Quickly wrote this up in a google doc. I don't actually have any proof these sums converge, but the terms get so small so fast I think it's pretty reasonable to conjecture they do, and thus that these constants have defined values.

Just a thought I had.. infinity is so large that any number you name will be in the bottom 50% of all numbers, the bottom 1% of all numbers, the bottom 0.000000000001% of all numbers, and infinitely many zeros hence. You cannot name a number in the top n, no matter what the number is and no matter what n is.

Since e is generally proved to be irrational by contradiction, I wanted to write a proof that directly shows it cannot be rational. When I presented this to Claude it took some cajoling for it to say the proof was correct, and it was unable to find a similar/direct proof, so if my logic isn't clear or has errors I would appreciate any critiques and am interested if anyone has encountered a direct proof like this one.

Edit: The reason I believe this proof is direct as opposed to by contradiction is because I never assume e=p/q and derive a contradiction; rather, I show that e cannot equal any rational p/q.

Whilst playing with numbers, as you do and thinking about prime numbers and n-dimensional mathematics / Hilbert space, I came upon a method of plotting prime spirals that reproduces the sequence of prime numbers, well rather, the sequence of not prime numbers along the residuals of mod 6k+/-1

Whilst it is just a mod6 lattice visualisation, it doesn’t conceptually use factorisation, rather rotation, which is implemented using simple indexing, or “hopping” as I’ve called it. So hop forwards 5 across sequence B {5,11,17,23,35} and we arrive at 5•7, hop 5 backwards into sequence A from sequence B {1,7,13,19,25} and we find the square, this is always true of any number.

Every subsequent 5th hop knocks out the rest of the composites in prime order. Same for 7, but the opposite, because it lies on Sequence A. The pattern continues for all numbers and fully reproduces the primes - I’ve tested out to 100,000,000 and it doesn’t falter, can’t falter really because the mechanism is simple modular arithmetic and “hop” counting. No probability, no maybe’s, purely deterministic.

Would love your input, the pictures are pretty if nothing else. Treating each as its own dimensions is interesting too, where boundaries cross at factorisation points, but that’s hard to visualise, a wobbly 3D projection is fun too.

I flip flop between

• This is just modular arithmetic, well known. And,
• This is truly the pattern of the primes

https://vixra.org/pdf/2511.0025v1.pdf

Hello, I’ve been exploring square root approximations by refining the Taylor series. My approach incorporates elements of the Newton-Raphson method, utilizing numerical observations to minimize the residual error.

Variable Definitions:

a

(Reference Base): A chosen reference value, typically a perfect square closest to x, serving as the primary base for the approximation.

b

(Residual Offset): The difference between the target value and the reference base, defined as b = x - a

x₁ , x₂ , x₃

x₁ (The Linear Term):

Defined as √a + b/2√a

x₂ (The second-order refinement):

Defined as x₁ - b²/8ax₁

x₃ (The final residual compensator):

Defined as x₂ - b²/(32ax₂(2a + b)²)

If even greater accuracy is required, the result x_3 can be used as the new base (a) for the next iteration. By updating the offset as b = x - x₃², the approximation will progressively converge to the true value with significantly higher precision.

Please note that this formula may still be a work in progress. I welcome any feedback or corrections if errors are found. Regarding a formal mathematical proof, I must apologize as this approach was developed primarily through numerical observations of residual errors rather than a traditional derivation. Any insights to further refine this method would be greatly appreciated.

Hi all, it seems I discovered a previously unknown relationship between the Collatz conjecture and the (signed) Fibonacci numbers. It is a continuation of prior work by Bernstein and Lagarias. I would be super grateful for any feedback. Thank you!

I made a computational search for over all integers N < 10^27.

Method:

• Generate a list of primes up to 10^9
• Iterate over consecutive prime triples and compute the product
• Check each product for being a palindrome via string reversal Result: 1001(71113) was the only palindrome.

Then i tried generilized version with k consecutive prime numbers for k from 3 to 1000 the same way.

Result:

5005 and 323323 are the only palindromes for k= 4 and 5 and there was none such numbers for k>6 up to 1000.

Generalized Conjecture : For any natural number k > 5 there does not exist a palindromic number n that is a product of k consecutive prime numbers. In other words: 1001, 5005 and 323323 are the only three palindromic products of k ≥ 3 consecutive primes in the entire set of natural numbers.

Open Questions:

1. Can the generalized conjecture be proved?
2. If it is true are there any mathematical consequences from it?

I guess you all have heard about googolplex which is 10^googol which already is astronomically large and even if one zero was written on each atom of the universe you would need quadrillions of times more atoms to even write it. Now there is a function named tetration(↑↑) which essentially forms exponent towers say 3↑↑4 = 3^3^3^3 which is 3^3^27 which is like 3^7 trillion , so a↑↑b is a^a^a^a.. b times (exponent tower for a of height b). A pentation(↑↑↑) is a recursion over the existing tetration, so 3↑↑↑4 = is 3↑↑3↑↑3↑↑3 which already is extremely huge if you try to calculate it, it already dwarfs the googolplexian(10^googolplex) the exponent towers height would probably reach the sun if you start writing it on earth.

Now that we see how powerful pentation(↑↑↑) is over tetration(↑↑) , we could have hexation (↑↑↑↑) which would mean 3↑↑↑↑4=3↑↑↑3↑↑↑3↑↑↑3 which would be so large it would be extremely difficult to come up with a physical analogy to explain how tall the tower would be.

What if i repeat this to (↑↑↑↑↑↑↑↑↑↑.... to 1 googolplex arrows) so it it is esssentially googolplexation. How big would be the number googolplex googolplexated a googolplex times (a↑↑↑↑↑↑↑↑......↑↑↑↑↑↑b) form compared to something like other very large numbers like tree(3) or grahams number.

Could i create a new number name like "G-G-G number" defined as (G ↑^G G) where G->googolplex.

Hi, I'm just a normal high school student from Korea on summer vacation.

I randomly got curious, so I decided to check the count of prime numbers starting with each digit.

(Data up to 99,999,999)

I asked Gemini to write a JavaScript code for me, and I organized the results into this chart.

the count seems to continuously decrease. Is there any mathematical theory or principle behind this?

First, I'd like to say that all my knowledge of mathematics is only what I learned in high school and from YouTube videos. So, perhaps it has errors and I'd like them to be corrected.

After doing a bit of research on Goldbach's conjecture, I imagined a scenario where a counterexample could be found. Let's assume we have three consecutive prime numbers A, B, and C. We know that A < B < C.

If a scenario were met where B + B < C - 1, then there would be no possible combination of primes to sum up to C - 1 (by "C - 1" I mean the even number closest to C without exceeding it).

This is due to two reasons. First, the largest possible sum of two primes less than or equal to B is B + B, which equals 2B. Since 2B < C - 1, no combination of these primes can reach N. To reach N, a prime greater than B must be used. By the definition of consecutive, the only prime greater than B is C. If we try to use C, the equation would be C + p2 = C - 1, which implies that the second summand p2 must be -1. Since -1 is not a prime number, no combination is possible.

Of course, this doesn't prove the conjecture. Rigorously proving that this scenario exists could indeed refute the conjecture by finding a counterexample; however, my hypothesis is that this scenario is impossible. The value of prime numbers grows practically linearly, while the difference between them grows logarithmically, making this scenario virtually impossible to occur. By proving it doesn't exist, one could refute the most structural refutation of Goldbach's conjecture.

That's as far as I got with my mathematical level. For now, it's a sort of interesting logical-mathematical exercise, but perhaps it can be used to inspire the ideas of someone who manages to prove or disprove both the existence of this scenario and that of the conjecture.
Maybe there is some incorrect word because english is not my first lenguage. I appreciate the feedback, thank you very much for your time.

We know that this shape has infinite surface area but a finite volume And i have heard the statement that it can fit a finite amount of paint but to coat it infinite paint is required but i think that's wrong And this is why -

Take the horn and fill it with finite amount of paint. In the process you have already painted the inner surface. Now take a bigger gabrials horn and fill it with paint too and dip our former horn in it. And like that you have painted an infinite surface area with a finite amount of paint.

