Finding the function of these numbers $1, 2, 5, 13, 34, 89, 233, 610$
$begingroup$
Firstly I used the differences between them but I found the numbers return again.
How can I find the function of these numbers
calculus
asked Dec 5 '14 at 17:35
E.H.EE.H.E
15.7k11966
$endgroup$
|
show 1 more comment
$begingroup$
Firstly I used the differences between them but I found the numbers return again.
How can I find the function of these numbers
calculus
asked Dec 5 '14 at 17:35
E.H.EE.H.E
15.7k11966
$endgroup$
1
$begingroup$
Think Fibonacci.
$endgroup$
– André Nicolas
Dec 5 '14 at 17:37
$begingroup$
also, maybe under it chart out the partial sums
$endgroup$
– turkeyhundt
Dec 5 '14 at 17:38
$begingroup$
$f(n) = 3f(n-1) - f(n-2)$
$endgroup$
– peterwhy
Dec 5 '14 at 17:39
$begingroup$
It's Pisot sequences.
$endgroup$
– Tacet
Dec 6 '14 at 0:29
$begingroup$
The best way to find a sequence like this is to just put the first whatever terms into oeis.org
$endgroup$
– GFauxPas
Dec 19 '14 at 19:15
|
show 1 more comment
$begingroup$
Firstly I used the differences between them but I found the numbers return again.
How can I find the function of these numbers
calculus
asked Dec 5 '14 at 17:35
E.H.EE.H.E
15.7k11966
$endgroup$
Firstly I used the differences between them but I found the numbers return again.
How can I find the function of these numbers
calculus
calculus
asked Dec 5 '14 at 17:35
E.H.EE.H.E
15.7k11966
asked Dec 5 '14 at 17:35
E.H.EE.H.E
15.7k11966
asked Dec 5 '14 at 17:35
E.H.EE.H.E
15.7k11966
asked Dec 5 '14 at 17:35
E.H.EE.H.E
15.7k11966
asked Dec 5 '14 at 17:35
E.H.EE.H.E
15.7k11966
15.7k11966
1
$begingroup$
Think Fibonacci.
$endgroup$
– André Nicolas
Dec 5 '14 at 17:37
$begingroup$
also, maybe under it chart out the partial sums
$endgroup$
– turkeyhundt
Dec 5 '14 at 17:38
$begingroup$
$f(n) = 3f(n-1) - f(n-2)$
$endgroup$
– peterwhy
Dec 5 '14 at 17:39
$begingroup$
It's Pisot sequences.
$endgroup$
– Tacet
Dec 6 '14 at 0:29
$begingroup$
The best way to find a sequence like this is to just put the first whatever terms into oeis.org
$endgroup$
– GFauxPas
Dec 19 '14 at 19:15
|
show 1 more comment
1
$begingroup$
Think Fibonacci.
$endgroup$
– André Nicolas
Dec 5 '14 at 17:37
$begingroup$
also, maybe under it chart out the partial sums
$endgroup$
– turkeyhundt
Dec 5 '14 at 17:38
$begingroup$
$f(n) = 3f(n-1) - f(n-2)$
$endgroup$
– peterwhy
Dec 5 '14 at 17:39
$begingroup$
It's Pisot sequences.
$endgroup$
– Tacet
Dec 6 '14 at 0:29
$begingroup$
The best way to find a sequence like this is to just put the first whatever terms into oeis.org
$endgroup$
– GFauxPas
Dec 19 '14 at 19:15
1
1
$begingroup$
Think Fibonacci.
$endgroup$
– André Nicolas
Dec 5 '14 at 17:37
$begingroup$
Think Fibonacci.
$endgroup$
– André Nicolas
Dec 5 '14 at 17:37
$begingroup$
also, maybe under it chart out the partial sums
$endgroup$
– turkeyhundt
Dec 5 '14 at 17:38
$begingroup$
also, maybe under it chart out the partial sums
$endgroup$
– turkeyhundt
Dec 5 '14 at 17:38
$begingroup$
$f(n) = 3f(n-1) - f(n-2)$
$endgroup$
– peterwhy
Dec 5 '14 at 17:39
$begingroup$
$f(n) = 3f(n-1) - f(n-2)$
$endgroup$
– peterwhy
Dec 5 '14 at 17:39
$begingroup$
It's Pisot sequences.
