Why is the 0th term of the padovan sequence 1 but for fibonacci it is 0?
Clash Royale CLAN TAG#URR8PPP
$begingroup$
padovan numbers are:
P(0)=1, P(1)=1, P(2)=1, P(3)=2, P(4)=2, P(5)=3, 4, 5, 7, 9, 12, 16, 21, etc
WHERE P(n) = P(n-2) + P(n-3)
fibonacci numbers are:
F(0)=0, F(1)=1, F(2)=1, F(3)=2, F(4)=3, F(5)=5, 8, 13, 21, etc
WHERE F(n) = F(n-1) + F(n-2)
Question
My question is, why not use P(0)=0 as follows:
P(0)=0, P(1)=1, P(2)=1, P(3)=1, P(4)=2, P(5)=2, P(6)=3, 4, 5, 7, 9, 12, 16, 21, etc
Since this satisfies P(n) = P(n-2) + P(n-3)
Also, when looking at the Fibonacci squares we see the first visible term of the sequence is F(1)
With the Padovan triangles the first visible term of the sequence is P(0)?
Which seems inconsistent to me. Can anyone give a mathematical explanation as to why P(0)=1 ... many thanks
sequences-and-series fibonacci-numbers
$endgroup$
add a comment |
$begingroup$
padovan numbers are:
P(0)=1, P(1)=1, P(2)=1, P(3)=2, P(4)=2, P(5)=3, 4, 5, 7, 9, 12, 16, 21, etc
WHERE P(n) = P(n-2) + P(n-3)
fibonacci numbers are:
F(0)=0, F(1)=1, F(2)=1, F(3)=2, F(4)=3, F(5)=5, 8, 13, 21, etc
WHERE F(n) = F(n-1) + F(n-2)
Question
My question is, why not use P(0)=0 as follows:
P(0)=0, P(1)=1, P(2)=1, P(3)=1, P(4)=2, P(5)=2, P(6)=3, 4, 5, 7, 9, 12, 16, 21, etc
Since this satisfies P(n) = P(n-2) + P(n-3)
Also, when looking at the Fibonacci squares we see the first visible term of the sequence is F(1)
With the Padovan triangles the first visible term of the sequence is P(0)?
Which seems inconsistent to me. Can anyone give a mathematical explanation as to why P(0)=1 ... many thanks
sequences-and-series fibonacci-numbers
$endgroup$
add a comment |
$begingroup$
padovan numbers are:
P(0)=1, P(1)=1, P(2)=1, P(3)=2, P(4)=2, P(5)=3, 4, 5, 7, 9, 12, 16, 21, etc
WHERE P(n) = P(n-2) + P(n-3)
fibonacci numbers are:
F(0)=0, F(1)=1, F(2)=1, F(3)=2, F(4)=3, F(5)=5, 8, 13, 21, etc
WHERE F(n) = F(n-1) + F(n-2)
Question
My question is, why not use P(0)=0 as follows:
P(0)=0, P(1)=1, P(2)=1, P(3)=1, P(4)=2, P(5)=2, P(6)=3, 4, 5, 7, 9, 12, 16, 21, etc
Since this satisfies P(n) = P(n-2) + P(n-3)
Also, when looking at the Fibonacci squares we see the first visible term of the sequence is F(1)
With the Padovan triangles the first visible term of the sequence is P(0)?
Which seems inconsistent to me. Can anyone give a mathematical explanation as to why P(0)=1 ... many thanks
sequences-and-series fibonacci-numbers
$endgroup$
padovan numbers are:
P(0)=1, P(1)=1, P(2)=1, P(3)=2, P(4)=2, P(5)=3, 4, 5, 7, 9, 12, 16, 21, etc
WHERE P(n) = P(n-2) + P(n-3)
fibonacci numbers are:
F(0)=0, F(1)=1, F(2)=1, F(3)=2, F(4)=3, F(5)=5, 8, 13, 21, etc
WHERE F(n) = F(n-1) + F(n-2)
Question
My question is, why not use P(0)=0 as follows:
P(0)=0, P(1)=1, P(2)=1, P(3)=1, P(4)=2, P(5)=2, P(6)=3, 4, 5, 7, 9, 12, 16, 21, etc
Since this satisfies P(n) = P(n-2) + P(n-3)
Also, when looking at the Fibonacci squares we see the first visible term of the sequence is F(1)
With the Padovan triangles the first visible term of the sequence is P(0)?
