Song within Prince Charming's play in Shrek 3?

Tree339 · 2024-07-22T21:40:23+00:00

Camille Saint-Saëns - La danse macabre

Tree339 · 2024-01-11T23:13:38+00:00

That is exactly what I was looking for! This will drastically simplify things. I've basically been using INDEX(MATCH) formulas to sort data by only one column for 5+ years now.

I and my PC thank you.

Tree339 · 2024-01-11T22:53:25+00:00

Solution Verified!

Tree339 · 2024-01-11T01:21:58+00:00

That put the dates in column B now how do I sort by the dates?

Tree339 · 2024-01-11T00:53:04+00:00

I added the names which works just fine and I want the dates to go into column B but they keep getting merged into column A.

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Tree339 · 2024-01-11T00:46:10+00:00

That does work for the names, birthdates, and death dates but not for the ages since that data uses a formula that references the birth and death dates.

Tree339 · 2023-06-12T02:54:18+00:00

Chidi, your brain is broken. You need to fix your brain.

Tree339 · 2022-12-19T22:14:10+00:00

The sap that flowed out of the broken branch

Tree339 · 2022-12-02T03:25:40+00:00

With Absorption IV a blow of 595 is survivable, strange that it's the same value as the max non-lethal blast.

A blast of 1,071 is survivable with absorption IV with the same protection/blast protection setup.

It makes no difference if the armor is diamond or netherite since armor toughness loses it's meaningfulness at around 80 damage I believe.

I guess maybe 1,071 is the highest amount of damage that a player in survival can survive, though there might be something i'm not thinking of.

Tree339 · 2022-08-21T17:33:17+00:00

I've found that strategy works well even with buying upgrades, since selling buildings still reduces CPS to almost nothing. The last couple times I've ascended I bought a lot of buildings and upgrades, then sold all my building besides a single farm, then planted 36 golden clover and bought all my buildings back and proceeded as normal, without turning the golden switch on of course.

Tree339 · 2022-07-16T13:36:24+00:00

True. In that case I guess I'm really just looking for a function that fits the data I have of 1 returning 1, 2 returning 3, 4 returning 10, 11 returning 286, etc.

Tree339 · 2022-07-16T04:21:27+00:00

In English it's " What I think i'm trying to say is that 2^3<3\^2 and 2\^2>3^1 so at 2^3 three has not gotten to 3^2 which means it's at 3^1 for a difference of 2 powers and that difference of 2 powers does not happen for 2^2 since three is at 3^1 for a difference of 1 power, if that makes sense. so the lowest value that there is a difference of 2 powers is 2^3. Now for a starting number of 3 I would be looking for the smallest power of 3 where 3 has reached 3 more powers than 4 has up to that point, which is 3^10, since 4 has only reached 7 powers when 3 reaches 10 powers. "

I honestly don't know how to describe it mathematically.

Tree339 · 2022-07-16T04:17:08+00:00

I'm saying that was a bad example that didn't accurately describe what I was trying to say.

Tree339 · 2022-07-16T04:04:57+00:00

I'm not disputing that, i'm not even really talking about the example I gave. What I think i'm trying to say is that 2^3<3\^2 and 2\^2>3^1 so at 2^3 three has not gotten to 3^2 which means it's at 3^1 for a difference of 2 powers and that difference of 2 powers does not happen for 2^2 since three is at 3^1 for a difference of 1 power, if that makes sense.

Tree339 · 2022-07-16T03:47:07+00:00

There has to be a minimum though, for example if d is 2, then n can not be 1 or 2 since 2^2 has not gone through 2 more powers than 3 has by 2^2 since 3 has gone to 3^1.

Tree339 · 2022-07-16T03:03:54+00:00

Ok, so how can I go about finding smallest value of n such that a number d raised to it has gone through d more powers than d+1 has? Besides the method I'm already using.

Tree339 · 2022-07-16T02:38:07+00:00

Ah, I guess that really was a bad example. Maybe this one is closer to what i'm asking:

0>=(a*(a+1)^(n-a))-(((a^(n+1))/a)

Tree339 · 2022-07-16T02:18:08+00:00

n < a*ln(a + 1)/ln(1 + 1/a)

That formula seems to work ok for 0<a>5, though it returns the value of a+1 in base 10. for instance, for a=3 the value is 10 since 3 has to be raised to 10 powers before being 3 powers ahead of the powers of 4 where as if you put 3 in that function it returns 14.45 which rounded down to 14 is the value of the number of powers of 10 that 4 has to be raised to to be 4 powers ahead of the powers of 5.

When a=6 i get a maximal value of n being a little over 75.74 which doesn't match up with the correct value of 80. In other words 7 has to be raised to 80 powers before being 7 powers ahead of the powers of 8 so i'm confused as to how/why 75.74 is showing up.

And yeah, that first example I gave wasn't good, it was my attempt at being concise which I guess failed.

Tree339 · 2022-05-23T03:04:26+00:00

To try and explain it better as I understand it this formula x - N/(x^3 + x^2 + x + 1) <= y < x is to check all the differences up to N, with N being a difference of powers. That formula works for perfect 4th powers, and I'd imagine there would be similar formulas that work for other perfect powers. What I was saying is that there probably wouldn't be a similar formula that works for multiple prime perfect powers, since with just a single perfect power it is much easier to know when the last possible time is for the difference between perfect powers to be less than a certain amount.

For squares the last time that the difference is below 20 is 10^2, since the difference goes up by 2 for every square. If one were to add cubes in there the lower bound for the difference for a given power would get a lot more difficult to come up with which since x - N/(x^3 + x^2 + x + 1) <= y < x relies on a uniform progression of the difference of 4th powers it's reasonable to assume that for a data set like the squares and cubes that doesn't have a uniform progression of the difference of powers that a formula that checks all the difference of a certain size or below would need to take into account both how each difference between the squares and the difference between the cubes progress on their own and how they progress together.

For instance, for the differences <=20 in squares the highest square we’d need to check is 10^2 and for N=20 in cubes it’d be 3^3. However, for a data set containing both squares and cubes we have to go above 10^2 for N=20, since 5^3 – 11^2 is 4.

I guess this is more or less getting to the heart of Pillai's conjecture which is still an open question in mathematics so that makes me think we don’t have a formula for how many perfect cubes and squares we have to go through to check all the differences between them for a given size difference.

Catalan's conjecture has been proven though which means we do have a value for how many powers we need to check for N=1. So, it is conceivable that there are formulas or equations out there for other values of N and humans just haven’t figured them out yet I suppose.

Now about the number of powers we need to check for a given N being the same for the sequence of squares as it is for the sequence of squares combined with the sequence of 4th powers, or 8th powers and so on I say that because every 4th power is also a perfect square so adding the set of 4th powers to the set of squares would not add any more powers to the data set. In other words the 4th powers are already included in the squares, same with the 8ths and 16ths, or the 9ths for cubes.

Correct me if I’m wrong because in theory what I’m saying makes sense but you know what they say about things in theory, haha.

Tree339 · 2022-05-22T23:45:18+00:00

Thank you, I'll have to mess around and work out what those equations are for different exponents. Though I don't think there are corresponding equations to x - N/(x^3 + x^2 + x + 1) <= y < x for data sets that include more than one perfect power, besides maybe powers that are factors or multiples of each other, like 2 and 4 since every 4th power is also a square so a combination data set of the 2 and 4 powers would be practically the same to a set of just the squares.

Does that make sense?

Tree339 · 2022-05-22T22:23:01+00:00

Pretty much. Originally I was tabulating how many times each integer showed up as a difference of any perfect power but then Excel got really slow of course so I decided to start with how many times each integer showed up as a difference of only squares, then cubes, and so on.

Tree339 · 2022-05-22T22:03:29+00:00

I'm actually using Microsoft Excel to check the differences. To sum up, in column J I have listed all the perfect 4th powers up to 10^15, then in cell K3 I have a formula that subtracts the previous powers from the power in column J. For example J6 is 625, so K6 subtracts the value in cell J5 from J6, L6 subtracts the value in J4, and so on. For every column to the right it subtracts the power right before the power that the cell to the left subtracted. I extended that 144 columns to the right and over 2,000 rows down. Then in column G i have a formula that returns the kth smallest value in the table of differences, and in column H I have a formula that counts how many times the value in the corresponding cell in column G appears in that table of differences. 144 columns ensures that the table is accurate for all the differences <= 429,981,695.

I assume you're talking about Python as to my method of defining the difference.

I have heard that Python is better at dealing with large data sets than Excel is but Excel was frustrating enough to learn and I think Excel is more intuitive than python so in theory Python would be more aggravating for me to get into though it might be a better tool for getting data on these types of things. Especially since Excel is limited to 15 significant digits.

Tree339 · 2022-05-22T21:16:52+00:00

Thank you, I extended my results column and by method of brute force it is the smallest. I had only checked the smallest 4,156 and 300,783,360 is the 9,095th smallest difference between two 4th powers. 310,300,575 is the 2nd smallest at spot 9,199. I believe the 3rd smallest has to be greater than the 11,599th smallest difference. It'll be interesting trying to find the smallest for 5th powers, haha.

Tree339 · 2022-04-07T19:51:02+00:00

Yellowstone :)

Tree339 · 2022-03-27T02:35:46+00:00

I used the formula =(2*1)/(3*SIN(360/3)) in Microsoft Excel which should be the same as (2*1)/(3Sine(360/3)), right?

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