27999: The End of a Millennium

A millennium is strictly speaking a period of 1000 years but in my case, dealing as I do in days rather than years, I'm regarding it as a period of 1000 days. Tomorrow I'll be 27999 days old and on Sunday, the 30th of November 2025, I'll turn \( \textbf{28000}\) days old, thus marking a transition from one millennium to another. It must fall on a Sunday of course since the number is divisible by 7, meaning I'll also be 4000 weeks old. 82 days after that I'll celebrate my first palindromic day of the new millennium: 28082.

One of the properties of 28000 is that it is a member of OEIS A085708: Arithmetic derivative of \(10^n\) where \(n=4\). Here is Gemini's account of how the arithmetic derivative is determined.

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Python

def arithmetic_derivative(n):
    if n < 0: return -arithmetic_derivative(-n)
    if n == 0 or n == 1: return 0
    
    # Simple trial division for factorization
    i = 2
    factors = []
    d = n
    while i * i <= d:
        if d % i:
            i += 1
        else:
            d //= i
            factors.append(i)
    if d > 1:
        factors.append(d)
        
    # Apply logarithmic derivative formula: n' = n * sum(1/p for p in factors)
    # Actually formula is sum( n/p for p in factors ) since p'=1
    # n' = n * sum( p'/p ) = n * sum( 1/p )
    # Let's use the explicit sum rule: (abc...)' = (abc...)(1/a + 1/b + 1/c ...) *if* they are primes? 
    # Let's stick to the definition: (ab)' = a'b + ab'
    # For n = p1 * p2 * ... * pk
    # n' = n * (p1'/p1 + p2'/p2 + ... + pk'/pk)
    # Since pi are primes, pi' = 1.
    # So n' = n * (1/p1 + 1/p2 + ... + 1/pk)
    # Where the factors list includes multiplicity.
    
    result = 0
    for p in factors:
        result += n // p
    return result

val = 10000
der = arithmetic_derivative(val)
print(f"{val=}, {der=}")

Code output

val=10000, der=28000

To find the arithmetic derivative of $10000$, we use the definition and properties of the arithmetic derivative function $n'$.

Rules of Arithmetic Derivative:

1. Prime Rule: For any prime number $p$, $p' = 1$.

2. Product Rule (Leibniz Rule): For any integers $a$ and $b$, $(ab)' = a'b + ab'$.

From these rules, we can derive the Power Rule for a prime $p$ raised to a power $k$:

$$(p^k)' = k p^{k-1}$$

---

Step-by-Step Calculation

Step 1: Prime Factorization

First, find the prime factorization of $10000$.

$$10000 = 10^4 = (2 \times 5)^4 = 2^4 \times 5^4$$

Step 2: Apply the Product Rule

Let $a = 2^4$ and $b = 5^4$. According to the product rule:

$$(2^4 \times 5^4)' = (2^4)' \times 5^4 + 2^4 \times (5^4)'$$

Step 3: Calculate Derivatives of the Factors

Using the power rule $(p^k)' = k p^{k-1}$:

• For $2^4$:

  $$(2^4)' = 4 \times 2^{4-1} \times (2)' = 4 \times 2^3 \times 1 = 4 \times 8 = 32$$

• For $5^4$:

  $$(5^4)' = 4 \times 5^{4-1} \times (5)' = 4 \times 5^3 \times 1 = 4 \times 125 = 500$$

Step 4: Substitute and Simplify

Substitute these derivatives back into the equation from Step 2:

$$10000' = (32) \times 5^4 + 2^4 \times (500)$$

Calculate the terms:

• First term: $32 \times 625 = 20000$

• Second term: $16 \times 500 = 8000$

Step 5: Final Sum

$$10000' = 20000 + 8000 = 28000$$

Alternative Method (Logarithmic Derivative Formula)

For a number $n = p_1^{e_1} \cdots p_k^{e_k}$, the formula is:$$n' = n \sum_{i=1}^k \frac{e_i}{p_i}$$For $10000 = 2^4 \times 5^4$:

$$10000' = 10000 \left( \frac{4}{2} + \frac{4}{5} \right)$$

$$10000' = 10000 \left( 2 + 0.8 \right) = 10000(2.8) = 28000$$

Answer:

The arithmetic derivative of 10000 is 28,000.

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Courtesy of Gemini, I also learned that the number shows up in the orbital speed of 28,000 km/h that is the specific speed required to maintain a Low Earth Orbit (LEO). 

• This is the cruising speed of the International Space Station (ISS). At this speed, the ISS circles the Earth once every 90 minutes.

• If it moved much slower, gravity would pull it down; much faster, and it would break orbit into deep space.

More disconcertingly, and again thanks to Gemini, the number shows up as the "28,000 Days" Concept in which the number is often used in philosophy and motivational literature to represent the average human lifespan.

• $\dfrac{28,000 \text{ days}}{365 \text{ days}} \approx 76.7 \text{ years}$.

• It is frequently used to visualize the finiteness of life e.g. "You only get 28,000 days".

Let's not forget that the two key digits in 28000 are the 2 and the 8, with the 0's acting as magnifiers of a sort. Thus the focus falls on 28 which of course is a perfect number, being equal to the sum of its proper divisors:$$ \begin{align} 28 &= 1 + 2 + 4 + 7 + 14 \\ 28000 &= 1000 + 2000 + 4000 + 7000 + 14000 \end{align} $$ \( \textbf{Here's what my SageMath algorithm generated:}\)

\(28000 = 2^5 \times 5^3 \times 7\)

There are 48 divisors

The divisors are [1, 2, 4, 5, 7, 8, 10, 14, 16, 20, 25, 28, 32, 35, 40, 50, 56, 70, 80, 100, 112, 125, 140, 160, 175, 200, 224, 250, 280, 350, 400, 500, 560, 700, 800, 875, 1000, 1120, 1400, 1750, 2000, 2800, 3500, 4000, 5600, 7000, 14000, 28000]

The sum of the divisors is \(78624 = 2^5 \times 3^3 \times 7 \times 13\)

The sum of the proper divisors is \(50624 = 2^6 \times 7 \times 113\)

28000 is abundant because its sum of proper divisors is greater than the number itself

Distinct prime factors are [2, 5, 7]

Sum of DISTINCT prime factors of 28000 is \(14 = 2 \times 7\)

28000 is NOT unprimeable because 28001 is prime

Unitary Divisors are [1, 7, 32, 125, 224, 875, 4000, 28000]

28000 is pseudoperfect meaning it is equal to the sum of a PROPER subset of its PROPER divisors

28000 is a Zumkeller number meaning that its divisors can be split into two disjoint sets with same sum

28000 is a gapful number

28000 is a practical number meaning that all smaller numbers can be written as sums of its divisors

The totient is \(9600 = 2^7 \times 3 \times 5^2\)

The cototient is \(28000 - 9600 = 18400 = 2^5 \times 5^2 \times 23\)

The sum of squares of the digits is \(68 = 2^2 \times 17\)

The sum of cubes of the digits is \(520 = 2^3 \times 5 \times 13\)

28000 is a plaindrome in base 6 : 333344

28000 is a nialpdrome in base 8 : 66540

28000 is a xenodrome in base 9 : 42361

Gray Code Equivalent:

\(28000 = 110110101100000 \rightarrow 101101111010000 = 23504 = 2^4 \times 13 \times 113\)

Difference between 28000 and its Gray Code is \(4496 = 2^4 \times 281\)

Binary complement of 28000 is \(4767 = 3 \times 7 \times 227\)

Difference between binary complement and 28000 is \(23233 = 7 \times 3319\)

No energetic or d-powerful classification for 28000.

number is an energetic number

Sum of Squares of the following tuples (if any):

28000 is the sum of the cubes of 10 and 30

Sum of digits is 10 and product of digits is 0

28000 is a Harshad number since it is divisible by its sum of digits 10

\(28000 + 10 = 28010 = 2 \times 5 \times 2801\)

\(28000 - 10  = 27990 = 2 \times 3^2 \times 5 \times 311 \)

\(28000 + 0 = 28000 = 2^5 \times 5^3 \times 7\)

\(28000 - 0 = 28000 = 2^5 \times 5^3 \times 7\)

number + SOD - POD sequence with zero counted is:

[28000, 28010, 28021, 28034, 28051, 28067, 28090, 28109, 28129, 27863, 25873, 24218, 24107, 24121, 24115, 24088, 24110, 24118, 24070, 24083, 24100, 24107]

number + SOD - POD sequence with zero NOT counted is:

[28000, 28010, 28005, 28020, 28000]

The Collatz trajectory requires 33 steps required to reach 1 with trajectory:

[28000, 14000, 7000, 3500, 1750, 875, 2626, 1313, 3940, 1970, 985, 2956, 1478, 739, 2218, 1109, 3328, 1664, 832, 416, 208, 104, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1]

The Aliquot Sequence for 28000 requires 79 steps and is:

[28000, 50624, 65200, 92404, 81840, 203856, 343728, 894288, 1494448, 1648208, 1649200, 3271120, 4585520, 6681616, 7404784, 7405776, 17989424, 17990416, 22007024, 25406608, 25867888, 25868880, 70077360, 154194000, 359582640, 859196496, 1455083952, 2979489360, 7116150192, 11860254288, 22402715440, 40354410704, 40354411696, 40737128272, 40819290362, 23229175558, 11614587782, 10147429498, 5602405262, 2801714530, 2259985694, 1216520482, 651851870, 521481514, 260870486, 144688954, 72344480, 111721864, 97756646, 60157978, 30189062, 17477938, 10281194, 9165718, 4582862, 2362594, 1500566, 848218, 640742, 325258, 162632, 153268, 114958, 58922, 34714, 20474, 11386, 5696, 5734, 3194, 1600, 2337, 1023, 513, 287, 49, 8, 7, 1, 0]

The Anti-Divisors of 28000 are:

[3, 11, 29, 33, 64, 320, 448, 1600, 1697, 1931, 2240, 5091, 8000, 11200, 18667]

The Arithmetic Derivative of 28000 is 90800

Difference between Arithmetic Derivative and 28000 is \(62800 = 2^4 \times 5^2 \times 157\)

The Maximum - Minimum Recursive Algorithm for 28000 produces:

[28000, 81972, 85932, 74943, 62964, 71973, 83952, 74943]

The Minimal Goldbach decomposition of 28000 is 3 and 27997

28000 requires 1 steps to reach the palindrome 28082 under the Reverse and Add algorithm

The number of steps required is to reach home prime is 3 :

[28000, 222225557, 293758449, 31931166247]

The additive digital root of 28000 is 1 and is derived as follows:

[28000, 10, 1]

The multiplicative digital root of 28000 is 0 and derived as follows:

[28000, 0]

The steps on the Descent to Zero are:

[28000, 16000, 6000, 0]

The number of steps required is 3

The multiplicative persistence of 28000 is as follows (NOT taking zeros into account):

[28000, 16, 6]

Trajectory of 28000 under ODD(+) and EVEN(-) algorithm is:

[28000, 27990, 28013, 28007, 28004, 27990]

28000 is a captive of the vortex beginning and ending with 27990

Trajectory of 28000 under EVEN(+) and ODD(-) algorithm is:

[28000, 28010, 28019, 28019]

28000 is a captive of the attractor 28019

Trajectory under the Primes(+) and Non_Primes(-) trajectory is:

[28000, 27994, 27981, 27972, 27981]

28000 is a captive of the vortex beginning and ending with 27981

Determinant is -60 for Circulant Matrix:

Difference between Determinant and 28000 is \(28060 = 2^2 \times 5 \times 23 \times 61\)

[ 2     8     0     0     0 ] 

[ 8     0     0     0     2 ]

[ 0     0     0     2     8 ]

[ 0     0     2     8     0 ]

[ 0     2     8     0     0 ]

Here is how the number holds up under \( \textbf{Conway's Game of Life} \):

Here's what Numbers Aplenty had to say about the number: