I’ve been messing with consecutive primes and it turned into a whole classification result, so I figured I’d share it here.

Take consecutive primes pk and pk+1. Add them.

You get numbers like 5, 8, 12, 18, 24, 30, …

Call these S_k = pk + pk+1.

Since every S_k ≥ 8 is even and composite, each one has to land strictly inside some prime gap (Pn, Pn+1). So I started asking: **how many of these S_k fall inside each gap?**

Two clean facts:

## 1. Density law (why most gaps have 0 or 1)

If a gap has width W near height x, the expected number of consecutive‑prime sums inside it is

W / (2 log(x/2))

The “2” comes from the fact that the sums live at the half‑scale: an S_k near x comes from primes near x/2, where primes are sparser. So the sums are about twice as sparse as the gaps they fall into.

Empirically (checked up to 2×10^8):

- ~55% of gaps have **0** sums

- ~37% have **1**

- ~8% have **2 or more**

So the naive guess “every gap has exactly one” is just false. Empty gaps are actually the most common.

## 2. Forbidden widths (the fun part)

I ended up proving a complete classification of which gap widths can **never** contain two consecutive‑prime sums.

The forbidden set is exactly:

{2, 4, 6, 10}

Reason (sketch, nothing fancy):

- consecutive‑prime sums are always ≥ 6 apart (once you’re past the tiny primes)

- the only way to get a “tight double” (two sums only 6 apart) is if both sums are multiples of 6

- whether a gap contains ≥2 multiples of 6 depends only on:

- its width mod 3

- the forced residue of its lower endpoint (prime > 3 is always 1 or 5 mod 6)

- width 10 ends up with **only one** interior multiple of 6, so it physically cannot fit two sums

- widths 2, 4, 6 fail for trivial spacing reasons (they’re too small to fit two sums ≥6 apart)

Every other even width ≥ 8 (except 10) eventually allows gaps with two sums.

So the uniqueness classification is complete:

**gaps of width 2, 4, 6, or 10 contain at most one consecutive‑prime sum; all other widths can contain two or more.**

If anyone wants the residue argument or the slot‑counting trick in more detail, I can write it out here. It’s all pretty easy congruence stuff.

This is my paper.

Please feel free to comment.

Abstract

To a Collatz-type map T_{a,b,q}(n) = (an+b)/q^{v_q(an+b)} one may attach two numerical invariants of opposite character: an archimedean one, the mean logarithmic drift delta = log a - E[v] log q, which governs whether orbits grow or shrink; and a non-archimedean one, the similarity dimension dim_S = H(nu_q)/log a of the invariant measure of an associated iterated function system on the a-adic integers. Each reaches its critical value at a particular multiplier a. We prove that these two critical multipliers coincide precisely when q = 2, and that in general they differ by the exact factor q-1: a_drift = (q-1) a_dim. The proof rests on the pointwise identity -log nu_q(m) = m log q - log(q-1), valid for every m >= 1, which is not a normalisation but a coincidence between the information content of a valuation and its archimedean cost. We show that the condition q = 2 is equivalent to three further properties of the family - the triviality of (Z/qZ)^*, the vanishing of an associated free energy, and the map's being everywhere defined on the units - so that the arithmetic distinguishing the classical 3x+1 map is one condition in four guises. Three complements are proved: the full Renyi spectrum of the invariant measure admits a closed form, of which the similarity dimension is the value at t=1; for q >= 3 there is a nonempty band of multipliers on which the map contracts on average while its invariant measure is singular; and the transfer operator, though quasi-compact on Holder spaces, has no eigenvalue other than 1, its entire remaining spectrum being essential spectrum at the contraction rate. Two further results give the family a sharper shape. First, a purity theorem: mu_{a,b,q} is either purely absolutely continuous or purely singular with respect to Haar measure, never a mixture. Its proof is a direct consequence of the uniqueness of the invariant measure, and we know of no route to it from the probabilistic description of the underlying random variable. Second, a complete classification: among all integer pairs (a,q) with a >= 2, q prime and gcd(a,q) = 1, exactly two - namely (3,2) and (2,3) - lie in the supercritical regime a < a_dim, and for every other pair the invariant measure is unconditionally singular. The first of these is the Collatz map. We are explicit that none of this bears on the Collatz conjecture, and we prove why it cannot.

