Convolutions of generalized $p-$Fibonacci sequences

Generalized $p-$Fibonacci numbers $U_{n}(p;a,b)$ are defined by eq. 1 for $p\in \NN0$, $b \neq 0$ and $a\neq 0$, together with the inputs $U_{j}(p;a,b):= a^j$ for $j=0,...,p$. For $p=1$ one recovers the generalized Fibonacci numbers $U_{n}(a,b)$ treated above.

Lemma 6: The generating function $U(p;a,b;x)$ for the generalized $p-$Fibonacci numbers is given by eq. 43.

Proof: From the recurrence with inputs given in eq. 1. $\square$

Lemma 7 (Riccati eq. for the generalized $p-$Fibonacci case):

If $D(p;a,b):= (p+1)^{p+1}\, b\, +\, a\,(a\,p)^p\,\neq\, 0$ (non-degenerate case) then $U(p;a,b;x)$ satisfies the Riccati eq. 45 with the polynomials $A_{p}(a,b;x)$ and $B_{p-1}(a,b;x)$ defined in eqs. 46 and  47.

Proof: Use the ansatz

$$U^{2}(p;a,b;x) \ =\ \Bigl{(}\sum_{i=0}^{p}\, A_{i}(p;a,b)\, x^i\B... ...,b;x)\ +\ \Bigl{(}\sum_{i=0}^{p-1}\, B_{i}(p;a,b)\, x^i \Bigr{)}\,U(p;a,b;x)\ ,$$ (60)

together with

$${\frac{\partial}{\partial\,x}}\, U(p;a,b;x)\ =\ (a\, +\, (p+1)\, b\, x^p)\,U^{2}(p;a,b;x)$$ (61)

from eq. 43. This implies

$$\Bigl{(}\sum_{i=0}^{p}\, A_{i}(a,b;x)\, x^i\Bigr{)}\,(a\, +\, (p+... ...^{p-1}\, B_{i}(a,b;x)\, x^i\Bigr{)}\, (1\, -\, a\,x \, -\, b\,x^{p+1})\ =\ 1\ .$$ (62)

Comparing first coefficients of the powers $x^k$, for $k=p+1,p+2,...,2\,p$, leads to the eqs. (remember that $b \neq 0$)

$$B_{i}(p;a,b)\ =\ (p+1)\,A_{i+1}(p;a,b)\ , \ \ {\rm for}\ i= 0,1,...,p-1.$$ (63)

With this result the coefficients of the powers $x^k$ for $k=0,1,...,p-1$ become, after iteration,

$$A_{j}(p;a,b)\ =\ \left({\frac{a\,p}{p+1}}\right)^{j-1}\, {\frac{1}{p+1}}\, (1\, -\, a\,A_{0}(p;a,b))\ {\rm for}\ j=1,2,...,p.$$ (64)

Therefore the eqs. 63 are now

$$B_{i}(p;a,b)\ =\ \left({\frac{a\,p}{p+1}}\right)^{i}\, (1\, -\, a\,A_{0}(p;a,b))\ , \ {\rm for} \ i= 0,1,...,p-1.$$ (65)

Finally, the coefficient of $x^p$ has to satisfy $(p+1)\, b\, A_{0}\, +\, a\, (A_{p}\, -\, B_{p-1}) \ =\ 0$ which implies, together with the above results,

$$A_{0}(p;a,b)\ =\ \left({\frac{a\,p}{p+1}}\right)^p\, {\frac{1}{(p+1)\, b\, +\, a\, \left({\frac{a\,p}{p+1}}\right)^p}}\ .$$ (66)

Inserting this into the above found expressions finally leads to

$$A_{j}(p;a,b) \ =\ \left({\frac{a\,p}{p+1}}\right)^{j-1}\, {\frac{b}{(p+1)\, b\, +\, a\, \left({\frac{a\,p}{p+1}}\right)^p}}\ , \ \ {\rm for}\ j=1,2,...,p,$$ (67)

$$B_{j}(p;a,b) \ =\ \left({\frac{a\,p}{p+1}}\right)^j\, {\frac{b\,(p+1)}{(p+1)\, b\, +\, a\,\left({\frac{a\,p}{p+1}}\right)^p}}\ , \ \ {\rm for}\ j=0,1,...,p-1,$$ (68)

$$A_{0}(p;a,b) \ =\ \left({\frac{a\,p}{p+1}}\right)^p\, {\frac{1}{(p+1)\, b\, +\, a\, \left({\frac{a\,p}{p+1}}\right)^p}}\ .$$ (69)

