# Recurrent sequences and Bernoulli-like numbers

Consider the Fibonacci polynomials defined by $$F_n(s)=F_{n-1}(s)+sF_{n-2}(s)$$ with initial values $F_0(s)=0$ and $F_1(s)=1$ and define a linear functional $L$ on the polynomials in $s$ by $$L(F_{2n})=\delta_{n,1}.$$ Then $$L(F_{2n+1})=(2n+1)B_n,$$ where $B_n$ are the Bernoulli numbers defined by $B_n={\sum{n\choose k} B_k\}$ for $n\ge2$ and $B_0=1.$ Choosing the linear functional $M$ defined by $$M(F_{2n+1})=\delta_{n,0},$$ gives $$M(F_{2n})=(-1)^n G_{2n},$$ where $G_{2n}$ are the Genocchi numbers $G_{2n}=(-1)^n 2 (1-4^n) B_{2n}.$

Finally let $H_n$ be a variant of the Hermite polynomials defined by $$H_n(s)=H_{n-1}(s)-(n-1)s H_{n-2}(s)$$ and the linear functional $N$ defined by $$N(H_{2n})=\delta_{n,0},$$ then we get $$N(H_{2n-1})=(-1)^{n-1} T_{2n-1},$$ where $T_{2n-1}=(-1)^{n-1} \frac{4^n (4^n-1)}{2n} B_{2n}$ are the tangent numbers.

My question is: Are these isolated results or special cases of a more general theorem? Does anyone know other such examples?

Edit. To make my question somewhat more precise: Define the Fibonacci polynomials by $F_n(x,s)=xF_{n-1}(x,s)+sF_{n-2}(x,s)$ and the Hermite polynomials by $H_n(x,s)=xH_{n-1}(x,s)-(n-1)s H_{n-2}(x,s).$ The above results follow from the identities $$(e^{xz} + 1)\sum {\frac{{F_{2n} (x,s)}}{{(2n)!}}z^{2n} =(e^{x z}-1) \sum {\frac{{F_{2n + 1} (x,s)}}{{(2n + 1)!}}z^{2n + 1} } }$$ and $$(e^{2xz} + 1)\sum {\frac{{H_{2n + 1} (x,s)}}{{(2n + 1)!}}z^{2n + 1} = (e^{2xz} - 1)\sum {\frac{{H_{2n} (x,s)}}{{(2n)!}}z^{2n} } }.$$

Thus a more precise question would be: Are there polynomial sequences which satisfy similar identities?

Further edit. A more precise question: Are there "naturally occurring sequences" $A_n(x,s)$ satisfying $A_n(x,s)=xA_{n-1}(x,s)+c(n,s)A_{n-2}(x,s)$ such that $$\sum {\frac{{A_{2n} (x,s)}}{{(2n)!}}z^{2n} =b(z,x) \sum {\frac{{A_{2n + 1} (x,s)}}{{(2n + 1)!}}z^{2n + 1} } },$$ where $b(z,x)$ does not depend on $s?$ The only examples I know besides the Fibonacci and Hermite polynomials are the Lucas polynomials $L_n(x,s)$ defined by $L_n(x,s)=xL_{n-1}(x,s)+sL_{n-2}(x,s)$ with initial values $L_0(x,s)=2$ and $L_1(x,s)=x.$

• Is there a nice continued fraction interpretation of these identities?
– j.c.
Dec 8, 2010 at 16:42
• Are you still thinking of $s$ as a formal variable for polynomials with $x$ as a parameter? That could make sense for $F_n$ but I don't see that $x$ does much of anything for $H_n$ Dec 8, 2010 at 20:14
• Yes, $x$ should be a fixed number and $s$ the variable in order that the polynomials with even resp. odd indices form a basis of the vector space. Dec 8, 2010 at 20:15
• Well then did you mean $H_n(x,s)=xH_{n-1}(x,s)-(n-1)s H_{n-2}(x,s)?$ Otherwise I am not sure how the $x$ comes in. With that interpretation, More generally, given functions $u(n)$ and $v(n)$ (and initial conditions) let $A_n=A_n(u,v,s)$ be $$A_n=u(n)A_{n-1}+sv(n)A_{n-2}$$ In your two examples $(u,v)=(x,1)$ and $(u,v)=(x,n-1)$. Dec 8, 2010 at 21:12
• Yes of course. This is a typo. Dec 9, 2010 at 5:39