$endgroup$
– Tacet
Dec 6 '14 at 0:29
$begingroup$
It's Pisot sequences.
$endgroup$
– Tacet
Dec 6 '14 at 0:29
$begingroup$
The best way to find a sequence like this is to just put the first whatever terms into oeis.org
$endgroup$
– GFauxPas
Dec 19 '14 at 19:15
$begingroup$
The best way to find a sequence like this is to just put the first whatever terms into oeis.org
$endgroup$
– GFauxPas
Dec 19 '14 at 19:15
|
show 1 more comment
6 Answers
6
active
oldest
votes
$begingroup$
Too long for a comment; the method for detecting polynomial sequences is the calculus of differences and well documented. Conway and Guy give (not their invention) a variant that detects sequences based on exponentials, including Fibonacci or every other Fibonacci as you have here. I know i photocopied a few pages giving the technique, I will see if I can find them. Note that the book is extremely informal.
answered Dec 5 '14 at 18:58
Will JagyWill Jagy
102k5101199
$endgroup$
add a comment |
$begingroup$
$a_0=1,a_1=2$ and for $ngeq2$ $a_n=3a_{n-1}-a_{n-2}$
$endgroup$
add a comment |
$begingroup$
Let $n_{1} = 1$. Then,
$n_{k+1} = n_{k}+ sum_{j=1}^{k} n_{j}, $ for $k geq 1$.
answered Dec 5 '14 at 17:39
Alex SilvaAlex Silva
2,70531332
$endgroup$
add a comment |
$begingroup$
1053253
${{
left({1+sqrt5}over2right)^n
-left({1-sqrt5}over2right)^n
}oversqrt5}$, where $n$ is of odd parity
answered Dec 6 '14 at 0:17
Senex Ægypti ParviSenex Ægypti Parvi
2,2031816
$endgroup$
add a comment |
$begingroup$
this is the positive numbers of the Fibonacci numbers in Z.
What I mean:
public class MySeedFibbo {
public static void main(String args) {
List<Integer> init = Arrays.asList(1, 1);
Stream<List<Integer>> negativeFib = Stream.iterate(init, l -> Arrays.asList(l.get(1) - l.get(0),l.get(0)));
negativeFib.limit(15).forEach(l->System.out.println(l.get(1)));
}
}
Result if you run this code in Java SE 8 :
0
1
-1
2
-3
5
-8
13
-21
34
-55
89
-144
233
-377
610
-987
1597
Modify the last line like this:
negativeFib.limit(20).forEach(l->{ if(l.get(1)>0) System.out.println(l.get(1));});
And we get only the positive numbers which form the sequence you're looking for.
answered Jan 9 at 1:34
moldoveanmoldovean
1114
$endgroup$
add a comment |
$begingroup$
I happened to notice each value is twice the previous value plus all the values before that. For example, $$13=2(5)+2+1$$ The sequence has $a_1=1$ and $$a_n=2a_{n-1}+sum_{k=1}^{n-2}a_k$$
But then $$a_{n-1}=2a_{n-2}+sum_{k=1}^{n-3}a_k$$ and subtracting left sides and right sides: $$a_n-a_{n-1}=2a_{n-1}-a_{n-2}$$ So $$a_n=3a_{n-1}-a_{n-2}$$ and $$a_{n-1}=a_{n-1}$$ So $$
begin{align}begin{bmatrix}a_n\ a_{n-1}end{bmatrix}&=begin{bmatrix}3&-1\1&0end{bmatrix}begin{bmatrix}a_{n-1}\ a_{n-2}end{bmatrix}\
&=begin{bmatrix}3&-1\1&0end{bmatrix}^{n-2}begin{bmatrix}a_{2}\ a_{1}end{bmatrix}\
&=begin{bmatrix}3&-1\1&0end{bmatrix}^{n-2}begin{bmatrix}2\ 1end{bmatrix}\
implies a_n&=begin{bmatrix}1&0end{bmatrix}begin{bmatrix}3&-1\1&0end{bmatrix}^{n-2}begin{bmatrix}2\ 1end{bmatrix}\
end{align}$$
If you want, you can diagonalize the matrix and express the formula for $a_n$ as a linear combination of powers of the eigenvalues.
