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I´ll be straightforward.

I encountered a pretty complex sum. I have tried to find a way of easily calculating Z($x,t$), but I haven´t been able to come up with anything. I will try to explain as clearly as I can the pattern of the sums.

Z1($x,t$) = ${0}\choose{0}$${x}\choose{1}$+${1}\choose{1}$${x+1}\choose{2}$+${2}\choose{2}$${x+2}\choose{3}$+${3}\choose{3}$${x+3}\choose{4}$+${4}\choose{4}$${x+4}\choose{5}$+...+${t-1}\choose{t-1}$${x+t-1}\choose{t}$

Z2($x,t$) = ${1}\choose{0}$${x+1}\choose{2}$+${2}\choose{1}$${x+2}\choose{3}$+${3}\choose{2}$${x+3}\choose{4}$+${4}\choose{3}$${x+4}\choose{5}$+${5}\choose{4}$${x+5}\choose{6}$+...+${t}\choose{t-1}$${x+t}\choose{t+1}$

Z3($x,t$) = ${2}\choose{0}$${x+2}\choose{3}$+${3}\choose{1}$${x+3}\choose{4}$+${4}\choose{2}$${x+4}\choose{5}$+${5}\choose{3}$${x+5}\choose{6}$+${6}\choose{4}$${x+6}\choose{7}$+...+${t+1}\choose{t-1}$${x+t+1}\choose{t+2}$

Z4($x,t$) = ${3}\choose{0}$${x+3}\choose{4}$+${4}\choose{1}$${x+4}\choose{5}$+${5}\choose{2}$${x+5}\choose{6}$+${6}\choose{3}$${x+6}\choose{7}$+${7}\choose{4}$${x+7}\choose{8}$+...+${t+2}\choose{t-1}$${x+t+2}\choose{t+3}$

...

I wrote them as binomial coefficients since I thought it would help. It´s easy to see how the diagonals of the first terms (the ones without x´s) form a pascal´s triangle.

The ${x+n}\choose{n+1}$ term could be seen as the nth summation of $x$, in the sense that

When n=0, ${x}\choose{1}$ = $x$

When n=1, ${x+1}\choose{2}$ = $\frac{x^2+x}{2}$ = S1($x$) = (summation of first $x$ integers)

When n=2, ${x+2}\choose{3}$ = $\frac{x^3+3x^2+2x}{6}$ = S2($x$) = (summation of the first $x$ summations of the first $x$ integers

...

In this sense ${x+n}\choose{n+1}$$-$${x-1+n}\choose{n+1}$=${x+n}\choose{n}$

= Sn($x$) $-$ Sn($x-1$) = Sn-1($x$)

Another way then of seeing the first table would be:

Z1($x,t$) = x + S1($x$) + S2($x$) + S3($x$) + S4($x$) + ... + St($x$)

Z2($x,t$) = S1($x$) + ($2$)S2($x$) + ($3$)S3($x$) + $(4)$S4($x$) +$(5)$S5($x$) + ... + ($t$)St+1($x$)

Z3($x,t$) = S2($x$) + $(3)$S3($x$) + $(6)$S4($x$) + $(10)$S5($x$) + $(15)$S6($x$) + ... + (S1($t$))St+2($x$)

Z4($x,t$) = S3($x$) + $(4)$S4($x$) + $(10)$S5($x$) + $(20)$S6($x$) + $(35)$S7$x$) + ... + (S2($t$))St+3($x$)

...

When $x = 1$

Z1($1,t$) = (1) + (1) + (1) + (1) + (1) + ... + (1) = Z2($1,t$)

Z2($1,t$) = (1) + (2) + (3) + (4) + (5) + ... + ($t$) = Z3($1,t$)

Z3($1,t$) = (1) + (3) + (6) + (10) + (15) + ... + S1($t$) = Z4($1,t$)

Z4($1,t$) = (1) + (4) + (10) + (20) + (35) + ... + S2($t$) = Z5($1,t$)

...

Some important facts that would be extremely helpful would be to know a function that approaches Zn($x,t$) as $t$ goes to infinity, or any sum that is similar to them.

Any thoughts would be really appreciated!

