Why is the 0th term of the padovan sequence 1 but for fibonacci it is 0?












2














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)



enter image description here



With the Padovan triangles the first visible term of the sequence is P(0)?



enter image description here



Which seems inconsistent to me. Can anyone give a mathematical explanation as to why P(0)=1 ... many thanks










share|cite|improve this question



























    2














    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)



    enter image description here



    With the Padovan triangles the first visible term of the sequence is P(0)?



    enter image description here



    Which seems inconsistent to me. Can anyone give a mathematical explanation as to why P(0)=1 ... many thanks










    share|cite|improve this question

























      2












      2








      2







      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)



      enter image description here



      With the Padovan triangles the first visible term of the sequence is P(0)?



      enter image description here



      Which seems inconsistent to me. Can anyone give a mathematical explanation as to why P(0)=1 ... many thanks










      share|cite|improve this question













      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)



      enter image description here



      With the Padovan triangles the first visible term of the sequence is P(0)?



      enter image description here



      Which seems inconsistent to me. Can anyone give a mathematical explanation as to why P(0)=1 ... many thanks







      sequences-and-series fibonacci-numbers






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      asked Jan 1 at 23:24









      danday74danday74

      1296




      1296






















          3 Answers
          3






          active

          oldest

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          6














          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.






          share|cite|improve this answer































            4














            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.






            share|cite|improve this answer

















            • 4




              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
              Jan 2 at 1:11








            • 1




              @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.
              – egreg
              Jan 2 at 10:10



















            3














            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.



            $$ left( x^{5} - x^{4} - 1 right) $$



            $$ left( x^{3} - x - 1 right) $$



            $$ left( x^{5} - x^{4} - 1 right) = left( x^{3} - x - 1 right) cdot color{magenta}{ left( x^{2} - x + 1 right) } $$






            share|cite|improve this answer























            • But this is true regardless of the indexing, what's the point?
              – user202729
              Jan 2 at 8:58











            Your Answer





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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            6














            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.






            share|cite|improve this answer




























              6














              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.






              share|cite|improve this answer


























                6












                6








                6






                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.






                share|cite|improve this answer














                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.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 2 at 3:38

























                answered Jan 1 at 23:53









                SomosSomos

                13.1k11034




                13.1k11034























                    4














                    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.






                    share|cite|improve this answer

















                    • 4




                      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
                      Jan 2 at 1:11








                    • 1




                      @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.
                      – egreg
                      Jan 2 at 10:10
















                    4














                    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.






                    share|cite|improve this answer

















                    • 4




                      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
                      Jan 2 at 1:11








                    • 1




                      @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.
                      – egreg
                      Jan 2 at 10:10














                    4












                    4








                    4






                    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.






                    share|cite|improve this answer












                    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.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 1 at 23:52









                    egregegreg

                    179k1485202




                    179k1485202








                    • 4




                      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
                      Jan 2 at 1:11








                    • 1




                      @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.
                      – egreg
                      Jan 2 at 10:10














                    • 4




                      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
                      Jan 2 at 1:11








                    • 1




                      @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.
                      – egreg
                      Jan 2 at 10:10








                    4




                    4




                    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
                    Jan 2 at 1:11






                    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
                    Jan 2 at 1:11






                    1




                    1




                    @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.
                    – egreg
                    Jan 2 at 10:10




                    @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.
                    – egreg
                    Jan 2 at 10:10











                    3














                    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.



                    $$ left( x^{5} - x^{4} - 1 right) $$



                    $$ left( x^{3} - x - 1 right) $$



                    $$ left( x^{5} - x^{4} - 1 right) = left( x^{3} - x - 1 right) cdot color{magenta}{ left( x^{2} - x + 1 right) } $$






                    share|cite|improve this answer























                    • But this is true regardless of the indexing, what's the point?
                      – user202729
                      Jan 2 at 8:58
















                    3














                    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.



                    $$ left( x^{5} - x^{4} - 1 right) $$



                    $$ left( x^{3} - x - 1 right) $$



                    $$ left( x^{5} - x^{4} - 1 right) = left( x^{3} - x - 1 right) cdot color{magenta}{ left( x^{2} - x + 1 right) } $$






                    share|cite|improve this answer























                    • But this is true regardless of the indexing, what's the point?
                      – user202729
                      Jan 2 at 8:58














                    3












                    3








                    3






                    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.



                    $$ left( x^{5} - x^{4} - 1 right) $$



                    $$ left( x^{3} - x - 1 right) $$



                    $$ left( x^{5} - x^{4} - 1 right) = left( x^{3} - x - 1 right) cdot color{magenta}{ left( x^{2} - x + 1 right) } $$






                    share|cite|improve this answer














                    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.



                    $$ left( x^{5} - x^{4} - 1 right) $$



                    $$ left( x^{3} - x - 1 right) $$



                    $$ left( x^{5} - x^{4} - 1 right) = left( x^{3} - x - 1 right) cdot color{magenta}{ left( x^{2} - x + 1 right) } $$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jan 2 at 1:57

























                    answered Jan 2 at 1:50









                    Will JagyWill Jagy

                    102k5100199




                    102k5100199












                    • But this is true regardless of the indexing, what's the point?
                      – user202729
                      Jan 2 at 8:58


















                    • But this is true regardless of the indexing, what's the point?
                      – user202729
                      Jan 2 at 8:58
















                    But this is true regardless of the indexing, what's the point?
                    – user202729
                    Jan 2 at 8:58




                    But this is true regardless of the indexing, what's the point?
                    – user202729
                    Jan 2 at 8:58


















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