## In Plato’s heaven, there exist an infinite number of bowls in a straight line.

Each bowl either contains some or none of a finite number of beans.

A child plays a game, which allows only one kind of move: removing two beans from any bowl, and putting one in each of the two adjacent bowls. The game ends when each bowl contains either one or no beans.

For example, consider two adjacent bowls containing $2$ and $3$ beans respectively, all other bowls being empty. The following eight moves will finish the game:

You are given the following sequences:

$\def\htmltext#1{\style{font-family:inherit;}{\text{#1}}}$

$

\begin{align}

\qquad t_0 &= 123456,\cr

\qquad t_i &= \cases{

\;\;\frac{t_{i-1}}{2},&$\htmltext{if }t_{i-1}\htmltext{ is even}$\cr

\left\lfloor\frac{t_{i-1}}{2}\right\rfloor\oplus 926252,&$\htmltext{if }t_{i-1}\htmltext{ is odd}$\cr}\cr

&\qquad\htmltext{where }\lfloor x\rfloor\htmltext{ is the floor function }\cr

&\qquad\!\htmltext{and }\oplus\htmltext{is the bitwise XOR operator.}\cr

\qquad b_i &= (t_i\bmod2^{11}) + 1.\cr

\end{align}

$

The first two terms of the last sequence are $b_1 = 289$ and $b_2 = 145$.

If we start with $b_1$ and $b_2$ beans in two adjacent bowls, $3419100$ moves would be required to finish the game.

Consider now $1500$ adjacent bowls containing $b_1, b_2, \ldots, b_{1500}$ beans respectively, all other bowls being empty. Find how many moves it takes before the game ends.

### This problem falls under the subjects of combinatorics, number theory and bitwise operations. Solving it requires significant mathematical theorems or techniques.

We have a sequence of numbers $t$ and a sequence of numbers $b$ derived from the $t$ sequence. It is implied that $t$ is used to generate the number of beans in each bowl, and then we are asked to find the number of moves needed to finish the game for 1500 bowls.

In order to calculate the number of moves for a given configuration of $b$ beans, let’s denote the resulting number of moves as $M(b_1, b_2, …, b_n)$. This function $M$ is yet unknown.

A key observation is that we can generalize the optimization problem. Instead of two beans from one bowl to its neighbors, we can consider moving one bean from it to its neighbor. Given initial number of beans $b_1$, $b_2$ for two bowls, the final situation would be $(b_1 – a, a)$ and $(b_2-a, a)$ where $a$ is the number of moves.

From these principles, we can have a general formula $M(b_1, b_2, …, b_n) = S(b_1, b_2, …, b_n) + M(max(b_1 – a, b_2 – a, …, b_n – a), …, min(b_1 – a, b_2 – a, …, b_n – a))$ where $S$ is the sum of all beans.

Using the formulas and principles above, you’ll have to write a code to calculate the movements for the 1500 cases, as doing it by hand isn’t feasible. Remember, each index in the $t$ sequence affects the $b$ sequence, and consequently, the number of moves.

Please note that this problem seems like it comes from a competitive programming problem due to its complexity. If you are working on improving your competitive programming skills, be sure to thoroughly understand each part of the problem and the underlying principles used to solve it.

This problem involves a deep understanding of bitwise operations, number theory and dynamic programming. Even with the solution insight, implementing the solution can be equally challenging.

##### More Answers:

Euler’s NumberCross Flips

Special Partitions