Introduction and Summary

The generating function for generalized Fibonacci numbers $\{F_{n}(a,b)\}_{0}^{\infty}$, defined by the three term recurrence relation

$$F_{n}(a,b)\ =\ a\, F_{n-1}(a,b)\ +\ b\,F_{n-2}(a,b),\ \ \ F_{0}(a,b)\ =\ 0,\ F_{1}(a,b)\ =\ 1,$$ (1)

with given real $a\neq 0$ and $b \neq 0$, is well-known. For arbitrary $a$ and $b$, $F_{n}(a,b)$ can be considered as a polynomial in two variables. If one considers the numbers, or polynomials, $U_{n}(a,b)\, :=\, F_{n+1}(a,b)$ one has from the recursion with input $U_{0}(a,b)=1$ and $U_{1}(a,b)= a$ (or $U_{-1}(a,b)=0$)

$$U(a,b;x)\ :=\ \sum_{n=0}^{\infty}\, U_{n}(a,b)\,x^n \ =\ {\frac{1}{1\, -\, a\,x \, -\, b\,x^2}}\ .$$ (2)

Similarly, for the generalized Lucas numbers $\{L_{n}(a,b)\}_{0}^{\infty}$ which satisfy the same recursion eq. 1 but with inputs $L_{0}(a,b)=2, L_{1}(a,b)=a$, one finds, with $V_{n}(a,b) \ :=\ L_{n+1}(a,b)/a$, remembering that $a\neq 0$,

$$V(a,b;x) \ :=\ \sum_{n=0}^{\infty}\, V_{n}(a,b)\,x^n \ =\ {\frac{1+2\,b\,x/a}{1\, -\, a\,x \, -\, b\,x^2}}\ .$$ (3)

The input is now $V_{0}(a,b)=1$ and $V_{1}(a,b)= (a^2 \, +\, 2\,b)/a$ (or $V_{-1}(a,b)=2/a$).

These (ordinary) generating functions can also be written in terms of the characteristic roots corresponding to the recursion relation eq. 1

$$\lambda_{\pm} \ \equiv \ \lambda_{\pm}(a,b)\ :=\ {\frac{1}{2}}\,(a\,\pm\,\sqrt{a^2+4\,b})$$ (4)

as follows.

$$U(a,b;x) \ =\ {\frac{1}{x\,(\lambda_{+}-\lambda_{-})}}\,\Bigl{(}{\frac{1}{1\, -\, \lambda_{+}\, x}}\ -\ {\frac{1}{1\, -\, \lambda_{-}\, x}} \Bigr{)}\ ,$$ (5)

$$V(a,b;x) \ =\ {\frac{1}{\lambda_{+}+\lambda_{-}}}\,\Bigl{(}{\frac{\lambda_{+}}{... ...lambda_{+}\, x}}\ +\ {\frac{\lambda_{-}}{1\, -\, \lambda_{-}\, x}} \Bigr{)} \ .$$ (6)

The corresponding Binet forms of the corresponding numbers are in the non-degenerate case $\lambda_{+}\neq\lambda_{-}$, i.e. $D(a,b):=a^2+4\,b\neq 0$,

$$U_{n}(a,b) \ =\ {\frac{\lambda_{+}^{n+1}\ -\ \lambda_{-}^{n+1}}{\lambda_{+}-\lambda_{-}}}\ ,$$ (7)

$$V_{n}(a,b) \ =\ {\frac{\lambda_{+}^{n+1}\ +\ \lambda_{-}^{n+1}} {\lambda_{+}+\lambda_{-}}} \ .$$ (8)

In the degenerate case one has

$$U_{n}(a) \ :=\ U_{n}(a,-{\frac{a^2}{4}})\ =\ (n+1)\, \Bigl{(}\frac{a}{2}\Bigr{)}^{n}\ ,$$ (9)

$$V_{n}(a) \ :=\ V_{n}(a,-{\frac{a^2}{4}}) \ =\ \Bigl{(}\frac{a}{2}\Bigr{)}^{n}\ .$$ (10)

