Birthday Digits

Gemini's Infographic Summary of the Content in this Post

I recently turned 28077 days old and with my 77th birthday rapidly approaching I wondered if it's possible for a person's birthday to fall in such a way that their age in years matched the last two digits of their diurnal age. I prompted Gemini:

Suppose someone is born on April 3rd 1949. Let's count this as day 0. April 4th 1949 then counts as day 1 and so on. This progressive number we term the person's diurnal age as opposed to their age in years. As the person ages (let's say from 1 to 100) is there a birthday such that the last two digits of their diurnal age equals their age in years. Can you write a program in SageMath to determine this. The program should run in SageMathCell or a Jupyter notework.

The program (permalink) came up with only one age, 14, and an associated diurnal age of 5114 for person's born within a certain date range. For example for persons born in 2012 who are turning 14 this year (2026), only those with birthdates between the 1st January and 28th February will satisfy. Let's check for somebody turning 14 today (16th February 2026) using Wolfram Alpha. See Figure 1.

Figure 1

As can be seen, someone turning 14 years old today will be 5114 days old, satisfying the requirement. I queried Gemini as to whether there might be ages other than 14 that satisfy and this was the response:

Yes, I am sure. Age 14 is the only age in a normal human lifespan (1–100+) where this mathematical coincidence occurs. We can prove this by checking the mathematical constraints for every single age.

The Logic: Why it's only 14

For a match to happen at Age $N$, the following equation must be true: $$(64 \times N) + \text{Leap Days} \quad \text{must be divisible by } 100$$

At Age 14: $64 \times 14 = 896$. The remainder is 96. To reach the next 100, we need 4 more. Can a 14-year-old have lived through 4 leap days? Yes (e.g., born Jan 1948). Result: Match.

At Age 39 (The next closest candidate): $64 \times 39 = 2496$. The remainder is 96. To reach the next 100, we again need 4 more. Can a 39-year-old have lived through only 4 leap days? No. In 39 years, you must experience 9 or 10 leap years. Result: Impossible.

Gemini also explained it earlier in a slightly different way so I'll include that here:

The Math Behind the Age 14 Match

For a match to occur, we need: $$\text{Total Days} \pmod{100} = \text{Age}$$

Let's look at Age 14:

Standard Days: $14 \text{ years} \times 365 \text{ days/year} = 5110 \text{ days}$.

The Remainder: $5110 \pmod{100} = 10$.

The Gap: The remainder is 10, but we need it to equal the age, 14.

The Solution: We need exactly 4 extra days (leap days) to bridge the gap from 10 to 14 ($5110 + 4 = 5114$).

Therefore, any person who experiences exactly 4 leap years by their 14th birthday will have a diurnal age of 5114. Since 5114 ends in 14, this is a perfect match.

Why April 3, 1949 Misses the Match

Let's apply this to your specific date: April 3, 1949.

Period: April 3, 1949 to April 3, 1963.

Leap Years Encountered: 1952, 1956, 1960.

Total Leap Days: 3.

Total Diurnal Age: $5110 + 3 = 5113$.

Result: $5113$ ends in 13. The age is 14. (Missed by 1 day).

Who Gets the Match?

To get 4 leap days in 14 years, you must be born just before a leap day, so that you catch the first one immediately, and then three more over the next 12 years.

This generally happens to people born in January or February of a Leap Year (or late in the year prior to a leap year).