answered Jan 9 at 8:28
alex.jordanalex.jordan
39k560120
$endgroup$
add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Too long for a comment; the method for detecting polynomial sequences is the calculus of differences and well documented. Conway and Guy give (not their invention) a variant that detects sequences based on exponentials, including Fibonacci or every other Fibonacci as you have here. I know i photocopied a few pages giving the technique, I will see if I can find them. Note that the book is extremely informal.
answered Dec 5 '14 at 18:58
Will JagyWill Jagy
102k5101199
$endgroup$
add a comment |
$begingroup$
Too long for a comment; the method for detecting polynomial sequences is the calculus of differences and well documented. Conway and Guy give (not their invention) a variant that detects sequences based on exponentials, including Fibonacci or every other Fibonacci as you have here. I know i photocopied a few pages giving the technique, I will see if I can find them. Note that the book is extremely informal.
answered Dec 5 '14 at 18:58
Will JagyWill Jagy
102k5101199
$endgroup$
add a comment |
$begingroup$
Too long for a comment; the method for detecting polynomial sequences is the calculus of differences and well documented. Conway and Guy give (not their invention) a variant that detects sequences based on exponentials, including Fibonacci or every other Fibonacci as you have here. I know i photocopied a few pages giving the technique, I will see if I can find them. Note that the book is extremely informal.
answered Dec 5 '14 at 18:58
Will JagyWill Jagy
102k5101199
$endgroup$
Too long for a comment; the method for detecting polynomial sequences is the calculus of differences and well documented. Conway and Guy give (not their invention) a variant that detects sequences based on exponentials, including Fibonacci or every other Fibonacci as you have here. I know i photocopied a few pages giving the technique, I will see if I can find them. Note that the book is extremely informal.
answered Dec 5 '14 at 18:58
Will JagyWill Jagy
102k5101199
answered Dec 5 '14 at 18:58
Will JagyWill Jagy
102k5101199
answered Dec 5 '14 at 18:58
Will JagyWill Jagy
102k5101199
answered Dec 5 '14 at 18:58
Will JagyWill Jagy
102k5101199
102k5101199
add a comment |
add a comment |
$begingroup$
$a_0=1,a_1=2$ and for $ngeq2$ $a_n=3a_{n-1}-a_{n-2}$
$endgroup$
add a comment |
$begingroup$
$a_0=1,a_1=2$ and for $ngeq2$ $a_n=3a_{n-1}-a_{n-2}$
$endgroup$
add a comment |
$begingroup$
$a_0=1,a_1=2$ and for $ngeq2$ $a_n=3a_{n-1}-a_{n-2}$
$endgroup$
$a_0=1,a_1=2$ and for $ngeq2$ $a_n=3a_{n-1}-a_{n-2}$
answered Dec 5 '14 at 17:38
user155385
add a comment |
add a comment |
$begingroup$
Let $n_{1} = 1$. Then,
$n_{k+1} = n_{k}+ sum_{j=1}^{k} n_{j}, $ for $k geq 1$.
answered Dec 5 '14 at 17:39
Alex SilvaAlex Silva
2,70531332
$endgroup$
add a comment |
$begingroup$
Let $n_{1} = 1$. Then,
$n_{k+1} = n_{k}+ sum_{j=1}^{k} n_{j}, $ for $k geq 1$.
answered Dec 5 '14 at 17:39
Alex SilvaAlex Silva
2,70531332
$endgroup$
add a comment |
$begingroup$
Let $n_{1} = 1$. Then,
$n_{k+1} = n_{k}+ sum_{j=1}^{k} n_{j}, $ for $k geq 1$.
answered Dec 5 '14 at 17:39
Alex SilvaAlex Silva
2,70531332
$endgroup$
Let $n_{1} = 1$. Then,
$n_{k+1} = n_{k}+ sum_{j=1}^{k} n_{j}, $ for $k geq 1$.