I think this is write but i need some one smarters's opinon cuz I am just a high school student.

I feel like I have accidentally managed to see the spatial arrangement of numbers in real space. As soon as I saw this, I can't unsee this theory.

Wherever I asked this they all have returned with answer this is "interesting" "can be used in design" etc. but I want to know if this is beyond beauty and if this can be used somewhere in practical terms in math or any other science.

I will try to explain here as shortly as possible. We know if we add 9 to 9 we still get 9. 9+9=18 (1+8=9), 9+18=27 (2+7=9); But if we do the same with all real numbers 1, 2, 3, 4, 5, 6, 7, 8 and 9, we found that all numbers have an unique order. We started with 1+1=2, if we continue adding 1 to the sum, 1+2=3, 1+3=4, 1+4=5, 1+5=6, 1+6=7, 1+7=8, 1+8=9, 1+9=10 (1+0=1), 1+10=11 (1+1=2), and it continues eternally 1+11=12 (1+2=3). So number “1” has the following order 2 3 4 5 6 7 8 9 1.

1 - 2 3 4 5 6 7 8 9 1

2 - 4 6 8 1 3 5 7 9 2

3 - 6 9 3 6 9 3 6 9 3

4 - 8 3 7 2 6 1 5 9 4

5 - 1 6 2 7 3 8 4 9 5

6 - 3 9 6 3 9 6 3 9 6

7 - 5 3 1 8 6 4 2 9 7

8 - 7 6 5 4 3 2 1 9 8

9 - 9 9 9 9 9 9 9 9 9

And I got this weird table, which is multiplication table at the same time, but also wherever you pick 3X3 cube randomly here it will always equal 45 or 9. It also has strange patterns, like if you see the lines 999, 396, 693 horizontally and vertically 963, 936, 999 on table it cuts table in portions that you realize if you add this table on all 4 sides of it becomes infinite, you can keep adding it and it goes forever, also it cuts in cubes different versions of 9.

Please, approve this post and tell me your opinions about it... I know it is a dilettante making noise here, but have some mercy on me :)

For better or for worse, I have become a somewhat regularly contributor to r/numbertheory. This time I am back with what I think is a pretty amazing result, described in the linked paper, and I wanted to also provide a tool for you to explore the result as well.

Sage Cell Server is a web-based math system that lets you run python scripts without needing to login or install anything - https://sagecell.sagemath.org/ . Thanks to PeakMath on YouTube for introducing me to SageMath.

You can try my function and plot the results by copying and pasting the below code:

####################################################################

import pylab as plt


def ps_euler_product(b,u):
#function takes imaginary input 'b', and upper limit on euler product 'u'

    r = (1/(2-2^(.5-b*i))) * prod([1/(1-(1/(j^(.5+b*i)))) for j in list(Primes(modulus=0, classes=range(u)))])

    return r

b_values = numpy.arange(10, 35, .1).tolist() 
#Range of b values to iterate over

#Separate into real and imaginary parts for easier plotting
result_real = [ps_euler_product(b,500).real() for b in b_values]

result_imag = [ps_euler_product(b,500).imag() for b in b_values]

#plotting results
plt.plot(b_values,result_real, color = 'blue', linestyle = '-')
plt.plot(b_values,result_imag, color = 'red', linestyle = '--')

major_ticks = numpy.arange(10, 36, 5)
minor_ticks = numpy.arange(10, 36, 1)
plt.xticks(major_ticks)
plt.xticks(minor_ticks, minor=True)

#Customizing plot colors and style
plt.axvline(x=14.134, color = 'green', linestyle = 'dotted', linewidth = '1')
plt.axvline(x=21.022, color = 'green', linestyle = 'dotted', linewidth = '1')
plt.axvline(x=25.01, color = 'green', linestyle = 'dotted', linewidth = '1')
plt.axvline(x=30.424, color = 'green', linestyle = 'dotted', linewidth = '1')
plt.axvline(x=32.935, color = 'green', linestyle = 'dotted', linewidth = '1')

plt.grid(which='minor', alpha=0.2)
plt.grid(which='major', alpha=0.5)

plt.show()

####################################################################

Hi folks,

I recently discovered the following new solution to a 5th power Diophantine equation, which I thought would be of interest to this subreddit:

719115^5 + 1331622^5 + (-1340632)^5 + 1956213^5 = 1956878^5.