Fibonacci From Prime Factors

Gemini's Infographic Summary of the Content in this Post

I noticed that the number (\( \textbf{28075} \)) associated with my diurnal age today has an interesting property relating to its prime factors:$$28075=5^2 \times 1123$$The number has distinct prime factors of \( \textbf{5}\) and \( \textbf{1123}\). If these two factors are written in reversed order and then concatenated, the number \( \textbf{11235}\) is formed with digits that form a Fibonacci sequence:$$1+1 \rightarrow 2 \text{ and } 2 + 3 \rightarrow 5$$This got me thinking about what other numbers have this property and so I set Gemini to work to find all such numbers in the range from 1 to 40000. It turns out that the following numbers qualify (permalink):

22, 26, 30, 44, 52, 60, 66, 70, 88, 90, 101, 104, 115, 120, 132, 140, 141, 150, 158, 167, 176, 180, 198, 203, 205, 208, 210, 240, 242, 253, 257, 264, 270, 280, 300, 301, 316, 330, 338, 347, 350, 352, 360, 396, 416, 420, 423, 427, 450, 480, 484, 490, 528, 540, 560, 575, 594, 600, 611, 617, 630, 632, 660, 676, 700, 704, 720, 726, 750, 771, 790, 792, 810, 832, 835, 840, 900, 960, 968, 980, 990, 1025, 1050, 1056, 1080, 1120, 1123, 1188, 1200, 1222, 1260, 1264, 1265, 1269, 1320, 1350, 1352, 1400, 1408, 1421, 1440, 1452, 1459, 1470, 1500, 1580, 1584, 1620, 1650, 1664, 1680, 1750, 1782, 1800, 1890, 1920, 1936, 1960, 1980, 2100, 2107, 2112, 2160, 2178, 2240, 2250, 2313, 2376, 2400, 2430, 2444, 2450, 2520, 2528, 2640, 2645, 2662, 2700, 2704, 2783, 2800, 2816, 2875, 2880, 2904, 2940, 2970, 2989, 3000, 3150, 3160, 3168, 3240, 3257, 3300, 3328, 3360, 3430, 3500, 3564, 3600, 3630, 3750, 3780, 3807, 3840, 3872, 3920, 3950, 3960, 4050, 4175, 4200, 4224, 4320, 4356, 4377, 4394, 4410, 4480, 4500, 4752, 4800, 4860, 4888, 4900, 4950, 5040, 5056, 5125, 5167, 5250, 5279, 5280, 5324, 5346, 5400, 5408, 5600, 5615, 5632, 5670, 5760, 5808, 5819, 5880, 5887, 5940, 6000, 6300, 6319, 6320, 6325, 6336, 6480, 6534, 6600, 6627, 6656, 6720, 6750, 6860, 6939, 7000, 7128, 7200, 7260, 7290, 7350, 7500, 7560, 7680, 7744, 7840, 7900, 7920, 7943, 7986, 8100, 8250, 8400, 8405, 8448, 8640, 8712, 8750, 8788, 8820, 8910, 8960, 9000, 9450, 9504, 9600, 9720, 9776, 9800, 9900, 9947, 10080, 10112, 10201, 10290, 10500, 10560, 10648, 10692, 10800, 10816, 10890, 11200, 11250, 11264, 11340, 11421, 11520, 11616, 11760, 11880, 12000, 12150, 12250, 12482, 12600, 12640, 12672, 12943, 12960, 13068, 13131, 13200, 13225, 13230, 13312, 13440, 13500, 13720, 13915, 14000, 14256, 14375, 14400, 14520, 14580, 14700, 14749, 14850, 15000, 15120, 15360, 15488, 15680, 15750, 15800, 15840, 15886, 15972, 16038, 16200, 16500, 16800, 16896, 17010, 17150, 17280, 17424, 17500, 17576, 17640, 17820, 17920, 18000, 18150, 18750, 18900, 19008, 19200, 19440, 19552, 19600, 19602, 19750, 19800, 19881, 20160, 20224, 20250, 20580, 20817, 20875, 20923, 21000, 21120, 21296, 21347, 21384, 21600, 21632, 21780, 21870, 22050, 22400, 22500, 22528, 22680, 23040, 23232, 23520, 23760, 23958, 24000, 24010, 24300, 24500, 24750, 24964, 25200, 25280, 25344, 25625, 25920, 26047, 26136, 26250, 26400, 26460, 26624, 26730, 26880, 27000, 27440, 27889, 28000, 28075, 28350, 28512, 28717, 28800, 29040, 29095, 29160, 29282, 29400, 29700, 30000, 30240, 30613, 30720, 30870, 30976, 31360, 31500, 31600, 31625, 31680, 31772, 31944, 32076, 32400, 32670, 33000, 33600, 33750, 33792, 34020, 34263, 34300, 34560, 34848, 35000, 35152, 35280, 35640, 35840, 36000, 36300, 36450, 36750, 37500, 37800, 38016, 38400, 38880, 39104, 39200, 39204, 39393, 39500, 39600, 39690, 39930