The explicit form of these polynomials is

$$U_{n}(a,b) \ =\ \sum_{l=0}^{\lfloor {\frac{n}{2}}\rfloor}\, {{n-l} \choose {l}}\,a^{n-2l}\,b^{l},$$ (11)

$$V_{n}(a,b) \ =\ \ \sum_{l=0}^{\lfloor {\frac{n+1}{2}}\rfloor}\, {\frac{n+1}{n+1-l}}\,{{n+1-l} \choose {l}}\,a^{n-2l}\,b^{l} .$$ (12)

This result for $U_{n}(a,b)$ follows from a combinatorial interpretation of the recurrence relation, and the one for $V_{n}(a,b)$ is due to the Girard -Waring formula in its simplest version (for this cf. [4], [3], also for original refs.).

The observation reported in this work is that the generating functions eq. 2 (or eq. 5) and eq. 3 (or eq. 6) are the unique solutions to the following special type of Riccati eqs. (simultaneously a special type of Bernoulli eq. For the history of such eqs. see [9], ch.I, 1.1)

$$(a\, +\, 2\,b\,x)\,{{\partial {}}\over{\partial {x}}} U(a,b;x)\ +\ 4\,b\,U(a,b;x) \ -\ (a^2\, +\, 4\,b)\, U^{2}(a,b;x) \ =\ 0,$$ (13)

with the initial condition $U(a,b;0)=1$. Similarly,

$$(1\, +\, 2\,{\frac{b}{a}}\, x)^2\,{{\partial {}}\over{\partial {x... ...rac{b}{a}}\,x)\,V(a,b;x) \ -\ (a\, +\, 4\,{\frac{b}{a}})\, V^{2}(a,b;x) \ =\ 0,$$ (14)

with the initial condition $V(a,b;0)=1$. Such Riccati eqs. can be transformed to inhomogeneous linear differential eqs. of first order. Our interest here is to use these eqs. in order to obtain information about $U^2$ and $V^2$. The degenerate case $D(a,b)\, :=\, a^2\, +\, 4\, b = 0$, for which the above given differential eqs. become linear, will be considered separately.

These Riccati eqs. immediately allow one to express convolutions of Fibonacci or Lucas sequences in terms of these numbers. In the non-degenerate case, $D(a,b)\neq 0$, one rewrites the Riccati eqs. as

$$U^{2}(a,b;x)\ =\ {\frac{1}{a^2\, +\, 4\, b}}\,\Bigl{(}(a\, +\, 2\... ...l {} \phantom{x} }\over{\partial {x} \phantom{}}}\ +\ 4\, b \Bigr{)}\,U(a,b;x),$$ (15)

and

$$V^{2}(a,b;x)\ =\ {\frac{1}{a\, +\, 4\,{\frac{b}{a}}}}\,\Bigl{(} (... ... {x}}}\ +\ 2\, {\frac{b}{a}}(1\, +\, 2\, {\frac{b}{a}}\,x ) \Bigr{)}\,V(a,b;x).$$ (16)

The (first) convolution of these sequences, which we call

$$U^{(1)}_{n}(a,b):= \sum_{q=0}^{n}\, U_{n-q}(a,b)\, U_{q}(a,b) \ \ \ {\rm and}\ \ \ V^{(1)}_{n}(a,b):= \sum_{q=0}^{n}\, V_{n-q}(a,b)\, V_{q}(a,b)\ ,$$ (17)

and whose generating functions are $U^{2}(a,b;x)$ and $V^{2}(a,b;x)$, respectively, satisfy therefore

$$U^{(1)}_{n}(a,b)\ =\ {\frac{1}{a^2\, +\, 4\, b}}\,\Bigl{(}a\,(n+1)\, U_{n+1}(a,b)\ +\ 2\,b\,(n+2)\,U_{n}(a,b)\Bigr{)},$$ (18)

and

$$V^{(1)}_{n}(a,b)\ =\ {\frac{1}{a\, +\, 4\,{\frac{b}{a}}}}\,\Bigl{... ...{a}}\,(2\,n+1)\,V_{n}(a,b)\ +\ 4\,{\frac{b^2}{a^2}}\,n\,V_{n-1}(a,b)\Bigr{)}\ .$$ (19)

