Sequence with coeffs in sequence

After doing some work (which I don't really know how to render visually with LaTeX), I found that the coefficients, $b_k$ , are given by the recurrence relation $b_k = 2^{k-1} + a_{k-1}$ $a_k = a_{k-1} + b_k$ $a_0 = 0$ Here, the index $k$ corresponds to the horizontal rows of the table starting from $k=1$ at the top. Essentially, I had to create a new column to the left of the existing triangle, given by the coefficients $a_k$ and starting at $k=0$ . Because of this, I was able to calculate all the values of $b_k$ without having to worry about the rest of the columns to the right of it.

Using this recurrence relation, I found the set of $b_k$ 's: $[1, 3, 8, 20, 48, 112, 256, 576, 1280, ...]$

So $S(x)$ is given by $S(x) = \displaystyle \sum_{k=1}^\infty b_k x^k$ Now, the actual problem: $S\left(\frac{1}{3}\right) = \displaystyle \sum_{k=1}^\infty b_k \left(\frac{1}{3}\right)^k$

Revisiting the recurrence relation, it takes some more work to see that: $a_k = n \cdot 2^{k-1} + 2^n a_{k-n} = k \cdot 2^{k-1}$ $b_k = 2^{k-1} + a_{k-1} = 2^{k-1} + (k-1)2^{k-2}$

Finally, we can plug this back into the summation above:

$S\left(\frac{1}{3}\right) = \displaystyle \sum_{k=1}^\infty (2^{k-1} + (k-1)2^{k-2}) \left(\frac{1}{3}\right)^k = \frac{1}{4} \displaystyle \sum_{k=1}^\infty \left(\frac{2}{3}\right)^k + \frac{1}{4} \displaystyle \sum_{k=1}^\infty k \left(\frac{2}{3}\right)^k$

The last equality was found by using some algebra to rearrange the terms. The first term is simply the taylor series $\displaystyle \sum_{k=1}^\infty x^k = \frac{x}{1-x}$

Plugging in $x=2/3$ , we get $\frac{1}{4} \displaystyle \sum_{k=1}^\infty \left(\frac{2}{3}\right)^k = \frac{1}{4} \frac{2/3}{1/3} = \frac{1}{2}$

The last term requires a bit more work. Notice $\frac{d}{dx} \left( \displaystyle \sum_{k=1}^\infty x^k \right) = \frac{d}{dx} \left( \frac{x}{1-x} \right)$ $\displaystyle \sum_{k=1}^\infty kx^{k-1} = \frac{x}{(1-x)^2} + \frac{1}{1-x}$ $\displaystyle \sum_{k=1}^\infty kx^{k} = \frac{x^2}{(1-x)^2} + \frac{x}{1-x}$

Plugging and chugging: $\frac{1}{4} \displaystyle \sum_{k=1}^\infty k \left( \frac{2}{3} \right)^k = \frac{1}{4} \left( \frac{4/9}{1/9} + 2 \right) = \frac{3}{2}$

Adding the two terms together and we finally find our answer: $S\left(\frac{1}{3}\right) = \frac{1}{4} \displaystyle \sum_{k=1}^\infty \left(\frac{2}{3}\right)^k + \frac{1}{4} \displaystyle \sum_{k=1}^\infty k \left(\frac{2}{3}\right)^k = \frac{1}{2} + \frac{3}{2} = \boxed{2}$

As a bonus, here's the closed form of $S(x)$ :

