Bean facts

We can figure out an explicit formula for $B_n$ by repeatedly using the given recursive definition:

$\begin{aligned} B_n &= nB_{n-1} \\ &= n(n-1) B_{n-2} \\ &= n(n-1)(n-2)B_{n-3} \\ &= \ldots \\ &= n(n-1)(n-2)\cdots(3)B_2 \\ &= n(n-1)(n-2)\cdots(3)(2)B_1 \\ &= n!\cdot B_1 \\ &= n! \cdot \dfrac{1}{1024}.\end{aligned}$ Therefore, $B_n$ is an integer if and only if $n!$ is divisible by $1024.$ To find the lowest possible $n,$ we note that $1024 = 2^{10}.$ Then, we look at the exponent of $2$ in the prime factorization of $n!.$ To do that, we can look at the first few even numbers and see how many factors of $2$ they have:

• $2$ provides $1$ factor.
• $4$ provides $2$ factors.
• $6$ provides $1$ factor.
• $8$ provides $3$ factors.
• $10$ provides $1$ factor.
• $12$ provides $2$ factors.

Since $1+2+1+3+1+2 = 10,$ the minimum possible $n$ is $\boxed{12}.$

Note that the explicit rule for $B_{n} = \dfrac{n!}{1024}$ . We wish to find the smallest $n$ such that $\left\lfloor\dfrac{n!}{2}\right\rfloor + \left\lfloor\dfrac{n!}{4}\right\rfloor + ... + \left\lfloor\dfrac{n!}{2^{k}}\right\rfloor + ... = 10$ for integer $k$ . We find by inspection that $\boxed{12}$ is the smallest that works

A brief inspecticon will show this sequence can be expressed as:

$B_{n} =\frac{n!}{1024}$ .

$1024 = 2^{10}$

We have to find the lowest factorial that is multiple of 1024 and so multiple of $2^{10}$ .

$1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 * 11 * 12 = 2^{10} * (...)$ .

so $12!$ is divisible by 1024.

B(n) = n*B(n-1)

B(n) = n(n-1)B(n-2) since B(n-1)=(n-1)*B(n-2)

...

B(n) = n(n-1)(n-2)...(n-k+1)*B(n-k)

If n - k = 1 then k = n - 1

Therefore B(n) = n(n-1)(n-2)...(n-(n-1)+1)B(1) = n!/1024

B(n) = n!/2^10

Now we search for a n! that is multiple of 2^10, but only look for even numbers since just they have as prime factor 2.

For n = 2: n! is 2 wich is multiple of 2^1

For n = 4: n! is 4*2 wich is multiple of 2^3

For n = 6: n! is 642 wich is multiple of 2^4

For n = 8: n! is 864*2 wich is multiple of 2^7

For n = 10: n! is 108642 wich is multiple of 2^8

For n = 12: n! is 1210864*2 wich is multiple of 2^10

Then for n = 12, n! is multiple of 2^10

gn: B1=1/2^10=2^-10 then,B1=9.76 10^-4 Bn=n(Bn-1)(n<=2) (gn: B2=2B(2-1)=2B1=2(9.76 10^-4)=1.95 10^-3 then B3=3(B2)=5.85 10^-3...... continuing we get the first +ve integer for n=12(B12=12(B11)=467775

B(n) = n*B(n-1)

B(n) = n (n-1) B(n-2) since B(n-1)=(n-1)*B(n-2)

...

B(n) = n (n-1) (n-2)...(n-k+1)*B(n-k)

If n - k = 1 then k = n - 1

Therefore B(n) = n (n-1) (n-2) ...(n-(n-1)+1) B(1) = n!/1024

B(n) = n!/2^10

Now we search for a n! that is multiple of 2^10, but only look for even numbers since just they have as prime factor 2.

For n = 2: n! is 2 wich is multiple of 2^1

For n = 4: n! is 4*2 wich is multiple of 2^3

For n = 6: n! is 6 4 2 wich is multiple of 2^4

For n = 8: n! is 8 6 4*2 wich is multiple of 2^7

For n = 10: n! is 10 8 6 4 2 wich is multiple of 2^8

For n = 12: n! is 12 10 8 6 4*2 wich is multiple of 2^10

Then for n = 12, n! is multiple of 2^10

B(n) can be express as B(n)=1.2.3.4.5.6...n/1024 The numerator must have 10 factors of 2 to neutralize the denominator 1024. Therefore n must at least be 12 so that B(n) can be an integer