Why is the 0th term of the padovan sequence 1 but for fibonacci it is 0?
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
asked 1 hour ago
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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
asked 1 hour ago
danday74
1093
New contributor
danday74 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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
asked 1 hour ago
danday74
1093
New contributor
danday74 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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 1 hour ago
danday74
1093
New contributor
danday74 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
danday74
1093
New contributor
danday74 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
danday74
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Check out our Code of Conduct.
asked 1 hour ago
danday74
1093
asked 1 hour ago
danday74
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1093
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danday74 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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2 Answers
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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.
answered 1 hour ago
Somos
12.9k11034
add a comment |
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 1 hour ago
egreg
178k1484201
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).
– Sjoerd
3 mins ago
add a comment |
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2 Answers
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2 Answers
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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.
answered 1 hour ago
Somos
12.9k11034
add a comment |
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.
answered 1 hour ago
Somos
12.9k11034
add a comment |
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.
answered 1 hour ago
Somos
12.9k11034
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.
answered 1 hour ago
Somos
12.9k11034
edited 38 mins ago
answered 1 hour ago
Somos
12.9k11034
answered 1 hour ago
Somos
12.9k11034
answered 1 hour ago
Somos
12.9k11034
12.9k11034
add a comment |
add a comment |
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 1 hour ago
egreg
178k1484201
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).
– Sjoerd
3 mins ago
add a comment |
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 1 hour ago
egreg
178k1484201
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).
– Sjoerd
3 mins ago
add a comment |
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 1 hour ago
egreg
178k1484201
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 1 hour ago
egreg
178k1484201
answered 1 hour ago
egreg
178k1484201
answered 1 hour ago
egreg
178k1484201
answered 1 hour ago
egreg
178k1484201
178k1484201
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).
– Sjoerd
3 mins ago
add a comment |
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).
– Sjoerd
3 mins ago
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).
– Sjoerd
3 mins ago
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).
– Sjoerd
3 mins ago
add a comment |
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