With

$$A_{p}(a,b;x) \ :=\ ((p+1)^{p+1}\, b\, +\, a\,\left({\frac{a\,p}{p+1}}\right)^p)\,\sum_{i=0}^{p}\, A_{i}(p;a,b)\, x^i\ ,$$ (70)

$$\ b\,(p+1)^2\, B_{p-1}(a;x) \ :=\ ((p+1)^{p+1}\, b\, +\, a\, \left({\frac{a\,p}{p+1}}\right)^p)\,\sum_{i=0}^{p-1}\, B_{i}(p;a,b)\, x^i\ , \$$ (71)

one thus finds eqs. 47 and  46. The assertion can now be proved directly with the found polynomials. $\square$

Note 4: i) If $p=0$, $U(0;a,b,;x)\, =\, 1/(1-(a+b)\, x)$ generates powers of $a+b$, and one has to put $A_{0}(a,b;x)\equiv 1$ and $B_{-1}(a;x)\equiv 0$. This means that one puts $(a\,p)^p \, =\, 1$ for $p=0$.

ii) For given non-vanishing $a$ and $b$ $A_{p}(a,b;x)$ is a polynomial in $x$ of degree $p$, and $B_{p-1}(a,b;x)$ is one of degree $p-1$. The sum in $B_{p-1}(a,b;x)$ can be evaluated to yield the second of eqs. 47 provided $p\neq 0$ .

Lemma 8 (Coefficient triangles of numbers for polynomials $A_{p}(a,b;x)$ and $B_{p}(a,b;x)$):

The coefficients of the polynomials defined in eqs. 48 are given by eqs. 49 and  50.

Proof: $B_{p}(a,b;x)$ from eq. 47 leads immediately to eq. 50, remembering that $B_{-1}(a;x)\equiv 0$. Then eq. 49 follows from eq. 46 and $A_{0}(a,b;x)\equiv 1$. $\square$

Proposition 5 (Uniqueness of Riccati solution; non-degenerate case):

If $D(a,b)\,\neq\, 0$ then $y\, \equiv \, U(p;a,b;x)\, =\, 1/(1\, -\, a\,x\, -\, b\,x^{p+1})$ is the unique solution of the Riccati eq. 45 with eqs. 46,  47 and initial value $U(p;a,b;0)\, =\, 1$.

Proof: As a special Bernoulli eq. the Riccati eq. $y^{\prime}\, +\, f(x)\,y\, +\, g(x)\, y^2\, =\, 0$ is equivalent to the inhomogeneous linear differential eq. for $z\, =\, 1/y$: $z^{\prime}\, \equiv \, F(x,z)\, =\, f(x)\,z\, +\, g(x)$. Because $F(x,z)$ is continuous in the strip $0\leq x \leq A<\infty$, $\vert z\vert\,<\,\infty$ and is there $(k\, =\, k(p;a,b;A))$-Lipschitz, the existence and uniqueness theorem for linear differential eqs. proves the assertion (see e.g.[8], §6,I, p.62ff). In order to find $k$ one uses the summed expression for $B_{p-1}$ from eq. 47 and repeated applications of the triangle inequality.$\square$

Lemma 9 (First convolution of $\{U_{n}(p;a,b)\}$; non-degenerate case):

The first convolution of the sequence $\{U_{n}(p;a,b)\}$, which is defined analogously to the first of eqs. 17, is given by eq. 51 with eq. 52.

Proof: One has to compute the coefficients of $x^n$ of the lhs of eq. 45, taking into account the $x-$dependence of $A_{p}$ and $B_{p-1}$ with the coefficients from eqs. 49 and  50. The case $p=0$ has to be considered separately. The recurrence eq. 1 cannot be used in order to simplify the sum in eq. 51. $\square$

Proposition 6 (Recurrence for $k+1$st power of the generating function $U(p;a,b;x)$; non-degenerate case):

For $D(p;a,b)\, :=\, b\, (p+1)^{p+1}\, +\, a\, (a\,p)^p \, \neq 0$ eq. 53 gives $U^{k+1}(p;a,b;x)$.

