Mysterious polynomial sequence












4












$begingroup$


Can someone identify this polynomial sequence? Is it known in mathematics? I'm interested in various properties of this sequence.

I'd like to find $P(n)$, $nin mathbb{Z}^+$



begin{align}
P(0)&= 1\
P(1)&= a\
P(2)&= a^2+b\
P(3)&= a^3+2ab\
P(4)&= a^4+3a^2b+b^2\
P(5)&= a^5+4a^3b+3ab^2\
P(6)&= a^6+5a^4b+6a^2b^2+b^3\
P(7)&= a^7+6 a^5 b+10 a^3 b^2+4 a b^3\
P(8)&= a^8 + 7 a^6 b + 15 a^4 b^2 + 10 a^2 b^3 + b^4\
P(9)&= a^9 + 8 a^7 b + 21 a^5 b^2 + 20 a^3 b^3 + 5 a b^4\
P(10)&= a^{10} + 9 a^8 b + 28 a^6 b^2 + 35 a^4 b^3 + 15 a^2 b^4 + b^5
end{align}



More steps upon request.



I'll be grateful for any hints!






sequences-and-series polynomials







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Essentially OEIS A011973 and OEIS A169803
    $endgroup$
    – Henry
    Dec 17 '18 at 18:13










  • $begingroup$
    Have a look at Lucas sequences. I've a feeling you'll find a rich vein of material there. Is there some pair of polynomials in a,b you can substitute into $P,Q$ here: en.wikipedia.org/wiki/Lucas_sequence#Examples
    $endgroup$
    – user334732
    Dec 17 '18 at 19:22
















4












$begingroup$


Can someone identify this polynomial sequence? Is it known in mathematics? I'm interested in various properties of this sequence.

I'd like to find $P(n)$, $nin mathbb{Z}^+$



begin{align}
P(0)&= 1\
P(1)&= a\
P(2)&= a^2+b\
P(3)&= a^3+2ab\
P(4)&= a^4+3a^2b+b^2\
P(5)&= a^5+4a^3b+3ab^2\
P(6)&= a^6+5a^4b+6a^2b^2+b^3\
P(7)&= a^7+6 a^5 b+10 a^3 b^2+4 a b^3\
P(8)&= a^8 + 7 a^6 b + 15 a^4 b^2 + 10 a^2 b^3 + b^4\
P(9)&= a^9 + 8 a^7 b + 21 a^5 b^2 + 20 a^3 b^3 + 5 a b^4\
P(10)&= a^{10} + 9 a^8 b + 28 a^6 b^2 + 35 a^4 b^3 + 15 a^2 b^4 + b^5
end{align}



More steps upon request.



I'll be grateful for any hints!






sequences-and-series polynomials







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Essentially OEIS A011973 and OEIS A169803
    $endgroup$
    – Henry
    Dec 17 '18 at 18:13










  • $begingroup$
    Have a look at Lucas sequences. I've a feeling you'll find a rich vein of material there. Is there some pair of polynomials in a,b you can substitute into $P,Q$ here: en.wikipedia.org/wiki/Lucas_sequence#Examples
    $endgroup$
    – user334732
    Dec 17 '18 at 19:22














4












4








4





$begingroup$


Can someone identify this polynomial sequence? Is it known in mathematics? I'm interested in various properties of this sequence.

I'd like to find $P(n)$, $nin mathbb{Z}^+$



begin{align}
P(0)&= 1\
P(1)&= a\
P(2)&= a^2+b\
P(3)&= a^3+2ab\
P(4)&= a^4+3a^2b+b^2\
P(5)&= a^5+4a^3b+3ab^2\
P(6)&= a^6+5a^4b+6a^2b^2+b^3\
P(7)&= a^7+6 a^5 b+10 a^3 b^2+4 a b^3\
P(8)&= a^8 + 7 a^6 b + 15 a^4 b^2 + 10 a^2 b^3 + b^4\
P(9)&= a^9 + 8 a^7 b + 21 a^5 b^2 + 20 a^3 b^3 + 5 a b^4\
P(10)&= a^{10} + 9 a^8 b + 28 a^6 b^2 + 35 a^4 b^3 + 15 a^2 b^4 + b^5
end{align}



More steps upon request.



I'll be grateful for any hints!






sequences-and-series polynomials







share|cite|improve this question











$endgroup$




Can someone identify this polynomial sequence? Is it known in mathematics? I'm interested in various properties of this sequence.

