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Who can continue (complete) the following sequence:

$$1,n-1,\frac{(n-2)(n-1)}{2},\frac{(n-4)(n-3)(n+1)}{6},\frac{(n-7)(n-4)(n-2)(n+3)}{24},\dots$$

This was emerging in the course of this question as crucial coefficients in the transformation of Fibonacci polynomials.

I am pretty sure I have seen this before in another context, but I cannot remember it exactly, I have the vague memory that it contains some more complicated structure than factorials, maybe something like rising factorials/Pochhammer-symbols.

Edit

The next sequence of this row is numerically $$a_6 = \{-4,12,-21,-24,-9,42,154,360,702,1232\dots\}$$

This is no 5$^{th}$ degree polynomial anymore and I am unable to detect the law.. If fact it is one, this can be seen from the fifth difference sequence which gets constant (when calculated correctly).

By interpolation and factorisation I finally got the law (see below).

Edit-2

Meanwhile I found an alternative way to generate these series, but however still got stuck at the 6$^{th}$ one.

They can be expressed as the coefficients of the series expansion of polynomial fractions $f_i(x)$ at $x=0$, the functions are in detail:

\begin{array}{c|c|c|c|c|c|c} i& 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline a_i&1 & n-1 & \frac{(n-2)(n-1)}{2}& \frac{(n-4)(n-3)(n+1)}{6}&\frac{(n-7)(n-4)(n-2)(n+3)}{24}&v.i.&v.i.\\ \hline f_i(x)&\frac{1}{x-1}& \frac{1}{(x-1)^2}&\frac{1}{(x-1)^3}&\frac{(x-2)x}{(x-1)^4}&\frac{(x-2)(1+x(x-3))}{(x-1)^5}&?&\\ \end{array}

Edit-3

So, the expression for the next sequences are:

$$ a_6 = \frac{(n-5)(n-4)(n-1)(n^2-5n-54)}{120} $$

$$ a_7 = \frac{(n-6)(n-3)(n^4-12n^3-71n^2+642n+160)}{720} $$

I am still unable to guess a general law for $f_i(x)$ or $a_i$.

Edit-4

I will try to get some recursion relation now from the consideration of the difference series.

Raphael J.F. Berger
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1 Answers1

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As you reference in another question, $$F_n(x) = \sum_{k=0}^n \binom{\frac{n+k-1}{2}}{k} x^k$$

If we consider $F_n(x+1)$ directly we get $$\begin{eqnarray} F_n(x+1) & = & \sum_{k=0}^n \binom{\frac{n+k-1}{2}}{k} (x+1)^k \\ & = & \sum_{k=0}^n \binom{\frac{n+k-1}{2}}{k} \sum_{i=0}^k \binom{k}{i} x^i \\ & = & \sum_{i=0}^n x^i \sum_{k=i}^n \binom{\frac{n+k-1}{2}}{k} \binom{k}{i} \\ \end{eqnarray}$$

If we consider $F_n(x+1)$ as a sum of weighted $F_m(x)$ we get $$\begin{eqnarray} F_n(x+1) & = & \sum_{m=0}^n \lambda_m \sum_{k=0}^m \binom{\frac{m+k-1}{2}}{k} x^k \\ & = & \sum_{k=0}^n x^k \sum_{m=k}^n \binom{\frac{m+k-1}{2}}{k} \lambda_m \\ \end{eqnarray}$$

By equating coefficients in $x^i$ we get

$$\sum_{m=i}^n \binom{\frac{m+i-1}{2}}{i} \lambda_m = \sum_{k=i}^n \binom{\frac{n+k-1}{2}}{k} \binom{k}{i}$$

On the left-hand side, discard the term $m=i$ (since $2i-1$ is odd) and extract the term $m=i+1$ from the sum, and the sum can be passed to the right:

$$\lambda_{i+1} = \left(\sum_{k=i}^n \binom{\frac{n+k-1}{2}}{k} \binom{k}{i}\right) - \sum_{m=i+2}^n \binom{\frac{m+i-1}{2}}{i} \lambda_m$$

And voilà: one recurrence.

NB your $a_j$ is $\lambda_{n-j+1}$, so

$$a_j = \left(\sum_{k=n-j}^n \binom{\frac{n+k-1}{2}}{k} \binom{k}{n-j}\right) - \sum_{m=n-j+2}^n \binom{\frac{n+m-j-1}{2}}{n-j} a_{n-m+1}$$

and it makes sense to reindex at least the second sum as

$$a_j = \left(\sum_{k=n-j}^n \binom{\frac{n+k-1}{2}}{k} \binom{k}{n-j}\right) - \sum_{m=1}^{j-1} \binom{n-\frac{m+j}{2}}{n-j} a_{m}$$

Peter Taylor
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  • Using your expression directly after the line "by equating coefficients in $x^i$", I receive for $k=n$ that $\lambda_n=1$, which is fine. But for $k=n-1$, I get $\lambda_{n-1}=2n-1$ but correct would be $\lambda_{n-1}=n-1$. I suspect somewhere must be an error. – Raphael J.F. Berger Apr 12 '18 at 07:28
  • @R_Berger, assuming that you mean for $i=n$, you should get empty sums on both sides. For $i=n-1$ it gives $\binom{n-1}{n-1} \lambda_{n} = \binom{n-1}{n-1} \binom{n-1}{n-1}$ ; for $i=n-2$ it gives $\binom{n-2}{n-2} \lambda_{n-1} = \binom{n-1}{n-1} \binom{n-1}{n-2}$. I suspect that you're not taking into account that $\binom{a+\tfrac12}{b} = 0$ for $a,b\in\mathbb{N}$. – Peter Taylor Apr 12 '18 at 07:53
  • I always thought ${1/2 \choose 1}=\frac{(1/2)!}{(-1/2)!1!}=\frac{\sqrt{\pi}/2}{\sqrt{\pi}1}=1/2$, no? – Raphael J.F. Berger Apr 12 '18 at 08:10
  • @R_Berger, if you want to work with the analytic continuation of the binomial function rather than the one which is defined only for integer arguments, it's a case of reworking the algebra starting from $F_n(x)=\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n-k-1}{k}x^{n-2k-1}$ – Peter Taylor Apr 12 '18 at 08:17
  • @PeterTaylor_ So you are saying that the definition of $F_n$ I refer to is specifically valid only for the not analytically continued factorials/binomials? In either case we might want to get rid of the zero terms then I guess. – Raphael J.F. Berger Apr 12 '18 at 08:20
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    I am indeed saying that. – Peter Taylor Apr 12 '18 at 08:39
  • @PeterTaylor_: Another question: Is it correct that in the coefficient comparison we have terms like \begin{eqnarray} x^0 (a_0 + a_1 + a_2 + \dots+ a_n) & = & x^0 (b_0 + b_1 + b_2 + \dots +b_n) \ x^1 (a_1 + a_2 + \dots +a_n) &=& x^1 (b_1 + b_2 + \dots +b_n) \ x^2 ( a_2 + \dots +a_n) &=& x^2 (b_2 + \dots b_n) \ &\dots& \ x^n a_n & = & x^n b_n \ \end{eqnarray} Don't we have then to conclude then that $a_i=b_i \forall i$, or am I missing something? – Raphael J.F. Berger Apr 12 '18 at 08:45
  • @R_Berger, not quite. Firstly, because it only goes up to $x^{n-1}$; and secondly because there's weighting. – Peter Taylor Apr 12 '18 at 10:01
  • Your expression works, eg. in mathematica implemented like: 
    `ClearAll[binomial ] binomial[n_Integer, m_Integer] := Binomial[n, m];
binomial[n_, m_] := 0; 

a[n_,j_]:=Sum[binomial[(n+k-1)/2,k]  binomial[k,n-j] ,{k,n-j,n}]- Sum[
binomial[n-(m+j)/2,n-j]  a[n,m],{m,1,j-1}]`

    it gives the correct coefficients a[n,j] , which are the $a_{j}$ for $F_n(x+1)$. Thank you!

    Along this line it would be nice to show how to get from the recursions to the recursion free, for the first $n$. And finally maybe a general expression?

     – Raphael J.F. Berger Apr 12 '18 at 12:11
  • Yet, another thing you are exclusively working on the coefficients $\lambda$ in $F_n(x+1)=\sum \lambda_{n,i} F_i (x)$. Strictly this was the question from the "linked question". This question deals with the diagonals and subdiagonals in the coefficient matrix. So the sequences I gave $1,n-1,(n-1)(n-2)/2$ are the $a_{n,i} = \lambda_{i+n,n}$. So your answer actually belongs to here. Do like to migrate it? – Raphael J.F. Berger Apr 12 '18 at 12:51
  • @R_Berger, to the extent that this is also an answer to your earlier question it's because this question duplicates that one. If you want to flag this question and ask the mods to merge it into the other one, feel free. But this answer directly addresses this question: "Who can continue (complete) the following sequence". That sequence is precisely $a_1$, $a_2$, $a_3$, $\ldots$. Note that the $a_j$ are polynomials in $n$. – Peter Taylor Apr 12 '18 at 13:38
  • When I calculate $a_j$ for $j=6$ using your recursion formula for $n$ from $1\dots 10$ I get ${0,-3,-2,0,0,-4,-12,-21,-24,-9}$, while it should be ${-4,-12,-21,-24,-9,42,154,360,702,1232}$, so we are almost there. It seems you are right again (even when I still don't understand everything completely). But anyway, thank you very much! – Raphael J.F. Berger Apr 12 '18 at 16:21
  • Could you maybe do some little edit? I cannot upvote it until there was an edit. Maybe that were you compare the coefficients of $x^i$ with those of $x^k$. It seems everything is correct there as well, but it could be irritating for some - at least for me. – Raphael J.F. Berger Apr 12 '18 at 17:11