Proof:

$${\frac{1}{D(p;a,b)}}\,A_{p}(a,b;x)\, {\frac{\partial}{\partial\,x... ...(p;a,b;x)\ =\ k\,U^{k-1} {\frac{A_{p}}{D}}\,{\frac{\partial}{\partial\,x}}\, U$$

$$\ =\ k\, U^{k}(p;a,b;x)\, (-b\,(p+1)^2\, B_{p-1}(a;x)\ +\ U(p;a,b;x))\ ,$$ (72)

due to eq. 45.

Proposition 7 ($k-$fold convolution of $\{U_{n}(p;a,b)\}$; non-degenerate case):

For $D(p;a,b)\,\neq\, 0$ the entry $U_{n}^{(k)}(p;a,b)$ of the $k-$fold convolution of the sequence $\{U_{n}(p;a,b)\}$ is given by eq. 54.

Proof: This follows immediately from Proposition 5 after comparing coefficients of $x^n$ using the definition $U^{k+1}(p;a,b;x)\, =:\, \sum_{n=0}^{\infty}\, U_{n}^{(k)}(p;a,b)\, x^n\ .$ $\square$

Lemma 10 (Degenerate case $D(p;a,b)\, =\, 0$):

If $D(p;a,b)\, :=\,(p+1)^{p+1}\,b \, +\, a\,(a\,p)^{p} \, =\, 0$ then $U(p;a;x)\, =\, 1/(1-a\,x-b(p;a)\,x^{p+1})\, =\,$

$1/(1-a\,x\,(p/(p+1)^p\,(a\,x)^p)$ satisfies the first order linear differential eq. 57.

Proof: One proves $(a\, +\, (p+1)\,b\, x^p)\, A_{p}(a,b;x) \ +\ b\, (p+1)^2\,(1-a\,x-b\,x^{p+1})\,B_{p-1}(a;x)\, =\, 0$ with eqs. 47 and  46 in the version where the sum has been evaluated (the case $p=0$ is treated separately). If one factors out $b/(p+1-a\,p\,x)$ one finds that all terms cancel provided one replaces $a\,(a\,p)^p$ by $-b\,(p+1)^{p+1}$.$\square$

Note 5: The solution $1/(1-a\,x-b\,x^{p+1})$ of this linear differential eq. with input $U(p;a,b;0)\, =\, 1$ is unique. The proof is analogous to the one of proposition 5.

Note 6: If $U(p;a,b;x)\, =\, 1/(1\, -\, a\,x\, +\, (((a\,p\,x/(p+1))^{p+1})/p)$ we do not have a formula for $U^{2}(p;a,b;x)$, valid for all $p$, like in the non-degenerate case. Therefore, we cannot derive results for convolutions along the line shown above.

Lemma 11 (Recurrence in the degenerate case):

If $D(p;a,b)\, :=\,(p+1)^{p+1}\,b \, +\, a\,(a\,p)^{p} \, =\, 0$ (and $b \neq 0$) then one can replace the recurrence eq. 1 which has depth $p+1$, by the following one with depth $p\in \mathbb{N}$.

$$U_{n+1}(p;a)\ =\ {\frac{a}{(p+1)\,(n+1)}}\, \sum_{j=1}^{p}\, \left({\frac{a\,p}{p+1}}\right)^{j-1} \,(n+p+2-j)\, U_{n+1-j}(p;a)\ ,$$ (73)

where one uses the inputs $U_{j}(p;a)\, =\, a^j$ for $j=0,1,...,p-1$.

Proof: This derives from the sum on the rhs of eq. 51 which now vanishes. If the coefficients $C_{j}$ from eq. 52 are used with the replacement of $a\,(a\,p)^{p}$ by $-b\,(p+1)^{p+1}$ one arrives at the desired recurrence, after the common factor $b$ has been dropped. The inputs are adopted from the original recurrence except that $U_{p}$ can now be computed to be $a^p$.$\square$