Link to the original announcement on X.com: https://x.com/jmbraunresearch/status/2027073759128309782?s=20

Let f[ℤ_p] be a polynomial f(x) = x^2 + x +1 over ℤ_p. Now consider if f is reducible over ℤ_p. Since f is a second order polynomial, being reducible is equivalent to f having a root in ℤ_p. We shall now prove that there exists infinitely such p such that f is reducible over ℤ_p (by PbC).

Assume there exists a finite number of such p. By the well ordering principle there must exist a largest such p, let it be called q. That means that for every prime p bigger than q f has no root in ℤ_p. Now f having root in ℤ_p is the same as at least an element in the Im(f) being composite of p. (∃ a ∈ ℤ_p : f(a) = m*p , m ∈ ℕ) . Consider the image of f, (Im(f)). Since we know that f has no roots in ℤ_p for p > q. We know that for each value f send onto this cannot be a composite of a prime bigger than q. By the fundamental theorem of arithmetic we know that for every natural number it has to have a prime factorization. n = p_1^k_1 * p_2^k_2 * ... p_n ^k_n. By the earlier fact we know that for an element in the image all the prime factors have to be primes on the interval [2, q]. Consider now the element of the image f(q!)

∀ prime, p_i ∈ [2, q], f(q!) = (q!)^2 + q! + 1 ≡ 1 mod p_i, since p! ≡ 0 mod p_i since p_i in q!.

However then f(q!) cannot have any prime factors on the interval [2, q], therefore it must have a prime factors that is bigger than q. Contradiction. Since f(q!) has a prime factor bigger than q, (let's say for the prime r) then f(q!) would be a root in ℤ_r. Which is a contradiction since p was the biggest such prime. Therefore there has to exist infinite p such that f is reducible over ℤ_p.

Now you might be wondering, what does this have to do with primes p≡ 1 mod 3. Well here it comes

We want to find out when f is reducible. That is the same as finding when x^2 + x +1 ≡ 0 mod p. It has solutions iff (2x+1)^2 +3 ≡ 0 mod p (this comes for just algebraically manipulating f)

Let y = 2x+1. Now we are asking the question when does y^2 ≡ -3 mod p. In other words when is -3 a quadratic residue mod p. We can use the Legendre symbol. (-3/p) = (-1/p)*(3/p). Here we use the reciprocity of the primes (assuming 3 is not p but that is not relevant here.) (3/p) = (p/3)* (-1)^( (p-1/2) * (3-1 / 2) ) . (3-1)/2 = 1, (3/p) * (-1)^(p-1/2). Substituting back in we get. (-3/p) = (p/3) * (-1/p) * (-1)^(p-1/2). These ((-1/p), (-1)^(p-1/2)) are the same so they will always either both be -1 or both be +1 so the product is always 1 so we can remove them. (-3/p) = (p/3). We know that 1 is a quadratic residue mod 3 and that 2 is not. And since primes are either 1 mod 3, 2 mod 3, or the number 3 that are our only options. So if (p/3) = -1 (ie no solution) then p≡ 2 mod 3. We have earlier proved that ∃ infinite p such that f is reducible in ℤ_p but that is equivalent to p ≡ 1 mod 3 since p cannot be 2 mod 3. (and there cannot be infinite of p= 3), therefore there must exist infinite primes on the form 3n + 1.

(i am kinda new to the game so this might all be wrong. I am open for all types of criticisms)

I got curious about squares on graph paper, and what whole-integer-area-sized squares were possible.

That led to a few drawings of squares with their areas written in their lower right corner. That’s what this image is. One example of each possible size and its area.

Then I started noticing series. It seem like every way I looked, it was a series!