answered Dec 5 '14 at 17:39
Alex SilvaAlex Silva
2,70531332
answered Dec 5 '14 at 17:39
Alex SilvaAlex Silva
2,70531332
answered Dec 5 '14 at 17:39
Alex SilvaAlex Silva
2,70531332
answered Dec 5 '14 at 17:39
Alex SilvaAlex Silva
2,70531332
2,70531332
add a comment |
add a comment |
$begingroup$
1053253
${{
left({1+sqrt5}over2right)^n
-left({1-sqrt5}over2right)^n
}oversqrt5}$, where $n$ is of odd parity
answered Dec 6 '14 at 0:17
Senex Ægypti ParviSenex Ægypti Parvi
2,2031816
$endgroup$
add a comment |
$begingroup$
1053253
${{
left({1+sqrt5}over2right)^n
-left({1-sqrt5}over2right)^n
}oversqrt5}$, where $n$ is of odd parity
answered Dec 6 '14 at 0:17
Senex Ægypti ParviSenex Ægypti Parvi
2,2031816
$endgroup$
add a comment |
$begingroup$
1053253
${{
left({1+sqrt5}over2right)^n
-left({1-sqrt5}over2right)^n
}oversqrt5}$, where $n$ is of odd parity
answered Dec 6 '14 at 0:17
Senex Ægypti ParviSenex Ægypti Parvi
2,2031816
$endgroup$
1053253
${{
left({1+sqrt5}over2right)^n
-left({1-sqrt5}over2right)^n
}oversqrt5}$, where $n$ is of odd parity
answered Dec 6 '14 at 0:17
Senex Ægypti ParviSenex Ægypti Parvi
2,2031816
answered Dec 6 '14 at 0:17
Senex Ægypti ParviSenex Ægypti Parvi
2,2031816
answered Dec 6 '14 at 0:17
Senex Ægypti ParviSenex Ægypti Parvi
2,2031816
answered Dec 6 '14 at 0:17
Senex Ægypti ParviSenex Ægypti Parvi
2,2031816
2,2031816
add a comment |
add a comment |
$begingroup$
this is the positive numbers of the Fibonacci numbers in Z.
What I mean:
public class MySeedFibbo {
public static void main(String args) {
List<Integer> init = Arrays.asList(1, 1);
Stream<List<Integer>> negativeFib = Stream.iterate(init, l -> Arrays.asList(l.get(1) - l.get(0),l.get(0)));
negativeFib.limit(15).forEach(l->System.out.println(l.get(1)));
}
}
Result if you run this code in Java SE 8 :
0
1
-1
2
-3
5
-8
13
-21
34
-55
89
-144
233
-377
610
-987
1597
Modify the last line like this:
negativeFib.limit(20).forEach(l->{ if(l.get(1)>0) System.out.println(l.get(1));});
And we get only the positive numbers which form the sequence you're looking for.
answered Jan 9 at 1:34
moldoveanmoldovean
1114
$endgroup$
add a comment |
$begingroup$
this is the positive numbers of the Fibonacci numbers in Z.
What I mean:
public class MySeedFibbo {
public static void main(String args) {
List<Integer> init = Arrays.asList(1, 1);
Stream<List<Integer>> negativeFib = Stream.iterate(init, l -> Arrays.asList(l.get(1) - l.get(0),l.get(0)));
negativeFib.limit(15).forEach(l->System.out.println(l.get(1)));
}
}
Result if you run this code in Java SE 8 :
0
1
-1
2
-3
5
-8
13
-21
34
-55
89
-144
233
-377
610
-987
1597
Modify the last line like this:
negativeFib.limit(20).forEach(l->{ if(l.get(1)>0) System.out.println(l.get(1));});
And we get only the positive numbers which form the sequence you're looking for.
answered Jan 9 at 1:34
moldoveanmoldovean
1114
$endgroup$
add a comment |
$begingroup$
this is the positive numbers of the Fibonacci numbers in Z.
What I mean:
public class MySeedFibbo {
public static void main(String args) {
List<Integer> init = Arrays.asList(1, 1);
Stream<List<Integer>> negativeFib = Stream.iterate(init, l -> Arrays.asList(l.get(1) - l.get(0),l.get(0)));
negativeFib.limit(15).forEach(l->System.out.println(l.get(1)));
}
}
Result if you run this code in Java SE 8 :
0
1
-1
2
-3
5
-8
13
-21
34
-55
89
-144
233
-377
610
-987
1597
Modify the last line like this:
negativeFib.limit(20).forEach(l->{ if(l.get(1)>0) System.out.println(l.get(1));});
And we get only the positive numbers which form the sequence you're looking for.
answered Jan 9 at 1:34
moldoveanmoldovean
1114
$endgroup$
this is the positive numbers of the Fibonacci numbers in Z.