I'd like to find $P(n)$, $nin mathbb{Z}^+$



begin{align}
P(0)&= 1\
P(1)&= a\
P(2)&= a^2+b\
P(3)&= a^3+2ab\
P(4)&= a^4+3a^2b+b^2\
P(5)&= a^5+4a^3b+3ab^2\
P(6)&= a^6+5a^4b+6a^2b^2+b^3\
P(7)&= a^7+6 a^5 b+10 a^3 b^2+4 a b^3\
P(8)&= a^8 + 7 a^6 b + 15 a^4 b^2 + 10 a^2 b^3 + b^4\
P(9)&= a^9 + 8 a^7 b + 21 a^5 b^2 + 20 a^3 b^3 + 5 a b^4\
P(10)&= a^{10} + 9 a^8 b + 28 a^6 b^2 + 35 a^4 b^3 + 15 a^2 b^4 + b^5
end{align}



More steps upon request.



I'll be grateful for any hints!






sequences-and-series polynomials




sequences-and-series polynomials






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 13:54







Ender

















asked Dec 17 '18 at 13:50









EnderEnder

868




868








  • 1




    $begingroup$
    Essentially OEIS A011973 and OEIS A169803
    $endgroup$
    – Henry
    Dec 17 '18 at 18:13










  • $begingroup$
    Have a look at Lucas sequences. I've a feeling you'll find a rich vein of material there. Is there some pair of polynomials in a,b you can substitute into $P,Q$ here: en.wikipedia.org/wiki/Lucas_sequence#Examples
    $endgroup$
    – user334732
    Dec 17 '18 at 19:22














  • 1




    $begingroup$
    Essentially OEIS A011973 and OEIS A169803
    $endgroup$
    – Henry
    Dec 17 '18 at 18:13










  • $begingroup$
    Have a look at Lucas sequences. I've a feeling you'll find a rich vein of material there. Is there some pair of polynomials in a,b you can substitute into $P,Q$ here: en.wikipedia.org/wiki/Lucas_sequence#Examples
    $endgroup$
    – user334732
    Dec 17 '18 at 19:22








1




1




$begingroup$
Essentially OEIS A011973 and OEIS A169803
$endgroup$
– Henry
Dec 17 '18 at 18:13




$begingroup$
Essentially OEIS A011973 and OEIS A169803
$endgroup$
– Henry
Dec 17 '18 at 18:13












$begingroup$
Have a look at Lucas sequences. I've a feeling you'll find a rich vein of material there. Is there some pair of polynomials in a,b you can substitute into $P,Q$ here: en.wikipedia.org/wiki/Lucas_sequence#Examples
$endgroup$
– user334732
Dec 17 '18 at 19:22




$begingroup$
Have a look at Lucas sequences. I've a feeling you'll find a rich vein of material there. Is there some pair of polynomials in a,b you can substitute into $P,Q$ here: en.wikipedia.org/wiki/Lucas_sequence#Examples
$endgroup$
– user334732
Dec 17 '18 at 19:22










2 Answers
2






active

oldest

votes


















9












$begingroup$

Hint. Note that the following recurrence holds: for $ngeq 2$,
$$P(n)=aP(n-1)+bP(n-2).$$
They are related to the Fibonacci polynomials. The wiki page gives a list of properties. For example we have that
$$P(n)=sum_{k=0}^{lfloor n/2rfloor}binom{n-k}{k}a^{n-2k}b^k.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @BarryCipra Yes it's better to stick to OP's notation. Thanks for pointing out.
    $endgroup$
    – Robert Z
    Dec 17 '18 at 14:10










  • $begingroup$
    Many thanks! :) I must study these properties to find if I find something useful
    $endgroup$
    – Ender
    Dec 17 '18 at 14:25



















2












$begingroup$

Try:



$$-frac{2^{-n} left(left(a-sqrt{a^2+4 b}right)^n-left(sqrt{a^2+4
b}+aright)^nright)}{sqrt{a^2+4 b}}$$






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    9












    $begingroup$

    Hint. Note that the following recurrence holds: for $ngeq 2$,
    $$P(n)=aP(n-1)+bP(n-2).$$
    They are related to the Fibonacci polynomials. The wiki page gives a list of properties. For example we have that
    $$P(n)=sum_{k=0}^{lfloor n/2rfloor}binom{n-k}{k}a^{n-2k}b^k.$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      @BarryCipra Yes it's better to stick to OP's notation. Thanks for pointing out.
      $endgroup$
      – Robert Z
      Dec 17 '18 at 14:10