Of course looking at these numbers it's not immediately apparent what the Fibonacci digit sequence is but the Gemini program creates a table to show this. I'll restrict the range to between 28000 and 29000. The result is shown in Figure 1:

Figure 1: permalink

The fact that we are only considering \( \textbf{distinct} \) prime factors helps the program run quickly and there are no problems using it with SageMathCell. However if we allow multiplicity of factors, the number of permutations increases dramatically and SageMathCell quickly times out even if we restrict the range to between 28000 and 29000. So I think working only with distinct prime factors is the way to go.

Generating Functions

Only yesterday in a post titled Linear Recurrence Revisited did I look at this recurrence in relation to my diurnal age of 28072:$$ \begin{align} &\text{a}(n+3) = 2\text{a}(n+2) - 3\text{a}(n+1) + 2\text{a}(n)\\ & \text{a}(0) = 1, \text{a}(1) = 1, \text{a}(2) = 0 \end{align}$$Today, a day later, I find that 28073 can also be linked to a linear recurrence via OEIS A157651:$$ \begin{align} &\text{a}(n) = 3\text{a}(n-1) -3\text{a}(n-2) +\text{a}(n-3) \\ &\text{a}(0) =57, \text{a}(1) =308, \text{a}(3)=759 \end{align}$$The initial terms are:

57, 308, 759, 1410, 2261, 3312, 4563, 6014, 7665, 9516, 11567, 13818, 16269, 18920, 21771, 24822, 28073, 31524, 35175, 39026, 43077, 47328, 51779, 56430, 61281, 66332, 71583, 77034, 82685, 88536, 94587, 100838, 107289, 113940, 120791, 127842

However, this time I want to look at ordinary generating functions and exponential generating functions in relation to this linear recurrence. The following was created by Gemini.

$$
\begin{array}{@{}l}



\textbf{Derivation of the Ordinary Generating Function } G(x) \\[12pt]


\text{Let } G(x) = \displaystyle\sum_{n=0}^{\infty} a_n x^n = a_0 + a_1 x + a_2 x^2 + \dots \\[12pt]
\text{The recurrence relation is: } a_n - 3a_{n-1} + 3a_{n-2} - a_{n-3} = 0 \text{ for } n \ge 3. \\[12pt]


\text{We multiply } G(x) \text{ by the polynomial } (1 - 3x + 3x^2 - x^3): \\[6pt]
(1 - 3x + 3x^2 - x^3)G(x) = \\[6pt]
\quad \phantom{-} (a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \dots) \\
\quad - \displaystyle 3x (a_0 + a_1 x + a_2 x^2 + \dots) \\
\quad + \displaystyle 3x^2 (a_0 + a_1 x + \dots) \\
\quad - \displaystyle x^3 (a_0 + \dots) \\[18pt]


\text{Grouping coefficients by powers of } x: \\[6pt]
\text{Constant: } a_0 = 57 \\[6pt]
x^1: a_1 - 3a_0 = 308 - 3(57) = 308 - 171 = 137 \\[6pt]
x^2: a_2 - 3a_1 + 3a_0 = 759 - 3(308) + 3(57) = 759 - 924 + 171 = 6 \\[6pt]
x^n \text{ (for } n \ge 3\text{): } a_n - 3a_{n-1} + 3a_{n-2} - a_{n-3} = 0 \\[18pt]


\text{Thus:} \\[6pt]
(1 - x)^3 G(x) = 57 + 137x + 6x^2 \\[12pt]


\text{The Ordinary Generating Function is:} \\[6pt]
G(x) = \displaystyle\frac{57 + 137x + 6x^2}{(1 - x)^3} \\[24pt]




\textbf{Derivation of the Exponential Generating Function } E(x) \\[12pt]


\text{Let } E(x) = \displaystyle\sum_{n=0}^{\infty} a_n \frac{x^n}{n!}. \\[12pt]
\text{The recurrence } a_n - 3a_{n-1} + 3a_{n-2} - a_{n-3} = 0 \\ \text{ becomes the differential equation:} \\[6pt]
E'''(x) - 3E''(x) + 3E'(x) - E(x) = 0 \\[12pt]


\text{The characteristic equation is } r^3 - 3r^2 + 3r - 1 = (r-1)^3 = 0. \\[6pt]
\text{This has a triple root at } r = 1. \\[12pt]


\text{The general solution for a triple root is:} \\[6pt]
E(x) = (C_1 + C_2 x + C_3 x^2)e^x \\[12pt]


\text{We find constants } C_1, C_2, C_3 \text{ using initial conditions } a_0=57, a_1=308, a_2=759. \\[6pt]
\text{Note that } a_n \text{ is the coefficient of } \frac{x^n}{n!}. \\[6pt]
\text{Expanding } E(x) = (C_1 + C_2 x + C_3 x^2) \displaystyle \sum \frac{x^n}{n!} \\ \text{ allows us to match coefficients directly.} \\[18pt]


\text{Solving for the constants (via derivatives at } x=0 \text{):} \\[6pt]
1. \quad E(0) = a_0 \implies C_1 = 57 \\[6pt]
2. \quad E'(0) = a_1 \implies C_1 + C_2 = 308 \implies 57 + C_2 = 308 \implies C_2 = 251 \\[6pt]
3. \quad E''(0) = a_2 \implies C_1 + 2C_2 + 2C_3 = 759 \\[6pt]
\phantom{3. \quad} 57 + 2(251) + 2C_3 = 759 \\[6pt]
\phantom{3. \quad} 559 + 2C_3 = 759 \implies 2C_3 = 200 \implies C_3 = 100 \\[18pt]


\text{The Exponential Generating Function is:} \\[6pt]
E(x) = (57 + 251x + 100x^2)e^x \\[24pt]




\textbf{Expansion of the Ordinary Generating Function} \\[12pt]


\text{The correct OGF is: } G(x) = \displaystyle\frac{57 + 137x + 6x^2}{(1-x)^3} \\[12pt]


\text{We use the binomial series expansion for } (1-x)^{-3}: \\[6pt]
(1-x)^{-3} = \displaystyle\sum_{n=0}^{\infty} \binom{n+2}{2} x^n = \displaystyle\sum_{n=0}^{\infty} \frac{(n+1)(n+2)}{2} x^n \\[12pt]


\text{Multiply this series by the numerator } (57 + 137x + 6x^2): \\[6pt]
a_n = 57 \cdot \binom{n+2}{2} + 137 \cdot \binom{n+1}{2} + 6 \cdot \binom{n}{2} \\[12pt]


\text{Let's verify the first few terms again with the correct formula:} \\[12pt]


\textbf{For } n=0: \\[6pt]
a_0 = 57 \cdot \frac{(1)(2)}{2} = 57(1) = \mathbf{57} \\[12pt]


\textbf{For } n=1: \\[6pt]
\text{The term } \binom{n}{2} \text{ is 0 for } n < 2. \\[6pt]
a_1 = 57 \cdot \frac{(2)(3)}{2} + 137 \cdot \frac{(1)(2)}{2} \\[6pt]
a_1 = 57(3) + 137(1) = 171 + 137 = \mathbf{308} \\[12pt]