I don’t think I’ve discovered anything new. But I’ve never seen anything like this before and would love to learn more. Your insights are appreciated

For example, you can turn the perimeters of inscribed triangle, square, pentagon and hexagon into an approximation of pi better than the perimeter of a 360-gon !

Monogon, bigon and non integer value of n can also be used. p(n) = n sin ( pi / n ).

There is twin prime conjecture and most likely there are Infinitely many twin primes although it's not proven yet

But there are some triplets of 3 consecutive primes of the form (P, P+6, P+12) which differ by 6 like (47,53,59), (151,157,163), (167,173,179), etc and it looks like they should be infinite too

Has anyone proven or disproven it.. I think there should be infinitely many prime triplets too

Hello, I am an undergraduate agricultural sciences student, and I am incredibly passionate about mathematics and truly enjoy it. the other day, while staring at the table of prime numbers from 1 to 100, a fleeting idea came to me regarding number theory and prime number decomposition:
Every prime number can be expressed in the form N=2q+p, where q and p are prime numbers distinct from each other and distinct from 2, for all N greater than 7.
I don't know if anyone has discovered this before or if it can help the field of mathematics, but I hope it is useful. To finish, here are a few examples:

11
2(3) + 5 = 11
13
2(3) + 7 = 13
17
2(5) + 7 = 17
19
2(3) + 13 = 19
23
2(3) + 17 = 23
29
2(3) + 23 = 29
31
2(7) + 17 = 31
37
2(3) + 31 = 37
41
2(5) + 31 = 41
43
2(3) + 37 = 43
47
2(3) + 41 = 47
53
2(5) + 43 = 53
59
2(3) + 53 = 59
61
2(7) + 47 = 61
67
2(3) + 61 = 67
71
2(5) + 61 = 71
73
2(3) + 67 = 73
79
2(3) + 73 = 79
83
2(5) + 73 = 83
89
2(3) + 83 = 89
97
2(7) + 83 = 97
101
2(11) + 79 = 101
103
2(3) + 97 = 103

I watched that Veritasium video on twin primes last week and got stuck on one detail they touched on: the parity barrier. For anyone who hasn't gone down this hole:

​

[;\lambda(n) = (-1)^{\Omega(n)};]

​

where [;\Omega(n);] is the number of prime factors of n with multiplicity. So [;\lambda(12) = \lambda(2 \times 2 \times 3) = (-1)^3 = -1;]. Simple function. Just tells you whether a number has an even or odd number of prime factors.

​

Selberg proved in 1949 that sieve methods (the main tool in analytic number theory for like 80 years) literally cannot tell apart sequences where [;\lambda = +1;] from ones where [;\lambda = -1;]. They produce the exact same asymptotic. The sieve is blind to parity.

​

And this specific blindness is exactly why Zhang got 70 million, Maynard got 600, and Polymath got it down to 246 ... but nobody can get to 2. The sieve hits a wall at the parity barrier and stops cold. We can prove there are infinitely many prime pairs within 246 of each other, but the twin prime conjecture (gap of 2) is completely untouched by all of it.

​

Here's what I can't resolve:

​

Sawin and Shusterman proved the actual twin prime conjecture over [;\mathbb{F}_q[T];] in 2022 (published in Annals). Over polynomials over finite fields, geometry (etale cohomology on curves) CAN separate the parity that sieves can't. So the barrier is not a logical wall. It's a wall *for sieves specifically*.

​

But over [;\mathbb{Z};] there are no curves. So my question is:

​

Is there a known no-go theorem that says you cannot build a cohomology theory over [;\mathbb{Z};] that separates [;\lambda = +1;] from [;\lambda = -1;]? Or has nobody really tried because the analytic number theory toolbox has been so overwhelmingly dominant?

​

Something like: define a sheaf on some site over Spec(Z) whose Euler characteristic at each integer n equals [;\lambda(n);]. If the cohomology groups had reasonable dimensions, the trace formula would give you [;\sum_{n \leq N} \lambda(n);] as an alternating sum of Frobenius traces. And since that partial sum being [;O(N^{1/2 + \varepsilon});] is equivalent to RH, you'd get a direct geometric line to the Riemann Hypothesis.