Which seems inconsistent to me. Can anyone give a mathematical explanation as to why P(0)=1 ... many thanks
sequences-and-series fibonacci-numbers
sequences-and-series fibonacci-numbers
asked Jan 1 at 23:24
danday74danday74
1296
1296
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The OEIS sequence A000931 is "Padovan sequence (or Padovan numbers): a(n) = a(n-2) + a(n-3) with a(0)=1, a(1)=a(2)=0." In contrast, your sequence $P(n) = A000931(n+5) = A134816(n+1).$ The OEIS entry also states
The following are basically all variants of the same sequence: A000931, A078027, A096231, A124745, A133034, A134816, A164001, A182097, A228361 and probably A020720. However, each one has its own special features and deserves its own entry.
The point is that it is a matter of convenience and choice of where to start the sequence and what index to use. The same holds for the Fibonacci numbers. Some people choose $F(0)=F(1)=1$. About your specific choice of $P(0)=P(1)=P(2)=1,$ that is exactly A134816 except offset differs by 1. You are welcome to submit your different offset sequence to the OEIS, but unlikely to succeed.
However, the English Wikipedia article Padovan sequence uses your notation, initial values, and contains the spiral of triangles. It mentions the OEIS sequence A000931 but neglects to mention the indexing difference.
$endgroup$
add a comment |
$begingroup$
Recurrence relations like those allow for going backwards; for instance we can define $F(-1)$ by $F(1)=F(0)+F(-1)$, so
$$
F(-1)=F(1)-F(0)=1
$$
Similarly, $F(-2)=F(0)-F(-1)=-1$ and so on. If we translate indices, we get a new Fibonacci-like sequence
$$
F'(0)=1,quad F'(1)=0,quad F'(n+2)=F'(n+1)+F'(n)
$$
and another one if we start from $-2$ and translate indices by $2$:
$$
F''(0)=-1,quad F'(1)=1,quad F''(n+2)=F''(n+1)+F''(n)
$$
If we apply the same idea to the Padovan sequence, we need $P(2)=P(0)+P(-1)$, so $P(-1)=P(2)-P(0)=0$ and we could define
$$
P'(0)=0,quad P'(1)=1,quad P'(2)=1,quad P'(n+3)=P'(n+1)+P'(n)
$$
Nothing different and no real mathematical explanation. Just history.
$endgroup$
4
$begingroup$
Note that F(-n) = (-1)^n * F(n) when n>=0, which makes the choice F(0) = 0 more natural, as it makes the sequence symmetrical (apart from the sign).
$endgroup$
– Sjoerd
Jan 2 at 1:11
1
$begingroup$
@Sjoerd I'm inclined to think this stems from usual problems in considering where to start sequences; it's rather like a religious war between those who consider 0 a natural number and those who don't.
$endgroup$
– egreg
Jan 2 at 10:10
add a comment |
$begingroup$
That's cute. The sequence also follows
$$ P(n) = P(n-1) + P(n-5) ; , ; $$
which directly explains the diagram of triangles. For example, from the diagram, $12 = 9+3$ and $16=12+4.$
Go Figure.
$$ left( x^5 - x^4 - 1 right) $$
$$ left( x^3 - x - 1 right) $$
$$ left( x^5 - x^4 - 1 right) = left( x^3 - x - 1 right) cdot colormagenta left( x^2 - x + 1 right) $$
$endgroup$
$begingroup$
But this is true regardless of the indexing, what's the point?
$endgroup$
– user202729
Jan 2 at 8:58
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The OEIS sequence A000931 is "Padovan sequence (or Padovan numbers): a(n) = a(n-2) + a(n-3) with a(0)=1, a(1)=a(2)=0." In contrast, your sequence $P(n) = A000931(n+5) = A134816(n+1).$ The OEIS entry also states
The following are basically all variants of the same sequence: A000931, A078027, A096231, A124745, A133034, A134816, A164001, A182097, A228361 and probably A020720. However, each one has its own special features and deserves its own entry.
The point is that it is a matter of convenience and choice of where to start the sequence and what index to use. The same holds for the Fibonacci numbers. Some people choose $F(0)=F(1)=1$. About your specific choice of $P(0)=P(1)=P(2)=1,$ that is exactly A134816 except offset differs by 1. You are welcome to submit your different offset sequence to the OEIS, but unlikely to succeed.
However, the English Wikipedia article Padovan sequence uses your notation, initial values, and contains the spiral of triangles. It mentions the OEIS sequence A000931 but neglects to mention the indexing difference.