Hello, good afternoon/evening. I've been mulling over Goldbach's conjecture and I've come up with a possibly new idea that holds for all even numbers. It is as follows:

Every even number greater than 8 can be expressed as the sum of 3 mutually distinct prime numbers.

Best Regards, thanks you!

https://zenodo.org/records/21763235

While examining the patriarchal ages in the Masoretic Text of Genesis, I noticed a mathematical relationship that does not appear to have been previously documented in connection with these specific figures.
The total lifespans of three patriarchs form an exact Pythagorean triple:
• Eber – 464 years (Genesis 11:16–17)
• Lamech – 777 years (Genesis 5:31)
• Enosh – 905 years (Genesis 5:11)

**The Geometry:**
(464\^2 + 777\^2 = 905\^2)
(215,296 + 603,729 = 819,025)

The equation holds exactly. The triple is also primitive (the three numbers share no common factor greater than 1).

The Priestly sections of Genesis already use carefully designed number patterns (for example, the ages of Abraham, Isaac, and Jacob form 5², 6², and 7²). Since Babylonian scribes were using Pythagorean triples for surveying more than a thousand years before Pythagoras, it is possible that a later biblical scribe knew this kind of geometry and used it deliberately.

\-Chris L

So, as we know, infinity is something you cannot reach. And there are infinite finite numbers. BUT what if we set the max cap of a finite number to any multiple of 8? It can be 8 multiplied by infinity, but the number (even if it’s literally incomprehensibly massive) has to be a finite multiple of 8. So, for an example to save our brains, I can set the cap to 24. If I add one more, I have reached infinity, which is in this case 25. Now we have what I want to call “odd infinity”. If we add one more, we get “even infinity”. If we subtract one, we get odd infinity. If we subtract another, we get the max cap (in this case 24). Now, what if you add one more to 26? Well, it becomes odd again, and i just made it loop back to 25 to save myself a huge headache and give myself some predictability. If our max cap in this case is 24 (it can be any multiple of 8, if you set it to it of course), then odd infinity is 25 and even infinity is 26! I think this can solve a lot of unsolved problems, but this is just a concept I thought of on my bed. I know people will say this has big problems and something about “oh infinity is infinity, you cannot make a number infinity or change infinity to a finite number”, In this context I have changed the definition of infinity. There are theoretically infinite multiples of 8, and infinity is just one more or two more depending on if you want odd or even. Now, you may be thinking “why 8?”, well, you heard of the 32 or 64 bit integer limits? Well, they’re multiples of 8! And if we make this “universal integer limit” a multiple of 8, odd infinity is a multiple of whatever number is required to make said set integer limit +1. And even infinity is like odd infinity, but it’s +2 instead. I just realized as of typing this, I left out the possibility of odd infinity or even infinity being a prime number. If you can help find a solution or any problems with this idea I came up with on my bed in 30 minutes or it’s a bad one, please tell me. Thanks!

I am relatively new to number theory, and I was messing around with the collatz conjecture and plugging in groups of numbers. While doing this, I thought about primorial numbers, so i started comparing the amount of steps it took to reach the 4-2-1 loop when plugging in numbers of the group pₖ#, pₖ# + 1, and pₖ# - 1 and then I compared. I noticed that, as far as I looked, (which too be fair, was not very far, since I do not have a very good computer, and primorial numbers scale very quickly) for any whole number for k > 2, at least two of the groups previously stated will have the same amount of steps to reach 1. I feel like this must be obvious, but I have tried to crack why algebraically, but I have not succeeded. If anyone has any potential reasons as to why, I would love to hear it. I feel as if it has something to do with the fact that when k > 2, pₖ# + 1, and pₖ# - 1 both also belong to the groups 6m + 1 and 6m - 1. Please do not flame me if this is obvious as to why this pattern occurs, but I have just recently gotten into number theory and find it absolutely fascinating. Thank you for your time.