    • $begingroup$
      Many thanks! :) I must study these properties to find if I find something useful
      $endgroup$
      – Ender
      Dec 17 '18 at 14:25
















    9












    $begingroup$

    Hint. Note that the following recurrence holds: for $ngeq 2$,
    $$P(n)=aP(n-1)+bP(n-2).$$
    They are related to the Fibonacci polynomials. The wiki page gives a list of properties. For example we have that
    $$P(n)=sum_{k=0}^{lfloor n/2rfloor}binom{n-k}{k}a^{n-2k}b^k.$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      @BarryCipra Yes it's better to stick to OP's notation. Thanks for pointing out.
      $endgroup$
      – Robert Z
      Dec 17 '18 at 14:10










    • $begingroup$
      Many thanks! :) I must study these properties to find if I find something useful
      $endgroup$
      – Ender
      Dec 17 '18 at 14:25














    9












    9








    9





    $begingroup$

    Hint. Note that the following recurrence holds: for $ngeq 2$,
    $$P(n)=aP(n-1)+bP(n-2).$$
    They are related to the Fibonacci polynomials. The wiki page gives a list of properties. For example we have that
    $$P(n)=sum_{k=0}^{lfloor n/2rfloor}binom{n-k}{k}a^{n-2k}b^k.$$






    share|cite|improve this answer











    $endgroup$



    Hint. Note that the following recurrence holds: for $ngeq 2$,
    $$P(n)=aP(n-1)+bP(n-2).$$
    They are related to the Fibonacci polynomials. The wiki page gives a list of properties. For example we have that
    $$P(n)=sum_{k=0}^{lfloor n/2rfloor}binom{n-k}{k}a^{n-2k}b^k.$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 17 '18 at 14:21

























    answered Dec 17 '18 at 13:52









    Robert ZRobert Z

    98.3k1067139




    98.3k1067139












    • $begingroup$
      @BarryCipra Yes it's better to stick to OP's notation. Thanks for pointing out.
      $endgroup$
      – Robert Z
      Dec 17 '18 at 14:10










    • $begingroup$
      Many thanks! :) I must study these properties to find if I find something useful
      $endgroup$
      – Ender
      Dec 17 '18 at 14:25


















    • $begingroup$
      @BarryCipra Yes it's better to stick to OP's notation. Thanks for pointing out.
      $endgroup$
      – Robert Z
      Dec 17 '18 at 14:10










    • $begingroup$
      Many thanks! :) I must study these properties to find if I find something useful
      $endgroup$
      – Ender
      Dec 17 '18 at 14:25
















    $begingroup$
    @BarryCipra Yes it's better to stick to OP's notation. Thanks for pointing out.
    $endgroup$
    – Robert Z
    Dec 17 '18 at 14:10




    $begingroup$
    @BarryCipra Yes it's better to stick to OP's notation. Thanks for pointing out.
    $endgroup$
    – Robert Z
    Dec 17 '18 at 14:10












    $begingroup$
    Many thanks! :) I must study these properties to find if I find something useful
    $endgroup$
    – Ender
    Dec 17 '18 at 14:25




    $begingroup$
    Many thanks! :) I must study these properties to find if I find something useful
    $endgroup$
    – Ender
    Dec 17 '18 at 14:25











    2












    $begingroup$

    Try:



    $$-frac{2^{-n} left(left(a-sqrt{a^2+4 b}right)^n-left(sqrt{a^2+4
    b}+aright)^nright)}{sqrt{a^2+4 b}}$$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Try:



      $$-frac{2^{-n} left(left(a-sqrt{a^2+4 b}right)^n-left(sqrt{a^2+4
      b}+aright)^nright)}{sqrt{a^2+4 b}}$$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Try:



        $$-frac{2^{-n} left(left(a-sqrt{a^2+4 b}right)^n-left(sqrt{a^2+4
        b}+aright)^nright)}{sqrt{a^2+4 b}}$$






        share|cite|improve this answer









        $endgroup$



        Try:



        $$-frac{2^{-n} left(left(a-sqrt{a^2+4 b}right)^n-left(sqrt{a^2+4
        b}+aright)^nright)}{sqrt{a^2+4 b}}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 17 '18 at 14:01









        David G. StorkDavid G. Stork

        11k41432




        11k41432






























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