After use of the recursion relation eq. 1 the last eq. can be written in the form

$$V^{(1)}_{n}(a,b)\ =\ {\frac{1}{a\,(a^2 \, +\, 4\,b)}}\,\Bigl{(}\b... ...n+1) \,\ +\ 4\,b\,n\bigr{]}\, V_{n+1}(a,b)\ +\ 2\, b\, a\,V_{n}(a,b)\Bigr{)}\ .$$ (20)

For $a=b=1$ one recovers well-known formulae for convolutions of ordinary Fibonacci, resp. Lucas numbers (e.g. [7], p.183, eqs. (98) and (99) (with corrected $L_{n-1}\to L_{n-i}$)). To see this, observe that $U^{(1)}_{n}(1,1)\, =\, F^{(1)}_{n+2}$ and $V^{(1)}_{n}(1,1)\, =\, L^{(1)}_{n+2}-4\,L_{n+2}$.

$$F^{(1)}_{n} \ =\ U^{(1)}_{n-2}(1,1)\ =\ {\frac{1}{5}}\, \Bigl{(}(n-1)\,F_{n}\ +\ 2\,n \,F_{n-1}\Bigr{)}\ =\ {\frac{1}{5}}\,(n\,L_{n}\,-\, F_{n})$$ (21)

$$L^{(1)}_{n} \, =\, V^{(1)}_{n-2}(1,1)\, +\, 4\,L_{n} \ =\ {\frac{1}{5}}\,\Bigl{(}(5\... ...L_{n}\, +\, 2\,L_{n-1} \Bigr{)}\, +\, 4\,L_{n}\, =\, (n+2)\, L_{n}\ +\ F_{n}\ .$$ (22)

For the $k-$th convolution $U^{(k)}_{n}(a,b)$ and $V^{(k)}_{n}(a,b)$, with $k\, =\, 1,2,...$, one employs in the case $D(a,b)\neq 0$ the following identities which follow from the Riccati eqs.

$$U^{k+1}(a,b;x)\ =\ {\frac{1}{(a^2\, +\, 4\, b)\,k}}\,\Bigl{(}(a\,... ...\,x)\,{{\partial {}}\over{\partial {x}}}\ +\ 4\,k\, b \Bigr{)}\,U^{k}(a,b;x)\ ,$$ (23)

and

$$V^{k+1}(a,b;x)\ =\ {\frac{a}{(a^2\, +\, 4\, b)\,k}}\,\Bigl{(} (1\... ...\ 2\,k\, {\frac{b}{a}}(1\, +\, 2\, {\frac{b}{a}}\,x ) \Bigr{)}\,V^{k}(a,b;x)\ .$$ (24)

These identities allow one to express the $k-$th convolution, generated by $U^{k+1}(a,b;x)\,=:\, \sum_{n=0}^{\infty} U_{n}^{(k)}(a,b)\, x^n$, in terms of the $k-1$-st one according to

$$U^{(k)}_{n}(a,b)\ =\ {\frac{1}{k\,(a^2\, +\, 4\,b)}}\,\Bigl{(}a\,(n+1)\, U^{(k-1)}_{n+1}(a,b)\ +\ 2\,b\,(n+2\,k)\,U^{(k-1)}_{n}(a,b)\Bigr{)},$$ (25)

with input $U^{(0)}_{n}(a,b)\, =\, U_{n}(a,b)$, and

$$V^{(k)}_{n}(a,b) \ =\ {\frac{1}{k\,a\,(a^2\, +\, 4\,b)}}\, \Bigl{(}(n+1)\,a^2\,V^{(k-1)}_{n+1}(a,b)\ +\ 2\,a\,b\,(2\,n \, +\, k)\, V^{(k-1)}_{n}(a,b)$$