The question is a bit vague (not that that is all bad) so my answer will be as well. There are (infinitely) many ways to define sequences and matrices of numbers and polynomials. Some of the simpler ones lead to sequences with special names. Simple transformations will send some of these nice ones to others. A sequence of polynomials gives an array of coefficients . Diagonals can give something else. For example the binomial coefficients have their recurrence $\binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k}$. If we change it to $B(n,k)=s\,B(n-1,k-1)+B(n-1,k)$ (with appropriate initial conditions) then we just have $B(n,k)=\binom{n}{k}s^k$ and each row can be viewed as a polynomial $(1+s)^n$. It is well known that the Fibonacci numbers can be viewed as sums of certain diagonals of Pascals triangle. If we use instead the B(n,k) then we get the Fibonacci polynomials which start your question. $1,1,s+1,2s+1,s^2+3s+1,3s^2+4s+1\dots$ Splitting these into even and odd terms gives us $1,s+1,s^2+3s+1,\dots$ and also $1,2s+1,3s^2+4s+1,\dots$ either is a basis of the space of polynomials as is $1,s,s^2,\dots$ sending any one of these to an integer sequence (starting with 1) sends the others to another. See what the linear transformations in your examples do to $1,s,s^2,s^3,\dots$

Here is another example (which I was happy to discover) My personal favorite recurrence is that for convergents to $\sqrt2$, $1/1,3/2,7/5,17/12,41/29,\dots$ The numerators $1,3,7,17,41,\dots$ and the denominators 1,2,5,12,29,... both satisfy the recurrence $a_n=2a_{n-1}+a_{n-2}$ just with different initial conditions. They could be called the Pell and Pell-Lucas numbers. If we say $A_n=2A_{n-1}+sA_{n-2}$ then we get a sequence $1,1,s+2,3s+4,s^2+8s+8,5s^2+20s+16,\dots$ These are closely related to Chebyshev polynomials of the first kind.(replace s by -s^2 and multiply every other one by s). Avoiding many digressions, Consider the two sequences

$$1,s+2,s^2+8s+8,s^3+18s^2+48s+32,\dots$$ $$1,3s+4,5s^2+20s+16,7s^3+56s^2+112s+64,\dots$$

If the first is mapped to 1,0,0,0,0...then the last is mapped to 1, 2/3, -8/15, 32/21, -128/15, 2560/33, -1415168/1365, 57344/3, -118521856/255 which is intriguingly close to the negatives of the cosecant numbers 1, -1/3, 7/15, -31/21, 127/15, -2555/33, 1414477/1365, -57337/3, 118518239/255

If the last one is mapped to 1,0,0,0,0 then the first is mapped to -2,16,-272,7936,... which are the tangent numbers (but with alternating signs).

later: The OEIS and associated Journal of Integer Sequences are full of sequences and transformations taking one to another. The example above was an illustration (the only one I tried) of what I am sure would generate many examples (probably some nicer than the one I gave). Start with an integer recurrence of the form $a_0=0$ $a_1=1$ and $a_m=u(m)a_{m-1}+v(m)a_{m-2}$. Replace this with $A_0=1$ $A_1=1$ and $A_m=u(m)a_{m-1}+sv(m)a_{m-2}$. Then (in general) $A_{2t}$ and $A_{2t+1}$ are both polynomials of degree $t$ so one has two bases for the additive vector-space of polynomials. Take the linear transformation which sends one to $1,0,0,0,\dots$ and see what it does to the other (or some other basis of that space). Look up the associated sequence (perhaps the numerators and denominators if it is a rational sequence) in the OEIS and see what you got. Examples (which I have not investigated) include powers of 2 ($a_n=a_{n-1}+2a_{n-2}$) and factorials ($a_n=(n-1)a_{n-1}+(n-1) a_{n-2}$). Those two are shifted, it would be nicer to start with $a_0=1$ and $a_1=1$ making appropriate adjustments. Starting the second example with $a_0=1$ and $a_1=0$ gives derangements. One could consider other initial conditions $A_0=c_0$ and $A_1=c_1$

There are many sequences of orthogonal polynomials like the Chebshev polynomials I mentioned above, (Legendre, Laguerre, Hermite, Jacobi,etc.) which are alternately even and odd. Dividing the odd ones by $x$ and then replacing $x$ with $\sqrt{x}$ (maybe taking absolute value of coefficients) gives more examples. Exponential generating functions might be more appropriate in some cases.

