The peculiar properties of 14732

Question

About a year ago, I delved into the properties of the number 14732. A number that have popped into my brain at random times for well over a decade.

Through a bit of experimenting with base10 and base2, I was able to extrapolate some properties of the number 14732, which I will list below.

Not knowing how to proceed to find other numbers with the same, or similar, properties, I sent my findings to Matt Parker, whom you might know from Numb3rphile on youtube.

He was suitable interested, and suggested that I post my findings here, if I hoped to find help in finding other such numbers, or determine if 14732 is unique.

It's been a busy year since, but I've finally gotten around to posting here, asking for help in finding other similar numbers, or even better, an explanation of why 14732 behaves in the way outlined below.

I would have liked to include the math formulas for each property, but since I am not skilled in complex algebra, I have been unable to do so.

Questions

• Is 14732 unique for all of the below six properties?
• If other numbers share all six properties, is there then a shared pattern that can be used to find other numbers?
• Are there any numbers that only satisfy the first five properties, but not the sixth?

First property

When representing the number (D1) in base2 (B1), it can be split into two halves of equal length (H1 and H2), each of which is a direct bit-for-bit opposite of the other.

Proof

D1 = 14732
B1 = 11100110001100

H1 = 1110011
H2 = 0001100

Second property

When H1 and H2 are added together in decimal (DH1 and DH2), they form a multiplicative (M) of D1.

Proof

DH1 = 1110011 => 115
DH2 = 0001100 => 12

M = DH1 + DH2
M = 115 + 12
M = 127

14732 / 127 = 116

Third property

When inverting B1, the result (B2) can be converted back to decimal (D2), which will also be a multiplicative of M.

Proof

B1 = 11100110001100
B2 = 00011001110011

D2 = B2 => 1651

1651 / 127 = 13

Fourth property

Subtracting D1 from D2 yields a third result (D3), which is also a multiplicative of M.

Proof

D3 = D1 - D2
D3 = 14732 - 1651
D3 = 13081

13081 / 127 = 103

Fifth property

Converting D3 to base2 (B3), inverting the bits (B4), and back to decimal (D4) results in 2D2.

Proof

B3 = 11001100011001
B4 = 00110011100110
D4 = B4 => 3302

D4 = 2D2
3302 = 2D2
3302 = 2 * (1651)

Sixth property

Reversing D2 to get D5, and subtracting D5 from D2, then multiplying by M, and adding D4 results in the starting number (D1)

Proof

D5 = 1651 => 1561

D1 = (D2 - D5) * M + D4
D1 = (1651 - 1561) * 127 + 2D2
D1 = (90) * 127 + 2 * (1651)
D1 = 11430 + 3302
D1 = 14732

Answer 1

Much too long for a comment: Let's try to write it algebraically. I'm not distinguishing between $d_1$ and $b_1$ because a number is the same number no matter how we choose to write it down.

Property 1. Let $n$ be the number of bits in each half of your number. Presumably $h_2$ must start with a zero such that your entire $d_1$ does not have a leading zero bit. That is, $$ 0 \le h_2 < 2^{n-1} $$ Whenever we choose a $h_2$ we can get everything else. Flipping the bits of $h_2$ produces $$ h_1 = (2^n-1)-h_2 $$ and then $$ d_1 = 2^nh_1 + h_2 = 2^n(2^n-1)-2^nh_2 + h_2 = (2^n-1)(2^n-h_2) $$

Property 2. Adding $h_1$ and its bit-for-bit flip produces a string of $n$ ones, which is $$ m = 2^n-1 $$ (This is how I computed $h_1$ above in the first place). As we see, this is automatically a factor of $d_1$.

Property 3. Inverting $d_1$ is the same as subtracting it from a string of $2n$ ones, which is $2^{2n}-1$ and also happens to equal $m(2^n+1)$ (because $a^2-b^2=(a-b)(a+b)$, here with $a=2^n$ and $b=1$). This produces $$ d_2 = m(2^n+1) - d_1 = m(2^n+1) - m(2^n-h_2) = m(h_2+1) $$ which automatically has $m$ as a factor.

Property 4. We get $$ d_3 = d_1 - d_2 = m(2^n-h_2) - m(h_2+1) = m(m-2h_2)$$ again of course a multiple of $m$ since it is the difference of two such multiples.

Property 5. Inverting $d_3$ is again a matter of subtracting it from $m(2^n+1)$: $$ d_4 = m(2^n+1) - m(m-2h_2) = m(2^n+1 - m + 2h_2) = m(2+2h_2) = 2m(h_2+1) = 2d_2 $$

So properties 2, 3, 4, 5 follow automatically once we have chosen $n$ and $h_2$ to create a $d_1$ that satisfies property 1 (as already noted by @achille hui).

Property 6. Here it gets complicated, because reversing the sequence of decimal digits of a number is not an algebraic operation. But let's see what we can do. You specify $$ d_1 = (d_2-d_5)m + d_4 $$ which first of all is the same as $$ m(2^n-h_2) = (m(h_2+1)-d_5)m + 2m(h_2+1) $$ so we can cancel the $m$s and rearrange to $$ d_5 = mh_2 + m + 3h_2+2 - 2^n $$ which simplifies further to $$ d_5 = (2^n+2)h_2 + 1 $$

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All in all, we're looking for $n$ and $h_2<2^{n-1}$ such that $$ d_2 = (2^n-1)(h_2+1) \qquad\text{and}\qquad d_5 = (2^n+2)h_2 + 1 $$ have reverse digit sequences as decimal representations.

In your number we have $n=7, h_2=12$, so $$ d_2 = 127\cdot 13 = 1651 \\ d_5 = 130\cdot12 + 1 = 1561 $$

Answer 2

I think we've got it mixed up then. When I'm writing about inverting the bits, I mean flipping them, so 0 becomes 1, and 1 becomes 0. The formula for that is this: $b_i = (2^{n_2}-1) - b$, where $n_2$ is the length of the binary number, and $2^{n_2}-1$ is converted to binary.

From the top:

Prop 1

$d_1 = 2$

$b_1 = 10$

$h_1 = 1$

$h_2 = 0$

Prop 2

$dh_1 = 1$

$dh_2 = 0$

$m = 1 + 0 = 1$

Prop 3

$b_2 = 2^{n_2}-1 - b_1 = 2^2-1 - 10 = 11 - 10 = 01$

$d_2 = 1$

Prop 4

$d_3 = d_1 - d_2 = 2 - 1 = 1$

Prop 5

$b_4 = (2^{n_2}-1) - b_3 = 2^1-1 = 1 - 1 = 0$

$d_4 = b_4 = 0$

$d_4 = 2(d_3)$

$0 = 2$