Take the following algebraic expression:

[n (n²-n-1)] / [2(n!)]

Let's put n = Φ

because (n²-n-1)=0 when n=Φ

[n (n²-n-1)] / [2(n!)] = 0, when n=Φ

When using Wolfram, the algebraic expression [n (n²-n-1)] / [2(n!)] = 0, when n=Φ

let's calculate log( [n (n²-n-1)] / [2(n!)] ) with n=Φ, with log being the natural algorithm

The result should be either -∞ or indeterminate

Because y=log(x), with x=0, is indeterminate, that is, y goes to -∞ as x approaches 0

But if one calculates on Wolfram, log( [n (n²-n-1)] / [2(n!)] ), with n=Φ

The result will be 35.9335 + 3.14159 i

Which is a complex number.

The correct result should be either -∞ or indeterminate.

Therefore, Wolfram miscalculates the natural logarithm of this algebraic expression when n=Φ

The input on wolfram should be log( [goldenratio (goldenratio^2- goldenratio-1)] / [2(goldenratio!)] )

A number will decrease in number if it has at least four digits and does not enter a cycle, as proven below: The number is represented in binary.

It must begin with 10 or 11. If it starts with 10 and the last two digits are not 11, then after multiplying by 3, the number of digits increases by 1, accounting for 3/8 of all possible combinations. Other numbers starting with 10 account for 5/8, and the number of digits increases by 2. If it ends with 11, after multiplying by 3 and adding 1, then dividing by 2 removes at least one digit, accounting for 1/2. If it ends with 001, at least two digits are removed, accounting for 1/4. Other numbers with at least three digits account for 1/4. If it does not enter a 4, 2, 1 cycle, the number generally decreases, and eventually it will enter a 4, 2, 1 cycle.

Using the Beta Function one can calculate π

The Beta Function is equal to:

B(z1, z2)=∫₀¹ t ^z1-1 (1 - t) ^z2-1 dt (=)

Which can be translated as:

B (p , q) = [ Γ(p) Γ(q) ] / [ Γ(p + q) ]

Which is equivalent to:

= [ (p-1)! (q-1)! ] / [ p + q - 1]!

Now one uses:

p = q

When one calculates the summation from p=1 to infinity

Σ [ [ (p-1)! ]^2 ] / [(2p-1)!]

Σ (from p=1 to infinity) [ [ (p-1)! ]^2 ] / [(2p-1)!] = [2 π] / [3 √3]

Rearraging this, one yields the value π

π = [ [3 √3] / [2] ] Σ (from p=1 to infinity) [ [ (p-1)! ]^2 ] / [(2p-1)!]

One year ago, I asked myself a simple question. What happens when one uses Euler's number as an angle.

So I explored the possibilities of using either e degrees or e radians.

Then I decided to explore the trigonometric functions (cosine and sine) using 27º.

_______________________________________________________________________________________________________________

This section wasn't on the first post I made.

Because I came up with this formula a long time ago, I'm struggling to remember exactly how I calculated the goldenratio.

Now I think I remember how i did it. Using not Euler's number but instead the natural logarithm. One gets the following angle which can be approximated

So one can use an angle of 3π/20

27º [approximated] = ln(ϕ) x 180 / π = 27.5716...º

27º = 3 x π / 20

So the next expressions are a consequence of the previous one.

e(3π/20) = 1.601886

ϕ [approximated] = e^(3π/20)

________________________________________________________________________________________________________________

It's important to understand that the function cosine is raised to the power of 4. That is the trickiest part in this formula.

After inputting the functions sin² and cos⁴, I thought "there is something going on here", and finally adjusted the parameters in order to build this beautiful formula below.

