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December 31

Generating a point on the Y axis from regular pentagon with point on X axis

For a $K$ consisting of points in R^2, define the function B such that $B(K)$ as the Union of $K$ and all points which can be produced in the following way. For each set of points A, B, C, & D from $K$ all different so that no three of A, B, C & D are co-linear. E is the point (if it exists) where ABE are colinear and CDE are co-linear.

If $K_{0}$ = the vertices of a regular Pentagon centered at 0,0 with one vertex at (1,0), does there exist N such that $B^{N}(K_{0})$ includes any point of the form (0, y)? (extending the question to any N-gon, with N odd) Naraht (talk) 05:16, 31 December 2024 (UTC)

I think you meant to write $B^{N}(K_{0}).$  --Lambiam 07:55, 31 December 2024 (UTC)

Changed to use the Math.Naraht (talk) 14:37, 31 December 2024 (UTC)

I'm not 100% sure I understand the problem, but try this: Label the vertices of the original pentagon, starting with (1, 0), as A, B, C, D, E. You can construct a second point on the x-axis as the intersection of BD and CE; call this A'. Similarly construct B', C', D', E', to get another, smaller, regular pentagon centered at the origin and with the opposite orientation from the the original pentagon. All the lines AA', BB', CC', DD', EE' intersect at the origin, so you can construct (0, 0) as the intersection of any pair of these lines. The question didn't say y could not be 0, so the answer is yes, with N=2.

There is some theory developed on "straightedge only construction", in particular the Poncelet–Steiner theorem, which states any construction possible with a compass and straightedge can be constructed with a straightedge alone if you are given a single circle with its center. In this case you're given a finite set of points instead of a circle, and I don't know if there is much theory developed for that. --RDBury (talk) 13:12, 1 January 2025 (UTC)

Here is an easy way to describe the construction of pentagon A'B'C'D'E'. The diagonals of pentagon ABCDE form a pentagram. The smaller pentagon is obtained by removing the five pointy protrusions of this pentagram.  --Lambiam 16:53, 1 January 2025 (UTC)

Here is one point other than the origin (in red)

If there is one such point, there must be an infinite number of them. catslash (talk) 22:52, 2 January 2025 (UTC)

Just to be clear, the black points are the original pentagon K, the green points are in B(K), and the red point is the desired point in B²(K); the origin is not shown. It would be nice to find some algebraic criterion for a point to be constructible in this way, similar to the way points constructible with a compass and straightedge are characterized by their degree over Q. --RDBury (talk) 01:45, 3 January 2025 (UTC)

Once you have a second one (such as the reflection of the red point wrt the x-axis), you have all intersections of the y-axis with the non-vertical lines through pairs of distinct points from $\textstyle {\bigcup _{n}B^{n}(K_{0})}.$  --Lambiam 16:22, 3 January 2025 (UTC)

RDBury *headslap* on (0,0) Any idea on y<>0? (← comment from Naraht)

See the above construction by catslash.  --Lambiam 16:12, 3 January 2025 (UTC)

The red point is at $x=0$, $y=-{\sqrt {\frac {5-{\sqrt {5}}}{10}}}$. catslash (talk) 16:18, 3 January 2025 (UTC)

January 1

What is the first number not contained in M136279841?

See (sequence A268068 in the OEIS), the first number not contained in M74207281 is 1000003, but what is What is the first number not contained in M136279841 (the currently largest known prime)? 61.224.131.231 (talk) 03:34, 1 January 2025 (UTC)

The corresponding sequence (11, 3, 8, 7, 6, 10, 4, 9, 1, 5, 25, 31, 39, ...) is not in OEIS. Finding the answer to your question requires an inordinate amount of computing power. The decimal expansion of this Mersenne prime has some 41 million digits, all of which need to be computed. If this is to be done in a reasonable amount of time, the computation will need the random access storage of at least some 22 million digits.  --Lambiam 10:10, 1 January 2025 (UTC)