$$\hskip 2cm \ +\ 4\, b^2\,(n\, +\, k \, -\, 1)\, V^{(k-1)}_{n-1}(a,b)\Bigr{)} \ ,\$$ (26)

with input $V^{(0)}_{n}(a,b)\, =\, V_{n}(a,b)$. The formula given in eq. 25 has been found earlier in [1] (p. 202, III and p.213, eq. (30)) without using the defining Riccati eq. for $U(a,b;x)$. The notations have to be translated with the help of $F_{n}^{(k)} \ \hat =\ U_{n}^{(k-1)}$, $a_{1}\ \hat =\ a$, and $a_{2}\ \hat =\ b$.

Before discussing iteration of these recursion relations we state the results for the degenerate case

$D(a,b)\, :=\, a^2\, +\, 4\,b\, =\, 0$. The Riccati eqs. 13 and  14 collapse to linear differential eqs. for $U(a;x)\, :=\, U(a,-a^2/4;x)$ and $V(a;x)\, :=\, V(a,-a^2/4;x)$

$$(1\, -\, {\frac{a}{2}}\,x)\, {{\partial {}}\over{\partial {x}}}U(a;x) \ =\ a\,U(a;x)\ \ \ , \ \ \ U(a;0)\, =\, 1,$$ (27)

$$(1\, -\, {\frac{a}{2}}\,x)\, {{\partial {}}\over{\partial {x}}} V(a;x) \ =\ {\frac{a}{2}}\, V(a;x)\ \ \ ,\ \ \ V(a;0)\, =\, 1 \ .$$ (28)

For the last eq. $x\neq 2/a$ was assumed. Because the solutions to these eqs. imply

$${{\partial {^2}}\over{\partial {x^2}}} U(a;x) \ =\ {\frac{3}{2}}\... ... \ {{\partial {}}\over{\partial {x}}} V(a;x) \ =\ {\frac{a}{2}}\, V^{2}(a;x)\ ,$$ (29)

the corresponding first ($k=1$) convolutions of these numbers $U_{n}(a)\, :=\, U_{n}(a,-a^2/4)$ and

$V_{n}(a)\, :=\, V_{n}(a,-a^2/4)$ are given by

$$U^{(1)}_{n}(a) \ =\ {\frac{2}{3\,a^2}}\,(n\, +\, 2)\,(n\, +\, 1)\... ...(a)\ \ \ , \ \ \ V^{(1)}_{n}(a) \ =\ {\frac{2}{a}}\,(n\, +\, 1)\, V_{n+1}(a)\ ,$$ (30)

with eqs. 9 and  10.

In order to derive the result for the $k-$th convolution one starts with identities which follow from the solutions of eqs. 27 and  28, namely

$$U^{k+1}(a;x) \ =\ {\frac{2}{a^2\,k\,(2\,k+1)}}\, {{\partial {^2}}\over{\partial {x^2}}} \Bigl{(} U^{k}(a;x)\Bigr{)}\ ,$$ (31)

$$\ V^{k+1}(a;x) \ =\ {\frac{2}{a\,k}}\, {{\partial {}}\over{\partial {x}}} \Bigl{(} V^{k}(a;x)\Bigr{)}\ . \$$ (32)

These identities imply for the $k-$th convolutions

$$U^{(k)}_{n}(a) \ =\ {\frac{2}{a^2\,k\,(2\,k+1)}}\, (n+2)\,(n+1)\, U^{(k-1)}_{n+2}(a)\ ,$$ (33)

$$\ V^{(k)}_{n}(a) \ =\ {\frac{2}{a\,k}}\, (n+1)\, V^{(k-1)}_{n+1}(a)\ ,\$$ (34)

with inputs $U^{(0)}_{n}(a)\, =\, U_{n}(a)\, =\, (n+1)\, (a/2)^n$ and $V^{(0)}_{n}(a)\, =\, V_{n}(a)\, =\, (a/2)^n$. See eqs. 9 and  10.

The iteration of these eqs. yields the final result, which for $k\in \NN0$, and in the degenerate case $b\, =\, -a^2/4$, is

$$U^{(k)}_{n}(a) \ =\ {{n+2\,k+1} \choose {2\,k+1}}\,\Bigl{(}{\frac{a}{2}} \Bigr{)}^n\ ,$$ (35)

$$\ V^{(k)}_{n}(a) \ =\ {{n+k} \choose {k}}\, \Bigl{(}{\frac{a}{2}}\Bigr{)}^n\ .\$$ (36)

Thus $V^{(2l+1)}_{n}(a)\, =\, U^{(l)}_{n}(a)$, and it suffices to treat $V^{(k)}_{n}(a)$. For even $a$ these are non-negative integer sequences. For $n,k\in\NN0$, $V^{(k)}_{n+k}(2\, l)$ constitutes a convolution triangle of numbers based on the $k=0$ column sequence $V^{(0)}_{n}(2\, l)\, =\, l^n$ (powers of $l$). See [5] for these triangles of numbers.

In the non-degenerate case the recursion relation eq. 24 can be iterated in order to express the $k-$th convolution of $U_{n}(a,b)$ as linear combination of these numbers according to

$$U^{(k)}_{n}(a,b)\ =\ {\frac{1}{k!\,(a^2 + 4\,b)^k}}\, \Bigl{(} AU... ...\,(n+1)\,a\, U_{n+1}(a,b)\ +\ BU_{k-1}(a,b;n)\,(n+2)\,b\, U_{n}(a,b)\Bigr{)}\ ,$$ (37)

with certain polynomials $AU_{k-1}(a,b;n)$ and $BU_{k-1}(a,b;n)$ of degree $k-1$ in the variable $n$, for arbitrary, but fixed, $a\neq 0$, $b \neq 0$, and $b\neq -a^2/4$.

The (mixed) recursion relations for these polynomials are deduced from eq. 24, and for $k=1,2,...,$ they are

$$AU_{k}(a,b;n) \ =\ a^2\,(n+2)\,AU_{k-1}(a,b;n+1)\ +\ 2\,b\,(n+2(k+1))\, AU_{k-1}(a,b;n)\ +\$$

$$b\, (n+3)\, BU_{k-1}(a,b;n+1) \ ,$$ (38)

$$BU_{k}(a,b;n) \ =\ a^2\,(n+1)\,AU_{k-1}(a,b;n+1)\ +\ 2\,b\,(n+2(k+1))\, BU_{k-1}(a,b;n)\ ,$$ (39)

with inputs $AU_{0}(a,b;n)=1$ and $BU_{0}(a,b;n)=2$.

In eqs. (26), resp. (27), of [1] one can find explicit results for $U^{(k)}_{n}(a,b)$ for the instances $k=2$, resp. $k=3$ (in eq. (26) of this ref. one has to multiply the lhs with $2!$, and in the second line of $N$ of eq.$(27)$ it should read $B(2,n+1)$).

For the case $a=1=b$ the triangles of the coefficients of these polynomials can be viewed under the nrs. A057995 and A057280 in [5]. For $a=2,\, b=1$ see A058402 and A058403.

Similarly, iteration of recursion eq. 25 results, with the help of the recursion relation eq. 1, in

$$V^{(k)}_{n}(a,b)\ =\ {\frac{1}{k!\,a\,(a^2 + 4\,b)^k}}\, \Bigl{(} AV_{k}(a,b;n)\, V_{n+1}(a,b)\ +\ BV_{k}(a,b;n)\,V_{n}(a,b)\Bigr{)}\ ,$$ (40)

with certain polynomials $AV_{k}(a,b;n)$ and $BV_{k}(a,b;n)$ of generic degree $k$ in the variable $n$, for fixed $a\neq 0$, $b \neq 0$, with $b\neq -a^2/4$.

The (mixed) recursion relations for these polynomials are found from eq. 26, and for $k=1,2,...$, they are

$$AV_{k}(a,b;n) \ =\ a^2\,(n+1)\,AV_{k-1}(a,b;n+1)\ +\ 2\,b\,(2\,n+k)\, AV_{k-1}(a,b;n)\ +\$$

$$a\, (n+1)\, BV_{k-1}(a,b;n+1)\ +\ 4\,{\frac{b}{a}}\,(n+k-1)\, BV_{k-1}(a,b;n-1) \ ,$$ (41)

$$BV_{k}(a,b;n) \ =\ 2\,b\,(2\,n+k)\,BV_{k-1}(a,b;n)\ -\ 4\,\,b\,(n+k-1)\, BV_{k-1}(a,b;n-1) \ +\$$

$$a\,b\,(n+1)\,AV_{k-1}(a,b;n+1)\ +\ 4\,{\frac{b^2}{a}}\,(n+k-1)\, AV_{k-1}(a,b;n-1)\ ,$$ (42)

with inputs $AV_{0}(a,b;n)=0$ and $BV_{0}(a,b;n)=a$.

For $a=1=b$ the triangles of coefficients of these polynomials in $n$ can be found under the nrs. A061188 and A061189 in [5]. Observe that $BV_{1}(1,1;n)$ is accidentally of degree 0. For $a=2, b=1$ see nrs. A062133 and A062134. For $a=2,\, b=1$ see A062133 and A062134.

Motivated by a recent paper [6] we consider also the following generalized $p-$Fibonacci numbers $U_{n}(p;a,b)$ defined for $p\in \NN0$ by the generating function

$$U(p;a,b;x)\ :=\ {\frac{1}{1\, -\, a\,x\, -\, b\,x^{p+1}}}\ =\ \sum_{n=0}^{\infty}\, U_{n}(p;a,b)\, x^n\ .$$ (43)

Of course, we assume $b \neq 0$ and also take $a\neq 0$. For $p=1$ these numbers reduce to the $U_{n}(a,b)$ treated above, and for $p=0$ they become the powers $(a+b)^n$. $U(p;1,1;x)$ appears in eq. 71 of [2]. The recursion relations are

$$U_{n}(p;a,b) \ =\ a\,U_{n-1}(p;a,b) \ +\ b\, U_{n-(p+1)}(p;a,b)\ ,$$ (44)

with inputs $U_{j}(p;a,b)\, =\, a^j$ for $j=0,1,...,p$. In order to derive expressions for the $k-$th convolution of these $p-$Fibonacci numbers consider first the following Riccati eq. satisfied by $U(p;a,b;x)$ written for the non-degenerate case $D(p;a,b)\, :=\,(p+1)^{p+1}\,b \, +\, a\,(a\,p)^{p}\, \neq\, 0$ if $p\in \mathbb{N}$, and $a+b\neq 0$ if $p=0$ (i.e. one puts $(a\,p)^p \, =\, 1$ if $p=0$).

$$U^{2}(p;a,b;x)\ =\ {\frac{1}{(p+1)^{p+1}\,b \, +\, a\,(a\,p)^p}}\... ...\over{\partial {x}}}\ +\ b\,(p+1)^2 \, B_{p-1}(a;x)\,\Bigr{\}}\, U(p;a,b;x) \ ,$$ (45)

with

$$A_{p}(a,b;x) \ =\ (a\,p)^p\ +\ b\,(p+1)\, x\, B_{p-1}(a;x)\ ,$$ (46)

$$B_{p-1}(a;x) \ =\ (p+1)^{p-1}\, \sum_{i=0}^{p-1}\,\Bigl{(} {\frac{a\,p}{p+1}}\Bigr{)}^i\,x^i \ =\ {\frac{(p+1)^p \, -\, (a\,p\,x)^p} {p+1\, -\, a\,p\,x}} .$$ (47)