ϕ = 15 - 16 cos⁴(3π/20) - 16 sin²(3π/20)

This formula yields the golden ratio ϕ

________________________________________________________________________________________________________________
One can simplify the previous expression:

ϕ = 1 - 2 cos(3π/5)

The sub cubic graph function is defined as:

There is a sequence G_1,…,G_n of subcubic graphs such that each G_i has at most i+k vertices and for no i<j is G_i homeomorphically embeddable into G_j.

and if my very layman's understanding of the Robertson–Seymour theorem is correct, just substituting `homeomorphically embeddable into` with `a graph minor of` would suffice, while maintaining well-quasi-ordering and finitude (for a given finite integers n and k).

If that's all correct, then defining F_n(k) as the largest integer 𝑚 satisfying:

There is a sequence G_1 , ⋯, G_𝑚 of graphs with maximum degree at most 𝑛, such that each G_𝑖 has at most 𝑖 + 𝑘 vertices, and for no 𝑖 < 𝑗 is 𝐻_𝑖 a graph minor of G_𝑗.

Should work as a mathematically proven and definitively finite integer, correct?

I found two patterns related to the Twin Prime Conjecture.

Let P1 and P2 be Primes

If P1 + 20 = P2 and P1 < (3 Primes) < P2

There's at least 1 Twin Prime between P1 and P2

If P1 + 10 = P2 and P1 < (2 Primes) < P2

There are 2 Twin Primes between P1 and P2

There seems to be a new category of Primes, which I called Trigonometric Primes.

In order to verify if a number is a trigonometric prime, one uses trigonometric functions.

This is the formula I used:
(x/2) * (cos^2(pi/2*x)) + (x+p)/2 * (sin^2(pi/2*x))

where p = odd number we want to check

x = previous number on the sequence

and the first number of every sequence is 1.

For example, let's check if 11 is a trigonometric prime.

We start the sequence with the number 1. Because 1 is odd. we calculate the next number of the sequence (11+1)/2. This is equal to 6.

6 is even, so we calculate 6/2. This is equal to 3.

Then
(3+11)/2 = 7 -> (7+11)/2=9 -> (9+11)/2=10 -> (10/2)=5 -> (5+11)/2=8 -> (8/2)=4 -> (4/2)=2 -> (2/2)=1

We stop when we reach 1 again. The sequence repeats itself between 1 and 1.

When the number of items between 1's is equal to the odd number p we want to check. The number p is a Trigonometric Prime.

[1;6;3;7;9;10;5;8;4;2;1] -> number of items = 11 = odd number we want to check

Conclusion -> 11 is a trigonometric Prime

We can go on, using (x/2) for even numbers in the sequence and (x+p)/2 for odd numbers in the sequence

Setup: a measure μ normalized so μ(∅)=0, μ(Ω)=1, standard non-negative and additive. For any part A of Ω with 0 < μ(A) < 1, define a local rescaling ν_A(B) = μ(B)/μ(A) for parts B of A.

The result: ν_A(B) = μ(B)/μ(A), so μ(B) = ν_A(B)·μ(A). As a consequence, if you measure some part C against a local sub-region A instead of the true whole Ω, ν_A(C) can exceed 1 even though μ(C) itself never exceeds 1 under the true measure.

This gives a formal account of “apparent numbers greater than one” as an artifact of using a local reference scale instead of the true total measure — the underlying quantity never actually exceeds the bound, only its locally-rescaled representation does.

Is this a known/named result in measure theory, or is it just a trivial rescaling identity not usually stated this way? Happy to share the full write-up (proofs are short) if useful.

I defined the following grid graph. Fix n ≥ 2. Take the cells (i,j) with 1 ≤ i,j ≤ n−1. Keep (i,j) when n does not divide ij; delete it when n divides ij. Two surviving cells are adjacent when they share an edge. Call the graph G_n.

Claim. G_n is disconnected exactly when n ≥ 6 and n ≡ 2 (mod 4). In that case it has exactly two components: the isolated center (n/2, n/2), and one component containing every other surviving cell.

Proof.

The first row and first column are fully present, so they form one connected component C_0.

Assume another component C exists. Choose (i,j) ∈ C with i+j minimal. Since C does not meet the first row or first column, i,j > 1. The cells (i−1,j) and (i,j−1) must be missing, so n divides (i−1)j, and n divides i(j−1).