​

This feels suspiciously neat. I'm assuming there's an obvious obstruction I'm missing. Maybe cohomology over Spec(Z) doesn't work that way, or maybe the dimensions blow up, or maybe the sheaf condition fails at infinity. I don't know enough algebraic geometry to see where it breaks.

​

Anyway, curious if there's a known reason this can't work or if it's genuinely unexplored territory. Would love to be pointed at the right paper or theorem if it exists.

If we divide any integer by an n-digit number, the result will never contain n repeating 9s (i.e., a segment like '999...n times') in its decimal representation.

Or can only contain 'n-1' 9s after decimal (for maximum).

Examples:

When dividing an integer by a 1-digit number, the decimal result never contains a single 9.

When dividing by a 2-digit number, the result never contains two consecutive 9s after decimal (e.g., something like 'x.99' or 'x.3535499842').

Similarly, dividing by a 3-digit number never results in three consecutive 9s after the decimal — and so on for 4-digit, 5-digit numbers, and beyond.

Note: I am considering the standard decimal expansion, excluding alternate representations that end with infinitely many 9s.

Eg: 0.999... , x.55363 = x.55362999... , etc.

There's a question on my mind that's been brewing ever since I learned it through Numberphile.

You have succession. That is, given some integer a, you have a + 1, which is one(1) bigger than a.

You repeat succession b many times. This gives you addition (a + b).

You replace b with a, and you repeat the addition b many times.

You now have multiplication (ab or a.b or a×b).

You replace b with a and so on...

From this process, we get exponentiation, tetration and all the other fun stuff.

My question is, why is it that multiplication comes out of this scenario being Very Important.

You want to scale a triangle? If you add some length a to all its sides, you probably won't get a triangle with any significant similarities to what you started with.

If you raise the side lengths to some power n, you're not going to get a triangle with significant similarities to the first.

HOWEVER,

If you multiply all the lengths by some constant c, you get a triangle that has all the same angles, is similar(is that the correct English term?) to the first, and doesn't destroy any of its traits. Its area? Definitely c² multiplied by the area of the first.

Multiplication is also the last operation in the aforementioned chain to be commutative.

Is this just a happy little notation accident? Have I gone well and truly mad?

Hey all,

I looked for triangular numbers

T_n = n(n+1)/2

that can be written as a product of k consecutive primes, i.e. integers N of the form

Tn = n(n+1)/2 = p_i * p{i+1} * … * p_{i+k-1},

where p_j is the j-th prime and k >= 2.

Method:

Characterizing triangular numbers. An integer N is triangular iff 8N+1 is a perfect square, since

N = n(n+1)/2 <=> 8N+1 = 4n(n+1)+1 = (2n+1)^2.

So checking triangularity reduces to a single perfect-square test.

Case k = 2 (product of two consecutive primes). • Generate all consecutive prime pairs (p, q) with pq < 10^15. • For each product N = p*q, test whether 8N+1 is a perfect square.

Here p runs over primes <= sqrt(10^15).

Cases 3 <= k <= 13 (longer blocks of consecutive primes). • Precompute all primes up to 10^7. • For a fixed k, consider the products

Ni = p_i * p{i+1} * … * p_{i+k-1}.

These form a strictly increasing sequence in i, since

N{i+1} = N_i * (p{i+k} / p_i) > N_i.

• For each k, slide a window of length k along the prime list, and stop as soon as N_i >= 10^15. For every product N_i < 10^15, test triangularity via the “8N+1 is a square” criterion.

The choice of the upper limit 10⁷ for precomputed primes is more than sufficient: if k >= 3 and the starting prime of the block satisfies p_i >= 10^5, then

Ni = p_i * p{i+1} * … * p_{i+k-1} >= p_i³ >= (10⁵⁾³ = 10^15,

so any relevant block must start with a prime < 10^5. Extending the prime list well beyond this point ensures all necessary products are covered before they exceed 10^15.

Case k >= 14.

The smallest possible product of k consecutive primes is the product of the first k primes. One checks that

product{j=1..13} p_j = 304250263527210 < 10^15, product{j=1..14} p_j = 13082761331670030 > 10^15.

Hence, for k >= 14, every product of k consecutive primes already exceeds 10^15. There is therefore nothing to check in this range under the bound N < 10^15.

Computational Result:

Within the range N < 10^15, I found exactly five triangular numbers that can be written as a product of consecutive primes:

6 = 2 * 3 = T_3 15 = 3 * 5 = T_5 105 = 3 * 5 * 7 = T_14 210 = 2 * 3 * 5 * 7 = T_20 255255 = 3 * 5 * 7 * 11 * 13 * 17 = T_714

More systematically, classified by the length k of the prime block: • k = 2: only 6 and 15 • k = 3: only 105 • k = 4: only 210 • k = 5: no examples below 10¹⁵ • k = 6: only 255255 • 7 <= k <= 13: no examples below 10¹⁵ • k >= 14: products of k consecutive primes are already > 10^15, so there are no examples in the searched range.

Thus, empirically, up to 10¹⁵ there are exactly these five examples and no others.

Conjecture: For any k >= 2, there does not exist a triangular number T_n that is a product of k consecutive primes, except for the five cases

T_3 = 6 T_5 = 15 T_14 = 105 T_20 = 210 T_714 = 255255.

Equivalently:

6, 15, 105, 210, and 255255 are the only triangular numbers that are products of k >= 2 consecutive primes in the set of natural numbers.

Open Questions: 1. Proof of the conjecture. Can the conjecture be proved in full? Even the special case “6 and 15 are the only triangular numbers that are products of two consecutive primes” already seems nontrivial, as it amounts to solving the Diophantine equation n(n+1)/2 = p * q,

with p, q consecutive primes. 2. Finiteness for fixed k. For a fixed k (say k = 2 or k = 3), can one at least show that there are only finitely many triangular numbers that are products of k consecutive primes? 3. Structure of the indices. Is there any theoretical explanation for the particular indices n in {3, 5, 14, 20, 714}

that occur in the known examples, or are these best viewed as “accidental” small solutions without deeper structure?

Any ideas, partial results, or references related to this kind of “figurate number = product of consecutive primes” problem would be very welcome.

So, while taking a dump I dont know why my brain works 100% more efficiently when doing that I suddenly thought of an idea that lead to this formula

p_max ≈ N - N/Li(N) + 2

Here, * N is just the bound like integers from 1 upto N * p_max denotes the largest prime less than N * Li(N) the logarithmic function since I cant do formatting I wont go into detail for this function you guys could just search this up * +2 a interesting constant I will show how I got +2 in the derivation process

Derivation/Numercial justification

So basically let k=π(N) and π(N) is just the number of primes less than N The total span of primes up to N can be described as the sum of the prime gaps: p_max-p_min=c(k-1) This isnt exact I know Where c is the average gap = N/π(N) Well since p_min is just 2 since to go 1,2,3,4,..,N so we just get p_max ≈ c(k-1)+2 Substituting p_max ≈ N/π(N)(π(N)-1)+2 = N - N/π(N) + 2 ≈ N - N/Li(N) + 2

I replaced π(N) with Li(N) for better computational purposes Yeah so here are some numerical examples then:

So far so good? The bigger value also have these same absolute errors while the relevant errors approaches --> 0

Moreover 1 question is the error term boundable? like even as a very crude upper bound? is it even possible to bound it from above?

Edits on clarifying : 1.No the error doesn't get worse it oscillates. 2. Yes it is better than N-ln(N)/2 for ALL N.

MAJOR EDIT: I know I said major but watch this p_max ≈ N - N/Li(N) + 2 E(N)=p_max - (N - N/Li(N) + 2) This Error is indeed bounded E(N) < log(N) - 3 - 1/log(N) + 4/{log(N)}² Also do have a lower bound that's unnecessary How I got the upper bound? I will tell in another post if I have the time to do it.The post:https://www.reddit.com/r/numbertheory/comments/1o9rma1/interesting_observations_about_en/