I'm not seeing that this question requires an inordinate amount of computing power to answer. 41 million characters is not a very large set of data. Almost all modern computers have several gigabytes of memory, so 41 million characters will easily fit in memory. I took the digits of M136279841 from https://www.mersenne.org/primes/digits/M136279841.zip and searched them myself, which took a few minutes on a consumer grade PC. If I have not made a mistake, the first number that does not appear is 1000030. The next few numbers that do not appear are 1000073, 1000107, 1000143, 1000156, 1000219, 1000232, 1000236, 1000329, 1000393, 1000431, 1000458, 1000489, 1000511, 1000514, 1000520, 1000529, etc. CodeTalker (talk) 03:59, 2 January 2025 (UTC)

To be fair, this depends on being able to find the digits on-line. To compute them from scratch just for this question would be more trouble than it's worth. But I take your point; it probably takes more computing power to stream an episode of NUMB3RS than to answer this question. My problem with the question is that it's basically a dead end; knowing the answer, is anyone going to learn anything useful from it? I'd question the inclusion of A268068 in OEIS in the first place simply because it might lead to this sort of boondoggle. But far be it for me to second guess the OEIS criteria for entry. --RDBury (talk) 01:13, 3 January 2025 (UTC)

OEIS includes similar sequences for the positions of the first location of the successive naturals in the decimal expansions of $e$ (A088576), $\pi$ (A032445) and $\gamma$ (A229192). These have at least a semblance of theoretical interest wafting over from the open question whether these numbers are normal.  --Lambiam 06:21, 3 January 2025 (UTC)

To compute them from scratch just for this question would be more trouble than it's worth.
Eh, I agree that the question is of little fundamental interest. However, it's not much work to compute M136279841. It is of course absolutely trivial to compute it as a binary number. The only real work is to convert it to decimal. I wrote a program to do this using the GNU Multiple Precision Arithmetic Library. It took about 5 minutes to write the program (since I've never used that library before and had to read the manual) and 29 seconds to run it. CodeTalker (talk) 18:06, 3 January 2025 (UTC)

Right, convert from binary, somehow I didn't think of that. Basically just divide by 10 41 million times, which would only be an issue if it was billions instead of millions. --RDBury (talk) 06:21, 4 January 2025 (UTC)

January 5

Reference request:coherence condition (adjoint functor)

Previously, in WPM (Coecke and Moore (2000)) taught me a statement of coherence condition for adjoint functors. This is sometimes called triangle identities or zigzag identities, and I'm looking for some references. I am also trying to find where to find the Wikipedia article that explains coherence conditions (adjoint functors). Also, I'm looking for a Wikipedia article that explains coherence conditions (adjoint functors). (e.g. coherence condition, adjoint functor, or new draft ?)

• "triangle identity". ncatlab.org.
• Ben-Moshe, Shay (2024). "Naturality of the ∞-categorical enriched Yoneda embedding". Journal of Pure and Applied Algebra. 228 (6). arXiv:2301.00601. doi:10.1016/j.jpaa.2024.107625.
• Borceux, Francis (1994). Handbook of Categorical Algebra: Basic category theory. Cambridge University Press. ISBN 978-0-521-44178-0.
• Coecke, Bob; Moore, David (2000). "Operational Galois adjunctions". arXiv:quant-ph/0008021.
• planetmath

SilverMatsu (talk) 02:41, 5 January 2025 (UTC)

Can you identify more precisely for which statement(s) you want a reference? Is it for the definition of counit–unit adjunction given in section Adjoint functors § Definition via counit–unit adjunction?

Explanations in mathematics can be of different kinds. One kind are explanations of definitions. Definitions are true by definition; an explanation generally means helping build up an intuition of the concept defined by showing familiar structures satisfying the definition. Another kind are explanations of statements. These typically offer a reformulation of a statements as an equivalent statement in terms of simpler concepts. Then there are explanations of proofs. These can include showing the proof "in action" on specific examples, using familiar structures as when explaining definitions. And they can assist in verifying the validity of proof steps, by unfolding definitions until the step becomes obvious. In general, all explanations can involve a combination of these.