$S(x) = \frac{x-x^2}{(1-2x)^2}$

If the numbers down the left-hand column are $a_0=1$ , $a_1=3$ , $a_2=8$ and so on, then $a_n$ is calculated as follows: Each integer $k+1$ , for $0 \le k \le n$ contributes $k+1$ to the sum $a_n$ a number of times. The number of times is the number of ways of getting from the position in the grid where $a_n$ lies to the position in the grid where $k+1$ lies, namely $\binom{n}{k}$ . Thus $a_n \; = \; \sum_{k=0}^n \binom{n}{k} (k+1)$ For example, $1$ contributes to $a_3$ $\binom{3}{0}=1$ time, $2$ contributes to $a_3$ $\binom{3}{1} = 3$ times, $3$ contributes $\binom{3}{2}= 3$ times, while $4$ contributes $\binom{3}{3}=1$ time, so that $20 \; = \; a_3 \; = \; 1 \times 1 + 3 \times 2 + 3 \times 3 + 1 \times 4$ Thus it is clear that $a_n \; = \; \sum_{k=0}^n \binom{n}{k}k + 2^n \; = \; n\sum_{k=1}^n \binom{n-1}{k-1} + 2^n \; = \; n2^{n-1} + 2^n = \frac12(n+2)2^n$ for all $n \ge 0$ , and hence, provided that $|x| < \tfrac12$ , $S(x) \; = \; x\tfrac12\sum_{n=0}^\infty (n+2)(2x)^n \; = \; \tfrac12x \frac{2x}{(1-2x)^2} + \frac{x}{1-2x} \; = \; \frac{x(1-x)}{(1-2x)^2}$ making $S(\tfrac13) = \boxed{2}$ .

Let S(x) = x + 3x^2 + 8x^3 + 20x^4 + 48x^5 + ....... The integral of S(x)dx = (1/2)x^2 + x^3 + 2x^4 + 4x^5 + ........ If |x| <1/2, we have an infinite convergent geometric progression with a = (1/2)x^2, r = 2x, and sum = a/(1 - r) = x^2/((2 - 4x).Differentiating, S(x) = (4x - 4x^2)/(4(1 - 2x)^2)), so S(1/3) = 2.0. Ed Gray

Part 1 find this sequences $1, 3, 8, 20, 48, 112, 256, 576$

Part 2 use Wolframalpha.

Answer $2$ .

Let $a_{1}$ =1 , $a_{2}$ =3 , $a_{3}$ =8 etc. If we let $a_{0}$ = $\frac{1}{4}$ , we obtain the recurrence relation $a_{n}$ = $2\times a_{n-1}+2^{n-2}$

Using this multiple times , we get
$a_{n}$ = $2^{n} \times a_{0} + n \times 2^{n-2}$ = $(n+1) \times 2^{n-2}$

We have to find

S( n )= $\sum_{i=1}^∞$ $a_{i}n^{i}$ = $\sum_{i=1}^∞$ $(i+1)2^{i-2}n^{i}$ .

Using summation formulas , this simplifies to $\frac{n(1-n)}{(1-2n)^{2}}$ for - $\frac{1}{2}$ < n < $\frac{1}{2}$

Also , substituting n = $\frac{1}{3}$ , we get S ( $\frac{1}{3}$ ) = $\boxed{2}$

Please, someone put a solution using limits

Taking the right border of the coefficients to be 1,3,2,5,3,7,4,9,5,11,6,13,7,15,8... and going down 1 and to the left 1 from each of these border coefficients gives successive right to left downward sloping lines of coefficients; e.g. 3,12,48 and 7,28,112 and 4,16,64,266. The coefficients in these successive downward sloping lines increase by 4 for each entry from the right border of the line, e.g. 7,28,112. The ends of the downward sloping lines form the coefficients 1,3 i8,20,48,112,256,576. For the odd entry number in this sequence, the series sum is x+sum from 2 to infinity (odd numbers only) of n [4^(n-1)] x^(2 n-1). The even entry number in the sequence, the sum is 3 [x^2]+the sum from 2 to infinity (even numbers only) of (2 n+1) [4^(n-1)]*[x^(2n)]. The sum of these odd and even series entries quickly converges to 2.

using $Mathematica$

x = 1/3; Sum[(n + 2)*2^(n - 1)*x^(n + 1), {n, 0, Infinity}]

returns $2$

Simple solution using Python :

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arr, seq, s, test, x = [1], [1], 0.0, 30, 1/3.0
for a in range(2,test): # define the sequence 1,3,8,20,48....
    arr.append(a)
    for b in reversed(range(a-1)): arr[b]+=arr[b+1]
    seq.append(arr[0])
for a in range(len(seq)): s+=seq[a]*x**(a+1) # calculate S(1/3)
print(s)

output: 1.9998709263974912

Hence $S(\frac{1}{3})=2$ .

This series (1,3,8,20,48,......) is

[(n+2)*{2^(n-1)},n=0,Inf].

Sum the infinite series terms in the infinite polynomial at x=(1/3) using WolframAlpha and obtain

Answer=2