What I mean:
public class MySeedFibbo {
public static void main(String args) {
List<Integer> init = Arrays.asList(1, 1);
Stream<List<Integer>> negativeFib = Stream.iterate(init, l -> Arrays.asList(l.get(1) - l.get(0),l.get(0)));
negativeFib.limit(15).forEach(l->System.out.println(l.get(1)));
}
}
Result if you run this code in Java SE 8 :
0
1
-1
2
-3
5
-8
13
-21
34
-55
89
-144
233
-377
610
-987
1597
Modify the last line like this:
negativeFib.limit(20).forEach(l->{ if(l.get(1)>0) System.out.println(l.get(1));});
And we get only the positive numbers which form the sequence you're looking for.
answered Jan 9 at 1:34
moldoveanmoldovean
1114
edited Jan 9 at 1:42
answered Jan 9 at 1:34
moldoveanmoldovean
1114
answered Jan 9 at 1:34
moldoveanmoldovean
1114
answered Jan 9 at 1:34
moldoveanmoldovean
1114
1114
add a comment |
add a comment |
$begingroup$
I happened to notice each value is twice the previous value plus all the values before that. For example, $$13=2(5)+2+1$$ The sequence has $a_1=1$ and $$a_n=2a_{n-1}+sum_{k=1}^{n-2}a_k$$
But then $$a_{n-1}=2a_{n-2}+sum_{k=1}^{n-3}a_k$$ and subtracting left sides and right sides: $$a_n-a_{n-1}=2a_{n-1}-a_{n-2}$$ So $$a_n=3a_{n-1}-a_{n-2}$$ and $$a_{n-1}=a_{n-1}$$ So $$
begin{align}begin{bmatrix}a_n\ a_{n-1}end{bmatrix}&=begin{bmatrix}3&-1\1&0end{bmatrix}begin{bmatrix}a_{n-1}\ a_{n-2}end{bmatrix}\
&=begin{bmatrix}3&-1\1&0end{bmatrix}^{n-2}begin{bmatrix}a_{2}\ a_{1}end{bmatrix}\
&=begin{bmatrix}3&-1\1&0end{bmatrix}^{n-2}begin{bmatrix}2\ 1end{bmatrix}\
implies a_n&=begin{bmatrix}1&0end{bmatrix}begin{bmatrix}3&-1\1&0end{bmatrix}^{n-2}begin{bmatrix}2\ 1end{bmatrix}\
end{align}$$
If you want, you can diagonalize the matrix and express the formula for $a_n$ as a linear combination of powers of the eigenvalues.
answered Jan 9 at 8:28
alex.jordanalex.jordan
39k560120
$endgroup$
add a comment |
$begingroup$
I happened to notice each value is twice the previous value plus all the values before that. For example, $$13=2(5)+2+1$$ The sequence has $a_1=1$ and $$a_n=2a_{n-1}+sum_{k=1}^{n-2}a_k$$
But then $$a_{n-1}=2a_{n-2}+sum_{k=1}^{n-3}a_k$$ and subtracting left sides and right sides: $$a_n-a_{n-1}=2a_{n-1}-a_{n-2}$$ So $$a_n=3a_{n-1}-a_{n-2}$$ and $$a_{n-1}=a_{n-1}$$ So $$
begin{align}begin{bmatrix}a_n\ a_{n-1}end{bmatrix}&=begin{bmatrix}3&-1\1&0end{bmatrix}begin{bmatrix}a_{n-1}\ a_{n-2}end{bmatrix}\
&=begin{bmatrix}3&-1\1&0end{bmatrix}^{n-2}begin{bmatrix}a_{2}\ a_{1}end{bmatrix}\
&=begin{bmatrix}3&-1\1&0end{bmatrix}^{n-2}begin{bmatrix}2\ 1end{bmatrix}\
implies a_n&=begin{bmatrix}1&0end{bmatrix}begin{bmatrix}3&-1\1&0end{bmatrix}^{n-2}begin{bmatrix}2\ 1end{bmatrix}\
end{align}$$
If you want, you can diagonalize the matrix and express the formula for $a_n$ as a linear combination of powers of the eigenvalues.