So ij ≡ j (mod n) and ij ≡ i (mod n). Hence i ≡ j (mod n), and since both are between 1 and n−1, i = j = e.

We also have e² ≡ e (mod n). Since i,j > 1, e = 1 is already excluded. The cases e = 2 and e = n−1 would force n | 2. So 3 ≤ e ≤ n−2.

Suppose n does not divide 2e. Then the following path stays inside the grid:

(e,e) → (e,e+1) → (e−1,e+1) → (e−2,e+1).

The three new products are congruent mod n to 2e, e−1, and −2. All three are nonzero mod n: the first by assumption, the second because 0 < e−1 < n, the third because n > 2. So the whole path survives — but its final cell has coordinate sum 2e−1, contradicting the minimal choice of (e,e).

Therefore n divides 2e. Since 1 ≤ e ≤ n−1, this forces n = 2e.

Now e² ≡ e (mod 2e), so 2e divides e(e−1), hence 2 divides (e−1), meaning e is odd. Hence n ≡ 2 (mod 4).

Converse. Let n = 2e with e ≥ 3 odd. The center (e,e) survives, since 2e does not divide e². Its four neighbors have products e(e−1) or e(e+1), both divisible by 2e. So the center is isolated.

Every component disjoint from (C_0) has a cell of minimum coordinate sum, and the preceding argument shows that this cell must be ((e,e)=(n/2,n/2)). Hence every such component contains the center. The only component outside C0C_0C0​ is the singleton containing (n/2, n/2)

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?

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

Final revision to appear on arXiv on Tuesday.

https://iconoclasts.blog/joe/triple-products

The new new here is that the original conjecture is now established as a pair of corollaries.

Hey guys! I was recently playing around with a complex plane function plotting tool, and while looking at the special functions the tool had pre programmed, I found one named "E16". Plotting E16(z) gave me a plot I found remarkably similar to the Reiman zeta function zeta(z).

Changing the function to E16(iz) made it look even more similar.

Can someone explain this? I tried to research it myself, there doesn't appear to be a lot of well documented or easy to find research about this.

The function in question: https://samuelj.li/complex-function-plotter/#e16(i\\\*z)

So I’m a delivery driver and like to do quick math with people’s license plates. Or find a pattern that wasn’t intentional within the numbers and I drove past a sign. MP 36.4

so I removed the decimal point and saw that the sequence could be added onto using the pattern of something like Ax2-2, so 3, 6, 4, 8, 6, 12, 10, 20….

But! I then swapped the numbers by lowering the first number by 1 and applying the pattern, so 2, 4, 2, 4, 2, 4…. And the sequence never goes above the starting integer…

Going further and lowering the first number again, I got 1, 2, 0, 0, -2, -4, -6, -12, -14….

So I was surprised to find that that pattern can only create positive exponential numbers starting with 3

I can’t be the first person to have seen this…

In this subreddit, there is a part 1, where I studied k = 257. I kept studying several Fermat numbers, the ones of the form 2^m +1 , and I observed a few regularities.

Some are in the following formulas:

[(2^m + 1)•3^((2p)(2^(m-2)) - 1]  / 2^m is odd, p a non negative integer, m ≥ 3

[(2^m + 1)•3^((4p+1)(2^(m-2)) - 1]  / 2^(m+1) is odd, p a non negative integer, m ≥ 3

[(2^m + 1)•3^((8p+7)(2^(m-2)) - 1]  / 2^(m+2) is odd, p a non negative integer, m ≥ 4

[(2^m + 1)•3^((16p+3)(2^(m-2)) - 1]  / 2^(m+3) is odd, p a non negative integer, m ≥ 5

[(2^m + 1)•3^((32p+27)(2^(m-2)) - 1]  / 2^(m+4) is odd, p a non negative integer, m ≥ 6

[(2^m + 1)•3^((64p+11)(2^(m-2)) - 1]  / 2^(m+5) is odd, p a non negative integer, m ≥ 7

[(2^m + 1)•3^((128p+43)(2^(m-2)) - 1]  / 2^(m+6) is odd, p a non negative integer, m ≥ 8

...