All that we are using of the polynomial structure is the (additive) book-keeping for the coordinates of a vector-space. Of course for famous sequences of polynomials that is still interesting. In general we could instead look at infinite lower triangular matrices. Then I think that your linear transformations correspond to the first column of the inverse matrix and the image on the companion sequence is simply the first column of a certain matrix product. Other columns might be of interest as well.

even later I did not say that it would give you what you wanted, only that I thought it might. Taking up my idea for powers of 2 gives two sequences of polynomials $1,2s+2,4s^2+10s+2,8s^3+36s^2+18s+2,\dots$ and $2,6s+2,16s^2+14s+2,40s^3+64s^2+22s+2,\dots$. Sending the first to $1,0,0,0,\dots$ sends the powers of $s$ to $1,-1,2,-7,38,-295,3098,\dots$ which look to be the absolute values of Genocchi numbers of second kind divided by 2^(n-1). This sends the other sequence to $2,-4,20,-172,2228,-40300,\dots$. The OEIS does not immediately identify that. Dropping the first term, still nothing. Since the remaining terms are multiples of 4 one divides that out to get $-1, 5, -43, 557, -10075, 241949, -7437547,\dots$ which does get identified as values B(2,n)/4 of Gandhi polynomials along with a reference to an article on combinatorial interpretations of Genochi numbers. I did not identify anything happening if we try to send the second sequence to $2,0,0,0,\dots$. I did not try for a functional equation.

• Thank you for your interesting comments. The numbers $M(s^n)$ are the so called mean Genocchi numbers. For the other results consider the more general Fibonacci polynomials defined by $F_n(s)=xF_{n-1}(s)+sF_{n-2}(s)$. They satisfy the identity $$(1+e^{x z}){\sum{\frac{F_{2n}(s)}{(2n)!} z^{2n}\}}= (e^{x z}-1) {\sum{\frac{F_{2n+1}(s)}{(2n+1)!}z^{2n+1}\}}$$ which is easily proved using the Binet formulas (cf. uk.arxiv.org/ftp/arxiv/papers/0908/0908.1219.pdf ). Dec 8, 2010 at 10:05
• @Aaron: Concerning your later remarks: My question is not about abstract generalities which are routine. I am interested in concrete cases where the functionals or the corresponding Seidel matrices give "famous" numbers as you call them or where the generating functions of the odd resp. even subsequences are connected via "simple" functions as in the examples I have given. Dec 8, 2010 at 19:48
• Thank you for your last remarks. This goes into the direction I am looking for. Your example is in fact a linear combination of Fibonacci polynomials. Your sequence is $a{(n,2s)}$ with $a(n,s) = F_n (1,s) + F_{n + 1} (1,s)$. The linear functionals give of course the same numbers in both cases. So it suffices to consider the sequence $a(n,s).$ Dec 9, 2010 at 11:28
• Define the functional $L$ by $L(a(2n,s)) = \delta_ {n,0}.$ Then $$L(a(2n+1,s)) = (-1)^n ( G_{2n + 2} + G_{2n + 4} ),$$ where ${(G_{2n})} = (1,1,3,17,155, \cdots)$ is the sequence of unsigned Genocchi numbers OEIS A110501. This gives the sequence $(2, - 4,20, - 172, \cdots )$. If we define the linear functional $M$ by $M(a(2n + 1,s)) = 2\delta_{n,0} ,$ then $$M(a(2n,s)) = (2n + 1)B_{2n} - (2n + 3)B_{2n + 2} .$$ Dec 9, 2010 at 11:29
• Sorry, it should be $$M(a(2n,s)) = 2((2n + 1)B_{2n} - (2n + 3)B_{2n + 2} ).$$ Dec 11, 2010 at 9:29