answered Jan 9 at 8:28
alex.jordanalex.jordan
39k560120
$endgroup$
add a comment |
$begingroup$
I happened to notice each value is twice the previous value plus all the values before that. For example, $$13=2(5)+2+1$$ The sequence has $a_1=1$ and $$a_n=2a_{n-1}+sum_{k=1}^{n-2}a_k$$
But then $$a_{n-1}=2a_{n-2}+sum_{k=1}^{n-3}a_k$$ and subtracting left sides and right sides: $$a_n-a_{n-1}=2a_{n-1}-a_{n-2}$$ So $$a_n=3a_{n-1}-a_{n-2}$$ and $$a_{n-1}=a_{n-1}$$ So $$
begin{align}begin{bmatrix}a_n\ a_{n-1}end{bmatrix}&=begin{bmatrix}3&-1\1&0end{bmatrix}begin{bmatrix}a_{n-1}\ a_{n-2}end{bmatrix}\
&=begin{bmatrix}3&-1\1&0end{bmatrix}^{n-2}begin{bmatrix}a_{2}\ a_{1}end{bmatrix}\
&=begin{bmatrix}3&-1\1&0end{bmatrix}^{n-2}begin{bmatrix}2\ 1end{bmatrix}\
implies a_n&=begin{bmatrix}1&0end{bmatrix}begin{bmatrix}3&-1\1&0end{bmatrix}^{n-2}begin{bmatrix}2\ 1end{bmatrix}\
end{align}$$
If you want, you can diagonalize the matrix and express the formula for $a_n$ as a linear combination of powers of the eigenvalues.
answered Jan 9 at 8:28
alex.jordanalex.jordan
39k560120
$endgroup$
I happened to notice each value is twice the previous value plus all the values before that. For example, $$13=2(5)+2+1$$ The sequence has $a_1=1$ and $$a_n=2a_{n-1}+sum_{k=1}^{n-2}a_k$$
But then $$a_{n-1}=2a_{n-2}+sum_{k=1}^{n-3}a_k$$ and subtracting left sides and right sides: $$a_n-a_{n-1}=2a_{n-1}-a_{n-2}$$ So $$a_n=3a_{n-1}-a_{n-2}$$ and $$a_{n-1}=a_{n-1}$$ So $$
begin{align}begin{bmatrix}a_n\ a_{n-1}end{bmatrix}&=begin{bmatrix}3&-1\1&0end{bmatrix}begin{bmatrix}a_{n-1}\ a_{n-2}end{bmatrix}\
&=begin{bmatrix}3&-1\1&0end{bmatrix}^{n-2}begin{bmatrix}a_{2}\ a_{1}end{bmatrix}\
&=begin{bmatrix}3&-1\1&0end{bmatrix}^{n-2}begin{bmatrix}2\ 1end{bmatrix}\
implies a_n&=begin{bmatrix}1&0end{bmatrix}begin{bmatrix}3&-1\1&0end{bmatrix}^{n-2}begin{bmatrix}2\ 1end{bmatrix}\
end{align}$$
If you want, you can diagonalize the matrix and express the formula for $a_n$ as a linear combination of powers of the eigenvalues.
answered Jan 9 at 8:28
alex.jordanalex.jordan
39k560120
answered Jan 9 at 8:28
alex.jordanalex.jordan
39k560120
answered Jan 9 at 8:28
alex.jordanalex.jordan
39k560120
answered Jan 9 at 8:28
alex.jordanalex.jordan
39k560120
39k560120
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1
$begingroup$
Think Fibonacci.
$endgroup$
– André Nicolas
Dec 5 '14 at 17:37
$begingroup$
also, maybe under it chart out the partial sums
$endgroup$
– turkeyhundt
Dec 5 '14 at 17:38
$begingroup$
$f(n) = 3f(n-1) - f(n-2)$
$endgroup$
– peterwhy
Dec 5 '14 at 17:39
$begingroup$
It's Pisot sequences.
$endgroup$
– Tacet
Dec 6 '14 at 0:29
$begingroup$
The best way to find a sequence like this is to just put the first whatever terms into oeis.org
$endgroup$
– GFauxPas
Dec 19 '14 at 19:15