[seqfan] Re: A005207+1 and A296516 as numbers of terms in some polynomials

[neil greubel]

It might be useful to show that P_{n} and Q_{n}, which lead to u_{n} and
v_{n}, are somehow related to the difference equations

u_{n} = 7*u_{n-1} - 18*u_{n-2} + 20*u_{n-3} - 6*u_{n-4} - 6*u_{n-5} +
5*u_{n-6} - u_{n-7}
v_{n} = 5*v_{n-1} - 7*v_{n-2} + v_{n-3} + 3*v_{n-4} - v_{n-5}.


Since
P_1 = x + y
P_2 = x + x*y + y
P_3 = x + y + xy + x^2 y + x y^2 = P_2 + xy P_1

Q_1 = x y
Q_2 = x^2 y + x y^2
Q_3 = (x^3 + (2x^2 + x^3) y + (x + x^3) y^2) y

then a prospective method may be to find a two variable generating
function, or triangle of powers. It appears that Q_{n} = y * q_{n}(x,y)
where q_{n} is a polynomial in y with coefficients of polynomials in x.

On Fri, Apr 13, 2018 at 4:06 PM, Luc Rousseau <luc_rousseau at hotmail.com>
wrote:

> Dears SeqFans,
>
> I'm studying two sequences u(n) and v(n) (n >= 1) defined by
> u(n) = the number of terms in the expanded form of bivariate polynomial
> P_n,
> v(n) = the number of terms in the expanded form of bivariate polynomial
> Q_n,
>
> where P_n and Q_n are defined by
> P_1 = P = x + y,
> Q_1 = Q = x * y,
> P_{n+1} = P(P_n, Q_n),
> Q_{n+1} = Q(P_n, Q_n).
>
> Example: Q_2 = (P_1) * (Q_1) = (x + y) * (xy) = x^2*y + x*y^2. Two terms
> ==> v(2) = 2.
>
> I am making no progress towards proving (or refuting) the conjectured
> formulas:
> u(n) = (F(2*n-1) + F(n+1))/2 + 1,
> v(n) = (F(2*n+1) + F(n+2))/2 - T(n-1) - 1,
> where F(n) denotes the n-th Fibonacci number and T(n) the n-th triangular
> number.
> Does anyone see a demonstration?
>
> Proved, the result would deserve:
> - a comment in long-existing A005207 : a(n) = u(n) - 1 = the number of "+"
> in P_n expanded,
> - a complete reformulation of A296516(n) =def v(n) (draft status),
> - crossref A005207 <-> A296516.
>
> Kind Regards,
>
> Luc Rousseau.
>
>
>
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