$endgroup$
add a comment |
$begingroup$
The OEIS sequence A000931 is "Padovan sequence (or Padovan numbers): a(n) = a(n-2) + a(n-3) with a(0)=1, a(1)=a(2)=0." In contrast, your sequence $P(n) = A000931(n+5) = A134816(n+1).$ The OEIS entry also states
The following are basically all variants of the same sequence: A000931, A078027, A096231, A124745, A133034, A134816, A164001, A182097, A228361 and probably A020720. However, each one has its own special features and deserves its own entry.
The point is that it is a matter of convenience and choice of where to start the sequence and what index to use. The same holds for the Fibonacci numbers. Some people choose $F(0)=F(1)=1$. About your specific choice of $P(0)=P(1)=P(2)=1,$ that is exactly A134816 except offset differs by 1. You are welcome to submit your different offset sequence to the OEIS, but unlikely to succeed.
However, the English Wikipedia article Padovan sequence uses your notation, initial values, and contains the spiral of triangles. It mentions the OEIS sequence A000931 but neglects to mention the indexing difference.
$endgroup$
add a comment |
$begingroup$
The OEIS sequence A000931 is "Padovan sequence (or Padovan numbers): a(n) = a(n-2) + a(n-3) with a(0)=1, a(1)=a(2)=0." In contrast, your sequence $P(n) = A000931(n+5) = A134816(n+1).$ The OEIS entry also states
The following are basically all variants of the same sequence: A000931, A078027, A096231, A124745, A133034, A134816, A164001, A182097, A228361 and probably A020720. However, each one has its own special features and deserves its own entry.
The point is that it is a matter of convenience and choice of where to start the sequence and what index to use. The same holds for the Fibonacci numbers. Some people choose $F(0)=F(1)=1$. About your specific choice of $P(0)=P(1)=P(2)=1,$ that is exactly A134816 except offset differs by 1. You are welcome to submit your different offset sequence to the OEIS, but unlikely to succeed.
However, the English Wikipedia article Padovan sequence uses your notation, initial values, and contains the spiral of triangles. It mentions the OEIS sequence A000931 but neglects to mention the indexing difference.
$endgroup$
The OEIS sequence A000931 is "Padovan sequence (or Padovan numbers): a(n) = a(n-2) + a(n-3) with a(0)=1, a(1)=a(2)=0." In contrast, your sequence $P(n) = A000931(n+5) = A134816(n+1).$ The OEIS entry also states
The following are basically all variants of the same sequence: A000931, A078027, A096231, A124745, A133034, A134816, A164001, A182097, A228361 and probably A020720. However, each one has its own special features and deserves its own entry.
The point is that it is a matter of convenience and choice of where to start the sequence and what index to use. The same holds for the Fibonacci numbers. Some people choose $F(0)=F(1)=1$. About your specific choice of $P(0)=P(1)=P(2)=1,$ that is exactly A134816 except offset differs by 1. You are welcome to submit your different offset sequence to the OEIS, but unlikely to succeed.
However, the English Wikipedia article Padovan sequence uses your notation, initial values, and contains the spiral of triangles. It mentions the OEIS sequence A000931 but neglects to mention the indexing difference.
edited Jan 2 at 3:38
answered Jan 1 at 23:53
SomosSomos
13.1k11034
13.1k11034
add a comment |
add a comment |
$begingroup$
Recurrence relations like those allow for going backwards; for instance we can define $F(-1)$ by $F(1)=F(0)+F(-1)$, so
$$
F(-1)=F(1)-F(0)=1
$$
Similarly, $F(-2)=F(0)-F(-1)=-1$ and so on. If we translate indices, we get a new Fibonacci-like sequence
$$
F'(0)=1,quad F'(1)=0,quad F'(n+2)=F'(n+1)+F'(n)
$$
and another one if we start from $-2$ and translate indices by $2$:
$$
F''(0)=-1,quad F'(1)=1,quad F''(n+2)=F''(n+1)+F''(n)
$$
If we apply the same idea to the Padovan sequence, we need $P(2)=P(0)+P(-1)$, so $P(-1)=P(2)-P(0)=0$ and we could define
$$
P'(0)=0,quad P'(1)=1,quad P'(2)=1,quad P'(n+3)=P'(n+1)+P'(n)
$$
Nothing different and no real mathematical explanation. Just history.
$endgroup$
4
$begingroup$
Note that F(-n) = (-1)^n * F(n) when n>=0, which makes the choice F(0) = 0 more natural, as it makes the sequence symmetrical (apart from the sign).