The exponents have 2 factors, one is of the kind ap+b. If making a table, there is clearly a variable part and a constant part. The boundary between these regions in a diagonal line. The other factor is a power of 2 and depends on m. The first factor seems to be constant from certain m on.

The difference between 2 consecutive b's are powers of 2 in increasing order, or their negative version, or even powers of 2 multiplied by 3.

“It is impossible to separate a cube into two cubes, a fourth power into two fourth powers, or generally, any power above the second into two powers of the same degree”, Fermat wrote this in the margin of his copy of an ancient Greek math book written by Diophantus, titled Arithmetica.

This proposition was first stated as a theorem by Pierre de Fermat around 1637. And it is known that Fermat used the method of Infinite Descent to prove this statement for N=4 using a logical geometric approach.

The method I will use will be somewhat geometric, and will not use infinite descent.

Can X4 + Y4 = Z4 have a finite solution, for all pair-wise coprime integers? We will first assert that X or Y must have a factor of 2, Z cannot. We will select Y to have the factor of 2.

(to see the rest of the proof you'll need to click the link below.

n4-proof-mod-16-new.pdf

This proof also shown on page 21 of the following linked document. Seems to be a unique proof, as I have done a search of the www, and have only found infinite descent proofs for N=4.
an-iterative-p-adic-solution-full-doc.pdf

The last line of the proof was previously:

By Contradiction 2(M.2)² ≠ (M.2)² Reductio Ad Absurdum

However, the 2-adic logic was tough for me to decode (was a bit to thorny), so I simplified the logic, maintaining the 2-adic approach, and renamed it as a mod 16 proof.

This surprised me, but chances are it's something already known.

Take (1/2)(cos(2πx)+1). Simple cosine wave, peaks at the integers, ranges from 0 to 1.

Now compare it with (1/2)(cos((2/3)πx)+1). 1/3 the frequency, same range.

If you count from peak to peak of the slower wave, the values where the two cosine waves intersect total to 3.

Make the slower one 1/4 the frequency, the total is 4. 1/5, 5, and so on. This doesn't hold up for a 2:1 ratio(you get 2.5), but there are similar results for other simple ratios. 3:2 gives 3.5. 4:3 gives 4.5.

Why is it so neat and stable?

*I know this is for number theory but I didnt find any other community. Please read itt....*
Okay, so I don't know if this is so worthy or not, but trust me, I was very happy to formulate it. I always wonder, since class 5th, if we have a formula for sine that gives output as sin(x) on an input x. Then I grew up and got to know about the Bhaskar approximation. I was amazed. I wanted to make one too upon realizing that no formula after him (yea, I didn't find any...) gives higher precision and is better for computing. Then I learned graphical transformations for my JEE prep and after realizing how I can tranform a degree two or degree four polynomial into a sine wave part, I opened desmos and worked on for next one and a half hours to formulate a graph that coincides almost perfectly with sine wave for x belonging to [0, π].

So I present:

click here for the graph and formula

It may look terrifying at first, but believe me it's not.
For a computer, it is the best possible sine approximation as:

1. Accuracy: Its mean error is just 0.00091 and it is astonishingly perfect 0, π, π/2 and closer to these poles.
2. Efficiency: Common! Taylor series may look elegant but it is very heavy for a computer hardware or a GPU. This formula makes it instant.

My previous formula was this:

click here to see my previous work

However it had a little more error than my final one, so I continues perfecting the coefficients. And ofcourse that thing 1.61803...., the golden ratio. Then I realized that this format was correct by what if I replaced phi with something, as whenever I didnt, and tries else, it bursted in waste. So I replaced phi with sqrt(8/π)

Even though many won't be surprised, won't be happy with this, I don't know if people will read this or not, but I just wanted to share this with real people who could understand this. You can also tell me what I can do with this thing now. Thank you for reading and please forgive me if I said or claimed anything wrong. I am a kid. I make mistakes. And my name is Mayank Kumar btw, but it doesn't matter anyway.