\textbf{For } n=2: \\[6pt]
a_2 = 57 \cdot \frac{(3)(4)}{2} + 137 \cdot \frac{(2)(3)}{2} + 6 \cdot \frac{(2)(1)}{2} \\[6pt]
a_2 = 57(6) + 137(3) + 6(1) \\[6pt]
a_2 = 342 + 411 + 6 = \mathbf{759} \\[12pt]


\textbf{For } n=3: \\[6pt]
a_3 = 57 \cdot \frac{(4)(5)}{2} + 137 \cdot \frac{(3)(4)}{2} + 6 \cdot \frac{(3)(2)}{2} \\[6pt]
a_3 = 57(10) + 137(6) + 6(3) \\[6pt]
a_3 = 570 + 822 + 18 = \mathbf{1410} \\[24pt]




\textbf{Expansion of the Exponential Generating Function} \\[12pt]


\text{We expand } e^x = \displaystyle\sum_{n=0}^{\infty} \dfrac{x^n}{n!} = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dots \\[12pt]


\text{Multiply by the polynomial } (57 + 251x + 100x^2): \\[6pt]
E(x) = 57 \displaystyle\sum \dfrac{x^n}{n!} + 251x \displaystyle\sum \dfrac{x^n}{n!} + 100x^2 \displaystyle\sum \dfrac{x^n}{n!} \\[12pt]


\text{To find } a_n, \text{ we look for the coefficient of } \frac{x^n}{n!} \text{ in the combined sum.} \\[6pt]
\text{Note that } x \cdot \dfrac{x^{n-1}}{(n-1)!} = n \dfrac{x^n}{n!} \text{ and } x^2 \cdot \dfrac{x^{n-2}}{(n-2)!} = n(n-1) \dfrac{x^n}{n!}. \\[12pt]


\text{This gives us the direct formula:} \\[6pt]
a_n = 57(1) + 251(n) + 100(n)(n-1) \\[6pt]
a_n = 100n^2 + 151n + 57 \\[12pt]


\text{Let's calculate the terms using this formula:} \\[12pt]


\textbf{For } n=0: \\[6pt]
a_0 = 0 + 0 + 57 = \mathbf{57} \\[12pt]


\textbf{For } n=1: \\[6pt]
a_1 = 100(1) + 151(1) + 57 = 251 + 57 = \mathbf{308} \\[12pt]


\textbf{For } n=2: \\[6pt]
a_2 = 100(4) + 151(2) + 57 = 400 + 302 + 57 = \mathbf{759} \\[12pt]


\textbf{For } n=3: \\[6pt]
a_3 = 100(9) + 151(3) + 57 = 900 + 453 + 57 = \mathbf{1410}
\end{array}
$$

I prefer to work with the ordinary generating function in which case the following simple code will generate the terms:

Linear Recurrence Revisited

It's been a while since I posted about linear recurrence relations. The only two posts were:

• Solving Linear Recurrence on the 16th of June 2020
  
• More on Linear Recurrence on the 26th December 2020

Today I turned \(28073\) days old and one of the properties of this number is that it's a member of OEIS A105578:$$ \begin{align}  &\text{a}(n+3) = 2\text{a}(n+2) - 3\text{a}(n+1) + 2\text{a}(n)\\ &a(0) = 1, a(1) = 1, a(2) = 0 \end{align}$$This linear recurrence of order 3 corresponds to the cubic equation:$$ \begin{align} &x^3 = 2x^2 -3x +2 \\ &x^3-2x^2+3x-2 = 0 \end{align}$$The polynomial \( \text{P}(x) = x^3-2x^2+3x-2 \) has three roots, one real (at \(x=1\) ) and two complex. See Figure 1.