For $p=0$ one has to use $A_{0}(a,b;x)\, =\, 1$ and $B_{-1}(a;x)\, =\, 0$. For given non-vanishing $a$ and $b$ these polynomials $A_{p}(a,b;x)$, resp. $B_{p-1}(a;x)$, in the variable $x$ of degree $p$, resp. $p-1$, have therefore the following explicit form.

$$A_{p}(a,b;x)\ =\ \sum_{m=0}^{p}\,A(a,b;p,m)\ x^m\ \ \ , \ \ \ B_{p-1}(a;x)\ =\ \sum_{m=0}^{p-1}\,B(a;p-1,m)\ x^m \ ,$$ (48)

with the coefficients

$$A(a,b;p,m) \ =\ \left\{ \begin{array}{ll} 0 & \mbox{if $m\,>\, p, $}\\ 1 & \mbox... ...frac{a\, p}{p+1}} \Bigr{)}^{m-1}$} & \mbox{if $m\geq 1\ . $} \end{array}\right.$$ (49)

$$B(a;p,m) \ =\ \left\{ \begin{array}{ll} 0 & \mbox{if $p\,<\, m \ {\rm or}\ p=-1... ... (p+1)}{p+2}} \Bigr{)}^m $} & \mbox{if $p \geq m\geq 0\ . $} \end{array}\right.$$ (50)

For $a\, =\, 1 \, =\, b$ these triangles of coefficients can be viewed under the numbers $A055858$ and $A055864$ in [5] where further details may be found.

Even though we cannot integrate the linear differential eq., which is equivalent to the Riccati eq.  45 for $p\neq 0, 1$, analytically, the above given solution is its unique one due to the existence and uniqueness theorem for the linear first order differential eq. satisfied by $Z(p;a,b;x)\, :=\, 1/U(p;a,b;x)$ with initial value $Z(p;a,b;0) \, =\, 1$.

The result for the first ($k=1$) convolution of the numbers $U_{n}(p;a,b)$ which flows from the Riccati eq. is

$$U_{n}^{(1)}(p;a,b)\ =\ {\frac{1}{b\,(p+1)^{p+1}\, +\, a\,(a\,p)^p}}\,\sum_{j=0} ^{p}\, C_{j}(n;p;a,b)\,U_{n+1-j}(p;a,b)\ ,$$ (51)

with

$$C_{j}(n;p;a,b)\ =\ \left\{ \begin{array}{ll} n+1 & \mbox{if $p = ... ...Bigr{)}^{j-1}$\ } & \mbox{if $p\geq 1$\ and $j = 1,...,p. $} \end{array}\right.$$ (52)

The $U_{n}(p;a,b)$ recursion cannot be used to simplify the sum in eq. 51.

This result can now be compared, after putting $a=1=b$, with a different formula for the same convolution found in [6], eq.(14). For given $p\in \NN0$ and $k=2,3,...,$ the recursion for $F^{(2)}_{p}(k)\ \hat =\ U_{k-2}^{(1)}(p;1,1)$ in [6] involves all $k-1$ terms $F_{p}(n)\ \hat =\ U_{n-1}(p;1,1)$, for $n=1,...,k-1$, whereas our result needs only $p+1$ terms for all $k$. For example, $F^{(2)}_{3}(7)\ \hat =\ U_{5}^{(1)}(3;1,1)$ is reduced to six terms involving $F_{3}(1)\ \hat =\ U_{0}(3;1,1)$,..., $F_{3}(6)\ \hat =\ U_{5}(3;1,1)$ in [6], but only to four terms, involving $U_{8}(3;1,1)\ \hat =\ F_{3}(9)$, $U_{7}(3;1,1)\ \hat =\ F_{3}(8)$,..., $U_{5}(3;1,1)\ \hat =\ F_{3}(6)$ in eq. 51.

For the $k-$th convolution one uses the following identity which derives from the Riccati eq. 45. It is written for the non-degenerate case, and is valid for $k\in \mathbb{N}$.

$$U^{k+1}(p;a,b;x)\ =\ {\frac{1}{k\,\Bigl{(}b\,(p+1)^{p+1}\, +\, a\... ...partial {x}}} \ +\ k\, b\,(p+1)^2 \, B_{p-1}(a;x)\,\Bigr{\}}\,U^{k}(p;a,b;x)\ .$$ (53)

For $p\in \NN0$ the corresponding recursion relation for the $k-$th convolution is (remember that one puts $(a\,p)^p \, =\, 1$ if $p=0$)

$$U_{n}^{(k)}(p;a,b)\ =\ {\frac{1}{k\,\Bigl{(}b\,(p+1)^{p+1}\, +\, ... ...Bigr{)}}}\, \sum_{j=0}^{p}\, C_{j}^{(k)}(n;p;a,b)\,U_{n+1-j}^{(k-1)}(p;a,b) \ .$$ (54)

with

$$C^{(k)}_{j}(n;p;a,b)\ =\ \left\{ \begin{array}{ll} n+1 & \mbox{if... ...Bigr{)}^{j-1}$\ } & \mbox{if $p\geq 1$\ and $j = 1,...,p. $} \end{array}\right.$$ (55)

Instead of showing the rather unwieldy formula for this iterated recursion relation after employing the fundamental recursion eq. 1 we prefer to state the result for the instance $p=2, k=2, a=1=b$, with the notation $U_{n}^{(1)}(2;1,1)\equiv U_{n}^{(1)}(2)$ and $U_{n}(2;1,1)\equiv U_{n}(2)$:

$$U_{n}^{(2)}(2) \ =\ {\frac{1}{2\cdot 31}}\,\Bigl{(}4\,(n+1)\, U_{n+1}^{(1)}(2)\ +\ 9\,(n+6)\,U_{n}^{(1)}(2)\ +\ 6\,(n+5)\, U_{n-1}^{(1)}(2)\,\Bigr{)}$$

$$\ =\ {\frac{1}{2\cdot 31^2}}\,\Bigl{(}\, (217\,n^2 \, +\, 1425\,n \, +\, 1922)\, U_{n}(2)\ +\ 2\,(n+2)\,(62\,n\, +\, 305)\,U_{n-1}(2)\ +\$$

$$\hskip 1.5cm 4\,(n+1)\,(31\,n+143)\, U_{n-2}(2)\, \Bigr{)}\ .$$ (56)

The recursion relation eq. 1 has been used twice.

In the degenerate case $D(p;a,b)\, :=\,(p+1)^{p+1}\,b \, +\, a\,(a\,p)^{p}\, \, =\, 0$ (where one puts $(a\,p)^{p}\, \equiv \, 1$ if $p=0$) one finds for $U(p;a,b=b(p;a);x)\, =:\,U(p;a;x)$, where $b(p;a)\, :=\, -p^p\,(a/(p+1))^{p+1}$, the linear differential eq.

$$\Bigl{\{}A_{p}(a,b(p;a);x)\, {{\partial {}}\over{\partial {x}}}\ +\ b(p;a)\,(p+1)^2 \, B_{p-1}(a;x)\,\Bigr{\}}\, U(p;a;x) \ =\ 0 \,$$ (57)

with $B_{p-1}$ and $A_{p}$ taken in their explicit form known from eqs. 47 and  46. For general $p$ and $D(p;a,b)=0$ we cannot say anything about convolution because we have no suitable expression for $U^{2}(p;a,b;x)$. The recurrence eq. 1 with depth $p+1$ can be replaced by one with only depth $p$. See eq. 73.

In the non-degenerate case one could also consider the other $p$ linear independent (Lucas-type) sequences defined by the recurrence eq.  and appropriate inputs but we will not do this here.

The remainder of this paper provides proofs for the above given statements.