$endgroup$
– Sjoerd
Jan 2 at 1:11
1
$begingroup$
@Sjoerd I'm inclined to think this stems from usual problems in considering where to start sequences; it's rather like a religious war between those who consider 0 a natural number and those who don't.
$endgroup$
– egreg
Jan 2 at 10:10
add a comment |
$begingroup$
Recurrence relations like those allow for going backwards; for instance we can define $F(-1)$ by $F(1)=F(0)+F(-1)$, so
$$
F(-1)=F(1)-F(0)=1
$$
Similarly, $F(-2)=F(0)-F(-1)=-1$ and so on. If we translate indices, we get a new Fibonacci-like sequence
$$
F'(0)=1,quad F'(1)=0,quad F'(n+2)=F'(n+1)+F'(n)
$$
and another one if we start from $-2$ and translate indices by $2$:
$$
F''(0)=-1,quad F'(1)=1,quad F''(n+2)=F''(n+1)+F''(n)
$$
If we apply the same idea to the Padovan sequence, we need $P(2)=P(0)+P(-1)$, so $P(-1)=P(2)-P(0)=0$ and we could define
$$
P'(0)=0,quad P'(1)=1,quad P'(2)=1,quad P'(n+3)=P'(n+1)+P'(n)
$$
Nothing different and no real mathematical explanation. Just history.
$endgroup$
4
$begingroup$
Note that F(-n) = (-1)^n * F(n) when n>=0, which makes the choice F(0) = 0 more natural, as it makes the sequence symmetrical (apart from the sign).
$endgroup$
– Sjoerd
Jan 2 at 1:11
1
$begingroup$
@Sjoerd I'm inclined to think this stems from usual problems in considering where to start sequences; it's rather like a religious war between those who consider 0 a natural number and those who don't.
$endgroup$
– egreg
Jan 2 at 10:10
add a comment |
$begingroup$
Recurrence relations like those allow for going backwards; for instance we can define $F(-1)$ by $F(1)=F(0)+F(-1)$, so
$$
F(-1)=F(1)-F(0)=1
$$
Similarly, $F(-2)=F(0)-F(-1)=-1$ and so on. If we translate indices, we get a new Fibonacci-like sequence
$$
F'(0)=1,quad F'(1)=0,quad F'(n+2)=F'(n+1)+F'(n)
$$
and another one if we start from $-2$ and translate indices by $2$:
$$
F''(0)=-1,quad F'(1)=1,quad F''(n+2)=F''(n+1)+F''(n)
$$
If we apply the same idea to the Padovan sequence, we need $P(2)=P(0)+P(-1)$, so $P(-1)=P(2)-P(0)=0$ and we could define
$$
P'(0)=0,quad P'(1)=1,quad P'(2)=1,quad P'(n+3)=P'(n+1)+P'(n)
$$
Nothing different and no real mathematical explanation. Just history.
$endgroup$
Recurrence relations like those allow for going backwards; for instance we can define $F(-1)$ by $F(1)=F(0)+F(-1)$, so
$$
F(-1)=F(1)-F(0)=1
$$
Similarly, $F(-2)=F(0)-F(-1)=-1$ and so on. If we translate indices, we get a new Fibonacci-like sequence
$$
F'(0)=1,quad F'(1)=0,quad F'(n+2)=F'(n+1)+F'(n)
$$
and another one if we start from $-2$ and translate indices by $2$:
$$
F''(0)=-1,quad F'(1)=1,quad F''(n+2)=F''(n+1)+F''(n)
$$
If we apply the same idea to the Padovan sequence, we need $P(2)=P(0)+P(-1)$, so $P(-1)=P(2)-P(0)=0$ and we could define
$$
P'(0)=0,quad P'(1)=1,quad P'(2)=1,quad P'(n+3)=P'(n+1)+P'(n)
$$
Nothing different and no real mathematical explanation. Just history.
answered Jan 1 at 23:52
egregegreg
180k1485202
180k1485202
4
$begingroup$
Note that F(-n) = (-1)^n * F(n) when n>=0, which makes the choice F(0) = 0 more natural, as it makes the sequence symmetrical (apart from the sign).
$endgroup$
– Sjoerd
Jan 2 at 1:11
1
$begingroup$
@Sjoerd I'm inclined to think this stems from usual problems in considering where to start sequences; it's rather like a religious war between those who consider 0 a natural number and those who don't.
$endgroup$
– egreg
Jan 2 at 10:10
add a comment |
4
$begingroup$
Note that F(-n) = (-1)^n * F(n) when n>=0, which makes the choice F(0) = 0 more natural, as it makes the sequence symmetrical (apart from the sign).