Figure 1

The roots are:

• \( x_1=1 \)
• \( x_2 = \dfrac{1}{2}(1-i \sqrt{7}) \)
• \( x_3 = \dfrac{1}{2}(1+i \sqrt{7}) \)

These roots can be used to find any term in the sequence because of the following relationship:$$a_n=A(x_1)^n+B(x_2)^n+C(x_3)^n \text{ with A, B and C constants}$$Substituting the initial conditions we get three equations in three unknowns:$$ \begin{align} A + B + C &= 1 \\ Ax_1+Bx_2+Cx_3 &=1 \\ A(x_1)^2+B(x_2)^2+C(x_3)^2 &=0 \end{align} $$In terms of the roots, the values for \( A, B \text{ and } C \) are:$$ \begin{align} A = \frac{x_1 - x_2 x_3 - (x_2 + x_3)}{(x_1 - x_2)(x_1 - x_3)} \\ B = \frac{x_2 - x_1 x_3 - (x_1 + x_3)}{(x_2 - x_1)(x_2 - x_3)} \\ C = \frac{x_3 - x_1 x_2 - (x_1 + x_2)}{(x_3 - x_1)(x_3 - x_2)} \end{align} $$We know that when \(n=32\), the term is 28072 (my diurnal age) so let's check if that works by substituting the values of \( x_1,x_2 \text{ and } x_3 \) into and summarising at the same time what we've found previously:$$
\begin{array}{l}
\textbf{Step 1: Determine the Roots of the Characteristic Equation} \\[6pt]
x^3 - 2x^2 + 3x - 2 = 0 \\[4pt]
(x-1)(x^2 - x + 2) = 0 \\[4pt]
\text{The roots are:} \\[4pt]
x_1 = 1 \\[4pt]
x_2 = \dfrac{1 + i\sqrt{7}}{2} \\[4pt]
x_3 = \dfrac{1 - i\sqrt{7}}{2} \\[18pt]


\textbf{Step 2: Find Constants A, B, and C} \\[6pt]
\text{Using the initial conditions } a_0=1, a_1=1, a_2=0: \\[6pt]
A = \dfrac{1}{2} \\[6pt]
B = \dfrac{1}{4} - \dfrac{i}{4\sqrt{7}} \\[6pt]
C = \dfrac{1}{4} + \dfrac{i}{4\sqrt{7}} \\[18pt]


\textbf{Step 3: Construct the Explicit Formula} \\[6pt]
a_n = A(x_1)^n + B(x_2)^n + C(x_3)^n \\[6pt]
a_n = \dfrac{1}{2}(1)^n + \left(\dfrac{1}{4} - \dfrac{i}{4\sqrt{7}}\right)\left(\dfrac{1 + i\sqrt{7}}{2}\right)^n + \left(\dfrac{1}{4} + \dfrac{i}{4\sqrt{7}}\right)\left(\dfrac{1 - i\sqrt{7}}{2}\right)^n \\[18pt]


\textbf{Step 4: Verify for } n=32 \text{ (Detailed Breakdown)} \\[12pt] \text{We need to sum three terms: } \\[4pt] a_{32} = \underbrace{A(x_1)^{32}}_{\text{Term 1}} + \underbrace{B(x_2)^{32}}_{\text{Term 2}} + \underbrace{C(x_3)^{32}}_{\text{Term 3}} \\[18pt] \textbf{1. Calculate Term 1 (The Real Root)} \\[6pt] \text{Since } x_1 = 1 \text{ and } A = 0.5: \\[6pt] \text{Term 1} = 0.5 \cdot (1)^{32} = \mathbf{0.5} \\[18pt] \textbf{2. Calculate Terms 2 and 3 (The Complex Roots)} \\[6pt] \text{Notice that } C \text{ is the complex conjugate of } B, \text{ and } x_3 \text{ is the conjugate of } x_2. \\[6pt] \text{This implies that Term 3 is the complex conjugate of Term 2.} \\[6pt] \text{Mathematical Rule: } Z + \bar{Z} = 2 \cdot \text{RealPart}(Z). \\[6pt] \text{Therefore, we only need to calculate Term 2 and double its real component.} \\[12pt] \text{Let } x_2 = \sqrt{2}e^{i\theta} \text{ (Polar form, where } \sqrt{2} \text{ is the magnitude)}. \\[6pt] (x_2)^{32} = (\sqrt{2})^{32} e^{i32\theta} = 2^{16} (\cos(32\theta) + i\sin(32\theta)) \\[6pt] (x_2)^{32} = 65536 (\cos(32\theta) + i\sin(32\theta)) \\[12pt] \text{When we multiply this by } B \text{ and take } 2 \times \text{Real Part, the result is exactly:} \\[6pt] \text{Term 2} + \text{Term 3} = \mathbf{28071.5} \\[18pt] \textbf{3. Final Total} \\[6pt] a_{32} = \text{Term 1} + (\text{Terms 2 \& 3}) \\[6pt] a_{32} = 0.5 + 28071.5 \\[6pt] a_{32} = \mathbf{28072} \end{array} $$

Getting back to OEIS A105578, a plot of the first 52 members is show in Figure 2:

1, 1, 0, -1, 0, 3, 4, -1, -8, -5, 12, 23, 0, -45, -44, 47, 136, 43, -228, -313, 144, 771, 484, -1057, -2024, 91, 4140, 3959, -4320, -12237, -3596, 20879, 28072, -13685, -69828, -42457, 97200, 182115, -12284, -376513, -351944, 401083, 1104972, 302807, -1907136, -2512749, 1301524, 6327023, 3723976, -8930069, -16378020, 1482119, 34238160, 31273923, -37202396

Figure 2: permalink

It can be seen that after a while the values between to fluctuate wildly between larger and larger positive and negative values.

There is a database for a great many linear recurrences of order 3 listed here.

The Revision Revisited

In my post titled A Revision, I outlined an alternative algorithm for building a sequence based on the factors of numbers considered with multiplicity. This is the algorithm:

Suppose we take any positive integer \(n \gt 1\) and apply the following rules to it:

• if a \(4k+1\) prime, double it and add 1: \(n \rightarrow 2n+1\)
• if a \(4k+3\) prime, subtract 1 and divide by 2: \(n \rightarrow (n-1)/2\)
• if composite, determine its number of factors \(f\) counted \( \textbf{with multiplicity}\)
• if \( n \pmod f \equiv 0\) then \(n \rightarrow \dfrac{n}{f} \)
• if \( n \pmod f \not\equiv 0 \) then \(n \rightarrow n \times f\)

Keep repeating this process until a loop is reached or call a stop after a fixed number of iterations. 

What I'm interested in looking at in this post are the record lengths of sequences generated over a range of numbers, noting as well the highest and lowest values that are reached by sequence members. Figure 1 shows the results in the range from one to one million (permalink):

Figure 1

Let's now consider the algorithm where multiplicity is ignored. This algorithm in as follows:

Suppose we take any positive integer \(n \gt 1\) and apply the following rules to it:

• if a \(4k+1\) prime, double it and add 1: \(n \rightarrow 2n+1\)
• if a \(4k+3\) prime, subtract 1 and divide by 2: \(n \rightarrow (n-1)/2\)
• if composite, determine its number of factors \(f\) counted \( \textbf{without multiplicity}\)
• if \( n \pmod f \equiv 0\) then \(n \rightarrow \dfrac{n}{f} \)
• if \( n \pmod f \not\equiv 0 \) then \(n \rightarrow n \times f\)

Keep repeating this process until a loop is reached or call a stop after a fixed number of iterations. 

Figure 2 shows the record lengths, highs and lows for the range from one to one million (permalink).

Figure 2

It can be noted that the record lengths of sequences are shorter when multiplicity is ignored and more end in 1 rather than a loop.

And Yet Another Weird Day

Since August of 2019 I've been taking note of weird numbers that pop up in the numbers associated with my diurnal age. See Figure 1 and Figure 2.

Figure 1: 25690 = 70 x 367

Figure 2: 27790 = 70 x 397

As can be seen, such numbers are few and far between but today is another such number. See Figure 3.

Figure 3: 28070 = 70 x 401

However, it will be 560 days before my next weird numbered day (28630). I've discussed weird numbers before in blog posts titled Weird Numbers and Another Weird Number so I won't repeat that here.

The point of this post is simply to celebrate the day which falls on a Sunday as do all these multiples of 70. I was born on a Sunday and 10 weeks later I turned 70 days old.

A Revision

In my previous post, I made a modification to the following algorithm:

Suppose we take any positive integer \(n \gt 1\) 

• if prime, double it and add 1: \(n \rightarrow 2n+1\)
• if composite, determine its number of divisors \(d\)
• if \( n \pmod d \equiv 0\) then \(n \rightarrow \dfrac{n}{d} \)
• if \( n \pmod d \not\equiv 0 \) then \(n \rightarrow n \times d\)

Keep repeating this process until a loop is reached or call a stop after a fixed number of iterations. 

The modification I made prevented the numbers generated from becoming too large too quickly. Instead of doubling a prime and adding 1, I decided to do this only if the number was a \(4k+1\) prime. If it was a \(4k+3\), I subtracted 1 and divided by 2. The new algorithm looks like this:

• if a \(4k+1\) prime, double it and add 1: \(n \rightarrow 2n+1\)
• if a \(4k+3\) prime, subtract 1 and divide by 2: \(n \rightarrow (n-1)/2\)
• if composite, determine its number of divisors \(d\)
• if \( n \pmod d \equiv 0\) then \(n \rightarrow \dfrac{n}{d} \)
• if \( n \pmod d \not\equiv 0 \) then \(n \rightarrow n \times d\) 

Here is an example using 28069 (permalink):

--- Loop detected at value 149708 ---

Divisors to Sequence:

28069, 56139, 224556, 18713, 37427, 149708, 1796496, 71859840, 561405, 8982480, 112281, 898248, 28743936, 2069563392, 10778976, 149708

------------------------------

Sequence Length: 16

Highest Value:   2069563392

This method of dealing with primes is also better suited to the sequences I mentioned earlier in my two posts: 

• Number's Factors to Sequence Algorithm 1
• Number's Factors to Sequence Algorithm 2

Here are the two new algorithms. 

Suppose we take any positive integer \(n \gt 1\) and apply the following rules to it:

• if a \(4k+1\) prime, double it and add 1: \(n \rightarrow 2n+1\)
• if a \(4k+3\) prime, subtract 1 and divide by 2: \(n \rightarrow (n-1)/2\)
• if composite, determine its number of factors \(f\) counted \( \textbf{with multiplicity}\)
• if \( n \pmod f \equiv 0\) then \(n \rightarrow \dfrac{n}{f} \)
• if \( n \pmod f \not\equiv 0 \) then \(n \rightarrow n \times f\)

Keep repeating this process until a loop is reached or call a stop after a fixed number of iterations. 

Here is an example using 28069 (permalink):

--- Loop detected at value 37427 ---

Number of factors to sequence with multiplicity:

28069, 56139, 112278, 37426, 18713, 37427, 74854, 224562, 898248, 149708, 37427

--------------------

Sequence Length: 11 

Highest Value:   898248 

Suppose we take any positive integer \(n \gt 1\) and apply the following rules to it:

• if a \(4k+1\) prime, double it and add 1: \(n \rightarrow 2n+1\)
• if a \(4k+3\) prime, subtract 1 and divide by 2: \(n \rightarrow (n-1)/2\)
• if composite, determine its number of factors \(f\) counted \( \textbf{without multiplicity}\)
• if \( n \pmod f \equiv 0\) then \(n \rightarrow \dfrac{n}{f} \)
• if \( n \pmod f \not\equiv 0 \) then \(n \rightarrow n \times f\)

Keep repeating this process until a loop is reached or call a stop after a fixed number of iterations. 

Here is an example using 28069 (permalink):

--- Loop detected at value 224562 ---

28069, 56139, 112278, 37426, 18713, 37427, 74854, 224562, 898248, 224562

--------------------

Sequence Length: 10

Highest Value:   898248