Sam
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  • May be of help: http://math.stackexchange.com/questions/1924327/possible-pattern-on-sum-of-first-n-l-th-powers/1924361#1924361 – Simply Beautiful Art Sep 23 '16 at 00:54
  • @Leo May we assume $x$ and $t$ are positive integers? – David H Oct 08 '16 at 23:41
  • @David H Yes. To know a formula in the case of $x$ and $t$ being integers will already be more than enough. Let me emphasize that also another function that is equal to $Zn(x,t)$ as $t$ goes o infinity would also be useful – Sam Oct 09 '16 at 06:11

2 Answers2

2

This is not an answer but it is too long for a comment.

If I properly looked at the problem, you are working $$Z_k(x,t)=\sum_{i=0}^t \binom{k-1}{i} \binom{x+k-1}{i+k}$$ It seems that we can obtain a "closed" form solution involving the generalized hypergeometric function and get for $Z_k(x,t)$ the following expression $$\frac{\Gamma (x+2 k-1)}{\Gamma (2 k) \,\Gamma (x)}-\binom{k-1}{t+1} \binom{x+k-1}{t+k+1} \, _3F_2(1,t-k+2,t-x+2;t+2,t+k+2;1)$$

I hope and wish that this could be of some help (but not sure).

Edit

$$Y_k(x)=\lim_{t\to \infty } \, Z_k(x,t)=\frac{\Gamma (x+2 k-1)}{\Gamma (2 k)\,\, \Gamma (x)}$$

If we consider large values of $x$, the asymptotics is $$Y_k(x)= \frac{x^{2k-1}}{\Gamma(2k)}\left(1+\frac{(k-1) (2 k-1)}{x}+\frac{(k-1) (2 k-3) (2 k-1) (3 k-2)}{6 x^2}+O\left(\frac{1}{x^3}\right)\right)$$

Claude Leibovici
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  • Could u provide a proof?? Sorry that it took me so long to comment – Sam Oct 09 '16 at 06:13
  • @Leo. Most of the beginning was obtained using an old CAS. As I wrote, this is not an answer. After your comment on today, I just finished with the limit. – Claude Leibovici Oct 09 '16 at 07:55
  • Are you sure you have the right equation?? As $t$ approaches infinity, $Zn$ approaches infinity for all $x$ greater than 1, but that does not happen with your equation. – Sam Oct 10 '16 at 20:47
  • @Leo. See my edit : for infinite $t$, $Z_k(x) \approx \frac{x^{2k-1}}{\Gamma(2k)}$ – Claude Leibovici Oct 11 '16 at 04:58
1

Not sure that it could be useful, but I would like to suggest an alternative approach that leads to a slightly different formula, providing the relative proof.

As I understood the problem, the summation of the product of the two bimomial coefficients should contain a term $i $ (ranging from $0$ to $t-1$) in all four quantities that define the binomial coefficients. In particular, it seems to me that the sums described in the OP should be expressed as

$$Z_k(x,t )=\sum_{i=0}^{t-1} \binom{i+k-1}{i} \binom{x+i +k-1}{i+k}$$

or equivalently

$$Z_k(x,t )=\sum_{i=0}^{t-1} \frac {(i+k-1)!}{i!(k-1)!} \frac {(x+i+k-1)!}{(i+k)!(x-1)!}$$

$$ =\frac {1}{(k-1)!(x-1)!} \sum_{i=0}^{t-1} \frac {(i+k-1)!}{i!} \frac {(x+i+k-1)!}{(i+k)!}$$

Because the sum can be written as a difference of hypergeometric series, we have

$$ Z_k(x,t )= \frac {1}{(k-1)!(x-1)!} \left[ \frac {(k-1)! (x+k-1)!}{k!} \, _2F_1 (k, x+k; k+1;1) - \frac {(k+t-1)! (x+k+t-1)!}{t!(k+t)!} \, _3F_2 (k+t, x+k+t, 1; k+t+1, t+1;1) \right] $$ $$= \binom{x+k-1}{k} \, \frac {\Gamma (k+1) \, \Gamma(1-x-k)} {\Gamma (1-x) } $$ $$ - \binom {k+t-1}{t} \binom {x+k+t-1}{k+t} \, _3F_2 (k+t, x+k+t, 1; k+t+1, t+1;1)$$

This last formula is confirmed by WA here.

Anatoly
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