answered Jul 27 at 9:18
fadedbee
1385
This is just a translation of @Somos' answer.
function f(n)
if (n <= 0) return ;
const m = Math.log2(n);
for (var j = 0, ret = ; j <= m; j++)
var i = j - 1;
if (i < 0)
ret[j] = 0;
else
ret[j] = Math.floor((n / 2**i) % 2 + i);
return ret;
for (var n = 0; n < 20; n++)
console.log(`n $n, series $f(n)`);
It outputs:
n 0, series
n 1, series 0
n 2, series 0,0
n 3, series 0,1
n 4, series 0,0,1
n 5, series 0,1,1
n 6, series 0,0,2
n 7, series 0,1,2
n 8, series 0,0,1,2
n 9, series 0,1,1,2
n 10, series 0,0,2,2
n 11, series 0,1,2,2
n 12, series 0,0,1,3
n 13, series 0,1,1,3
n 14, series 0,0,2,3
n 15, series 0,1,2,3
n 16, series 0,0,1,2,3
n 17, series 0,1,1,2,3
n 18, series 0,0,2,2,3
n 19, series 0,1,2,2,3
answered Jul 27 at 9:18
fadedbee
1385
answered Jul 27 at 9:18
fadedbee
1385
answered Jul 27 at 9:18
fadedbee
1385
answered Jul 27 at 9:18
fadedbee
1385
1385
add a comment |Â
add a comment |Â
1
The pattern looks like: if there are no pairs of repeated digits add a zero to the left, if there are double digits increment the right most digit and set the left of the pair to zero. Am I right in thinking then that $f(9)=[0,1,1,2],f(10)=[0,0,2,2]$? I can't see what to then do for f(11) because there are two pairs of repeated digits, it could be $[0,0,0,3]$ or $[0,1,2,2]$, I wouldn't know how to deal with a triple digit, so I guess it would be $[0,1,2,2]$.
– Benedict W. J. Irwin
Jul 26 at 12:59
1
Following that rule through large numbers seem to end in long strings of [...,0,3,0,3,0,3,0,3] etc.
– Benedict W. J. Irwin
Jul 26 at 13:09
1
@BenedictWilliamJohnIrwin Thanks for your interest, I've posted an update.
– fadedbee
Jul 26 at 13:36
@Winther Yes, it is well defined. I've posted a javascript function which generates the series in an update.
– fadedbee
Jul 26 at 14:05