For example, in our article Adjoint functors, the ramifications of the definition in the lead section, basically the existence of a bijection $\mathrm {hom} _{\mathcal {C}}(Fd,c)\cong \mathrm {hom} _{\mathcal {D}}(d,Gc)$ that is natural in $c$ and $d,$ are not immediately obvious, but by carefully unfolding the definitions, starting with the required naturality of the two morphisms making up the bijection, leads to the equivalent counit–unit definition.

I am not sure what kind of explanation of "coherence conditions (adjoint functors)" you are seeking. Perhaps studying this article, "Adjunctions", will answer this question.  --Lambiam 12:18, 5 January 2025 (UTC)

Thanks for the reply. I will read the article you recommended and study from it. Specifically, I am looking for some references to add the examples given in the WPM to the examples of coherence condition. Also, I think counit–unit addition is itself might be a topic on which to create a standalone article. SilverMatsu (talk) 05:30, 6 January 2025 (UTC)

My immediate reaction is that it is better to leave the treatment of the counit–unit definition to the article Adjoint functors. The terms "hom-set adjunction" and "counit–unit adjunction" suggest these are special kinds of adjunction, but this terminology is IMO misleading. What is defined there is each time precisely the same concept of adjunction as in the other definitions. All three definitions are fully equivalent. They are just different ways of looking at the same situation. The parable of the blind men and an elephant could be the story of the mathematicians and the adjoint situation.  --Lambiam 11:43, 6 January 2025 (UTC)

Concatenation of first 10 digits and last 10 digits of a number

Let a(n) be the concatenation of first 10 digits and last 10 digits of n, then we know that a(2^136279841-1) = 88169432759486871551, however:

1. Can a(2^n) take all 20-digit values which are multiples of 1024?
2. Can a(3^n) take all 20-digit values which are odd?
3. Can a(n^2) take all 20-digit values which end with 0, 1, 4, 5, 6, 9?
4. Can a(n^3) take all 20-digit values?
5. Can a(prime number) take all 20-digit values which end with 1, 3, 7, 9?
6. Can a(lucky number) take all 20-digit values which are odd?
7. Can a(Fibonacci number) take all 20-digit values?
8. Can a(partition number) take all 20-digit values?
9. What is a(9^9^9^9)?
10. What is a(9^^9), where ^^ is tetration?
11. What is a(9^^^9), where ^^^ is pentation?
12. What is a(Graham's number)?
13. What is a(TREE(3))?
14. If we know a(x), assume that x has at least 20 digits, can we also know a(2*x), a(3*x), etc.?
15. If we know a(x), assume that x has at least 20 digits, can we also know a(x^2), a(x^3), etc.?
16. If we know a(x) and a(y) as well as the number of digits of x and y, can we also know a(x+y)?
17. If we know a(x) and a(y) as well as the number of digits of x and y, can we also know a(x*y)?
18. If we know a(x) and a(y) as well as the number of digits of x and y, can we also know a(x^y)?

220.132.216.52 (talk) 08:33, 5 January 2025 (UTC)

The answer to 2 is easily "no". Powers of 3 are not divisible by 5, so their decimal representations cannot end on a 5. Five seconds of considering the question should suffice to figure this out.

The answer to 14ff is also no:

Put $x=111111111102222222222$ and $y=111111111152222222222.$

Then $a(x)=a(y),$ but $a(2x)\neq a(2y).$

Again, I think a beginning recreational mathematician should be able on their own to come up with this easy counterexample.

I do not find these particular questions interesting. Still, let me reflect on how questions 1–8 might be approached.

To start, the formation of a $20$-digit string is a red herring. It is conceptually cleaner to consider pairs of $10$-digit strings, such as (8816943275, 9486871551).

One can pose two related questions for each sequence under consideration:

1. Ini_$10$: does each $10$-digit string appear as an initial segment of the decimal expansion of some sequence element?
2. Fin_$10$: does each $10$-digit string, with the exclusion of some as specified, appear as a final segment of its decimal expansion?

These are necessary but not sufficient conditions for the joint occurrence of all combinations, but easier to study. If either one can be shown not to hold, the answer is no. Otherwise, if no proof of any of the two can be found, the question cannot be settled. Finally, if both can be shown to hold, it is at least plausible that the proofs will provide enough material to settle the original question.

As to the Ini question, it is unlikely that an affirmative answer depends on the particular number $10$. If Ini_$10$ holds for a given well-known mathematical sequence, Ini_$L$ will almost certainly hold for any length $L.$ This is then equivalent to the statement that a given sequence $(s_{n})_{n}$ satisfies $s_{n{+}1}{-}s_{n}=o(s_{n}).$ (For the notation $o(\cdot ),$ see Little-o notation.)

Similarly, if Fin_$10$ holds for a given well-known mathematical sequence, Fin_$L$ will plausibly hold for any length $L.$ The Fin question can therefore tentatively be generalized to:

Are, for any given modulus $m,$ the residue classes modulo $m$ of sequence $(s_{n})_{n}$ all classes not already excluded by the specified exclusions?

Note, though, that the generalization may fail to hold for some modulus while Fin_$10$ holds. For example, $n^{3}\not \equiv 2~~{\text{mod}}~4$ for any $n.$ So it is no longer a necessary condition and may need further finessing before application.  --Lambiam 00:28, 6 January 2025 (UTC)

Sorry, the first 4 questions should be:

1. Can a(2^n) take all 20-digit values which are multiples of 1024 and not multiples of 5?
2. Can a(3^n) take all 20-digit values which end with 1, 3, 7, 9?
3. Can a(n^2) take all 20-digit values which are quadratic residues mod 10¹⁰?
4. Can a(n^3) take all 20-digit values which are neither == 2, 4, 6 mod 8 nor == 5, 10, 15, 20, 25, 30, …, 115, 120 mod 125?

220.132.216.52 (talk) 01:14, 6 January 2025 (UTC)

12. It seems no one knows what the first digit of Graham's number is, let alone the first 10.

13. Similar situation.

GalacticShoe (talk) 15:53, 6 January 2025 (UTC)

9./10./11. I would wager that finding these would be extremely difficult. Normally if one has a large number $A$, then one can hope to find the first few digits by taking $10^{frac(\log _{10}(A))}$, with the hopes that $\log _{10}(A)$ will be small/amenable to calculation, with a precise fractional part. For example, $9^{9^{9}}$ has log $9^{9}\log _{10}(9)$, which is small ($369693099.63157035\ldots$) and thus has a precise fractional part ($0.63157035\ldots$) which, when fed back into the power of 10, yields us the first few digits $4281247731\ldots$ without having to calculate the 369 million-digit $9^{9^{9}}$ itself. Unfortunately, $9^{9^{9^{9}}}$ is not such a case; the log itself now has 369 million digits, so calculating the fractional part accurately would be a nightmare. For similar reasons, 10. and 11. would also seem to be extraordinarily difficult. GalacticShoe (talk) 00:47, 7 January 2025 (UTC)

1./2. All numbers that are multiples of 1024 and not of 5 are the last 10 digits of some power of 2, as for all $a$ coprime to 5, there is some $b$ such that $2^{b}\equiv a\!\!\!{\pmod {5^{10}}}$. Moreover, for $n\geq 10$ the modulos with respect to $10^{10}$ are periodic with some minimal period, say $Y$, so that all applicable modulos appear once in the period of $Y$ consecutive powers of 2.

Suppose now that we fix two distinct 10-digit strings $A$ and $B$, the former to represent the first ten digits we want of our desired power of 2, and the latte - divisible by 1024 and not by 5 - to represent the last ten digits. Let $X$ be the minimal power of $2$ that has $B$ as its last ten digits; all the numbers we are looking for are then of the form $2^{X+nY}$ where $n\geq 0$. What it means for our first ten digits to be $A$ then, is that there is some $k\geq 10$ such that:

$A\leq {\frac {2^{X+nY}}{10^{k}}}<A+1$.

This is equivalent to:

$k+\log _{10}(A)-X\log _{10}(2)\leq nY\log _{10}(2)<k+\log _{10}(A+1)-X\log _{10}(2)$.

Notice that $Y\log _{10}(2)$ is irrational. It is a bit overkill, but we can apply the equidistribution theorem here and note that there will always be some value of $n$ with a corresponding value of $k$ large enough such that the aforementioned equality holds, and also $k+\log _{10}(A)-X\log _{10}(2)>0$. Consequently, there must be some values of $k$ and $n$ that yield a power of 2 starting with A and ending with B. The same logic applies to $3^{n}$. GalacticShoe (talk) 00:26, 7 January 2025 (UTC)

3./4. If we look at squares, then we have $(X+nY)^{2}$ instead of $2^{X+nY}$, and the inequality becomes

$k+\log _{10}(A)\leq 2\log _{10}(X+nY)<k+\log _{10}(A+1)$.

While logs are not equidistributed modulo 1, this does not present a problem; because $2\log _{10}(X+(n+1)Y)-2\log _{10}(X+nY)$ tends to 0 as n goes to infinity, there will always be some point where, informally, $2\log _{10}(X+nY)$ cannot "jump the gap" between $k+\log _{10}(A)$ and $k+\log _{10}(A+1)$. When that happens, we get a square that starts and ends with our desired string. The same logic applies to cubes and all functions $A$ (yielding the desired ending substring) where $\lim _{n\rightarrow \infty }\log(A(n+1))-\log(A(n))=0$ or equivalently $\lim _{n\rightarrow \infty }A(n+1)/A(n)=1$. GalacticShoe (talk) 01:37, 7 January 2025 (UTC)

7. The approach with exponentials can actually help us with Fibonacci numbers, as

$F(n)={\frac {\phi ^{n}-(1-\phi )^{n}}{\sqrt {5}}}\approx {\frac {\phi ^{n}}{\sqrt {5}}}$.

All we have to do is use the Pisano period and add a fixed amount of leeway for the minuscule $(1-\phi )^{n}$ (which tends to 0, so any tiny threshold suffices), at which point the equidistribution theorem again applies. GalacticShoe (talk) 01:51, 7 January 2025 (UTC)

5. Based on our previous conversation, yes.

6. If our previous conversation on lucky numbers is anything to go by, it seems unlikely that enough is known to even ascertain whether there are infinitely many lucky numbers satisfying congruence mod 10^10.

8. Like the last question involving partition functions, and like with lucky numbers, without further information on their congruences it seems unlikely that we'll have a satisfactory answer to this.

GalacticShoe (talk) 02:19, 7 January 2025 (UTC)

To summarize:

1, 2, 3, 4, 5, 7: yes

6, 8: unclear

14, 15, 16, 17, 18: no

9, 10, 11, 12, 13: too large to calculate

GalacticShoe (talk) 02:22, 7 January 2025 (UTC)

For polynomials you can use $P(n{+}1){-}P(n)=o(P(n)).$  --Lambiam 02:39, 7 January 2025 (UTC)

So for 9 to 13, the first 10 digits are not known, but are the last 10 digits known? 220.132.216.52 (talk) 04:22, 7 January 2025 (UTC)

I found the answer of 9, it is 21419832941045865289, see (sequence A243913 in the OEIS). 220.132.216.52 (talk) 05:43, 7 January 2025 (UTC)

Another question: Are a(2^n) and a(3^n) periodic sequences (or eventually periodic)? 220.132.216.52 (talk) 05:45, 7 January 2025 (UTC)

No, this is excluded by the equidistribution theorem. Precisely because $(\{n\log _{10}2\})_{n}$ is dense in $[0,1),$ there will be pseudoperiodicities. For example, $\log _{10}2=0.301029...$ is approximated by $28/93=0.301075...,$, so there will be a similarity between the initial digits of $2^{n},2^{n{+}1},2^{n{+}2},...$ and those of $2^{n{+}93},2^{n{+}94},2^{n{+}95},...\,.$ The first two digits of $2^{27},2^{28},2^{29},...$ go like

$13,26,53,10,21,42,85,17,34,68,13,27,54,10,21,43,87,17,35,...,$

while those of $2^{120},2^{121},2^{122},...$ go like

$13,26,53,10,21,42,85,17,34,68,13,27,54,10,21,43,87,17,34,...\,.$

Eventually there will be arbitrarily long initial strings repeated arbitrarily often, but each run will eventually end with a discrepancy.  --Lambiam 12:38, 7 January 2025 (UTC)