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My task is to find formula for generating function of sequence $a_0, a_1...$ defined with following recurence $a_0=1$ and $a_n=\sum_{i=0}^{n-1} (n-i)a_i$.

I rewrote the expression $a_n=\sum_{i=0}^{n-1} (n-i)a_i=n+(n-1)a_1+(n-2)a_2+...+a_{n-1}$ and I counted few members of sequence $a_1=1,a_2=3,a_3=8,a_4=21$ etc.

There are two answears below-wchich one i correct? Is there any way how to do some backward examination to show that it gives correct solution?

KarmaL
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  • Note that $\sum_{i=0}^{n-1}(n-i)a_i$ is the convolution of the sequences $a_i$ and $1,2,3,4,...$, with generating function $\sum_k kx^k$. – logarithm May 31 '19 at 17:45
  • Setting $f(x) = \sum_{n\ge 0}a_nx^n$ we have by definition $$f(x) = f(x)\sum_{n\ge 0}nx^n=f(x)\cdot\frac x{(1-x)^2}\ .$$ – dan_fulea May 31 '19 at 17:48
  • This is one in between(alternative ) Fibonacci sequence such that $a_n=F_{n+2}$ where F denotes Fibonacci number. For example $a_3=F_5=8$, or $a_5=F_7=55$ so you can use famous formula for Fibonacci sequence for n+2 for $a_n$. – sirous May 31 '19 at 18:39

2 Answers2

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Consider the relation at $n+1$: $$ a_{n+1}=\sum_{i=0}^{n}(n-i+1)a_i\tag1 $$ Letting $A(x)=\sum_{n\ge 0}a_nx^n$, the LHS is the coeficient of $x^{n}$ in $(A(x)-a_0)/x$, while the the RHS is a convolution, so it is the coefficient of $x^n$ in $$A(x)\cdot \sum_{k\ge 0}(k+1)x^k=A(x)\cdot (1-x)^{-2}.$$ Therefore, multiplying both sides by $x^n$ in $(1)$ and summing over $n\ge 0$, you get $$ (A(x)-1)/x=A(x)\cdot{(1-x)^{-2}}\implies \boxed{A(x)=\frac{(x-1)^2}{1-3x+x^2}.} $$ Here is a Wolfram Alpha sanity check.

Mike Earnest
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If we compare this with Fibonacci sequence, we have:

$F_x= 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . .$

$.x=1, 2, 3, 4, 5, 6, 7, .8, .9, 10, 11, . . .$

$a_n=. . . 1, . .3, .. .8,. .. 21,. . ..55, . . .$

$n=.. . .1 . . .2. . ..3. . .. .4. . . . .5$

We find that $x=2n$and $c_a=c_F^2$ where $c_F$ is golden ratio and $c_a$ is the ratio of sequence $a_n$. So we may conclude that :

$a_{n+1}=c^2 a_n$

$$a_n=F_x=\frac{1}{\sqrt5}[\frac{1}{1-(\frac{1+\sqrt 5}{2})x}-\frac{1}{1-(\frac{1-\sqrt 5}{2})x}]$$

$x=2n$; $n∈N$.

sirous
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  • You say $a_n=F_{n+2}$, but $a_4=21\neq 8=F_{4+2}$. I believe the correct expression is $a_n=F_{2n}$, with the only exception being $a_0=1\neq F_0$. – Mike Earnest May 31 '19 at 20:16
  • Your sequence of k is shifted. $F_0=0$ not that $F_1=0$ but as it is written above $a_n=F_{2n}$ works. Is there a way how to show it is correct? And I also think that there shouldn't be the equation $a_n=f(k)$. – KarmaL Jun 02 '19 at 14:46
  • @KarmaL, you and Mike are right. – sirous Jun 02 '19 at 16:57