$endgroup$
– Sjoerd
Jan 2 at 1:11
1
$begingroup$
@Sjoerd I'm inclined to think this stems from usual problems in considering where to start sequences; it's rather like a religious war between those who consider 0 a natural number and those who don't.
$endgroup$
– egreg
Jan 2 at 10:10
4
4
$begingroup$
Note that F(-n) = (-1)^n * F(n) when n>=0, which makes the choice F(0) = 0 more natural, as it makes the sequence symmetrical (apart from the sign).
$endgroup$
– Sjoerd
Jan 2 at 1:11
$begingroup$
Note that F(-n) = (-1)^n * F(n) when n>=0, which makes the choice F(0) = 0 more natural, as it makes the sequence symmetrical (apart from the sign).
$endgroup$
– Sjoerd
Jan 2 at 1:11
1
1
$begingroup$
@Sjoerd I'm inclined to think this stems from usual problems in considering where to start sequences; it's rather like a religious war between those who consider 0 a natural number and those who don't.
$endgroup$
– egreg
Jan 2 at 10:10
$begingroup$
@Sjoerd I'm inclined to think this stems from usual problems in considering where to start sequences; it's rather like a religious war between those who consider 0 a natural number and those who don't.
$endgroup$
– egreg
Jan 2 at 10:10
add a comment |
$begingroup$
That's cute. The sequence also follows
$$ P(n) = P(n-1) + P(n-5) ; , ; $$
which directly explains the diagram of triangles. For example, from the diagram, $12 = 9+3$ and $16=12+4.$
Go Figure.
$$ left( x^5 - x^4 - 1 right) $$
$$ left( x^3 - x - 1 right) $$
$$ left( x^5 - x^4 - 1 right) = left( x^3 - x - 1 right) cdot colormagenta left( x^2 - x + 1 right) $$
$endgroup$
$begingroup$
But this is true regardless of the indexing, what's the point?
$endgroup$
– user202729
Jan 2 at 8:58
add a comment |
$begingroup$
That's cute. The sequence also follows
$$ P(n) = P(n-1) + P(n-5) ; , ; $$
which directly explains the diagram of triangles. For example, from the diagram, $12 = 9+3$ and $16=12+4.$
Go Figure.
$$ left( x^5 - x^4 - 1 right) $$
$$ left( x^3 - x - 1 right) $$
$$ left( x^5 - x^4 - 1 right) = left( x^3 - x - 1 right) cdot colormagenta left( x^2 - x + 1 right) $$
$endgroup$
$begingroup$
But this is true regardless of the indexing, what's the point?
$endgroup$
– user202729
Jan 2 at 8:58
add a comment |
$begingroup$
That's cute. The sequence also follows
$$ P(n) = P(n-1) + P(n-5) ; , ; $$
which directly explains the diagram of triangles. For example, from the diagram, $12 = 9+3$ and $16=12+4.$
Go Figure.
$$ left( x^5 - x^4 - 1 right) $$
$$ left( x^3 - x - 1 right) $$
$$ left( x^5 - x^4 - 1 right) = left( x^3 - x - 1 right) cdot colormagenta left( x^2 - x + 1 right) $$
$endgroup$
That's cute. The sequence also follows
$$ P(n) = P(n-1) + P(n-5) ; , ; $$
which directly explains the diagram of triangles. For example, from the diagram, $12 = 9+3$ and $16=12+4.$
Go Figure.
$$ left( x^5 - x^4 - 1 right) $$
$$ left( x^3 - x - 1 right) $$
$$ left( x^5 - x^4 - 1 right) = left( x^3 - x - 1 right) cdot colormagenta left( x^2 - x + 1 right) $$
edited Jan 2 at 1:57
answered Jan 2 at 1:50
Will JagyWill Jagy
102k5101199
102k5101199
$begingroup$
But this is true regardless of the indexing, what's the point?
$endgroup$
– user202729
Jan 2 at 8:58
add a comment |
$begingroup$
But this is true regardless of the indexing, what's the point?
$endgroup$
– user202729
Jan 2 at 8:58
$begingroup$
But this is true regardless of the indexing, what's the point?
$endgroup$
– user202729
Jan 2 at 8:58
$begingroup$
But this is true regardless of the indexing, what's the point?
$endgroup$
– user202729
Jan 2 at 8:58
add a comment |
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StackExchange.helpers.onClickDraftSave('#login-link');
);
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StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
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Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown