10,964 Pages

## On last digits of tetrations

Could you please give us a source of the last digits of tetrations? If they are your own results, could you tell us explicit algorithms which gave the results? Since the original description to compute the last digits in the article was wrong, I am afraid that you added wrong estimations following the article. Although I reverted your contributions, please rewrite them if you have sources available. Thanks.

p-adic 05:01, December 18, 2019 (UTC) I still don't really get why the recursive modular exponentiation method shouldn't work in base 10. I have verified that it indeed doesn't work in most other numeral bases (such as 7, 11, 13, or 19),  but using Mathematica, PowerMod[3, 5597484987, 10^10] gives the same result (6100739387) as PowerMod[3, 7625597484987, 10^10], and so the last digits of 3↑↑4 are indeed ...6100739387.  Similarly, 3 to the power of any number ending in ...6100739387 ends in ...9660355387, thus those are indeed the last 10 digits of 3↑↑5. I could keep going and show that 3↑↑↑3, 3↑↑↑↑3, and even Graham's number do indeed end in ...2464195387. Allam948736 (talk) 19:53, December 30, 2019 (UTC)

I have since devised another method for computing the last digits of $$^yx$$ in base b which should work across all bases, not just decimal:

$$s_1 = x$$

$$s_{n+1} = (x^{s_n} \mod lcm(\phi(b^d), b^d)$$

Continue until you reach $$s_y$$, then take the final value mod $$b^d). The reason why this should work is because the last \(d$$ base-b digits of $$x^y$$ can repeat after no more than $$\phi(b^d)$$, and if they repeat sooner, the period is always a divisor of $$\phi(b^d)$$. Taking the least common multiple preserves the result of the exponentiation mod both $$b^d$$ and $$\phi(b^d)$$. Using this, I obtain ...98,615,075,353,432,948,736 as the last 20 digits of $$2 \uparrow\uparrow 22$$ and beyond, for instance. I still can't prove that this method produces the same results for b=10 as the old method that was believed to be correct on here up to 3 months ago, however. Allam(2^^n mod 10^6 for n >= 8) (talk) 14:02, March 11, 2020 (UTC)

Oh... Have you forgotten that your proof here does not make sense...?
p-adic 14:57, March 11, 2020 (UTC)

I fixed most of the issues, but still have to figure out how the last equality relates to the first one (the a in the proof is equivalent to the x in the initial statement, the b in the proof is equivalent to the y, the c in the proof is equivalent to the b in the initial statement, and d is unchanged). Allam(2^^n mod 10^6 for n >= 8) (talk) 16:32, March 11, 2020 (UTC)

> I fixed most of the issues
I do not think so, because you always ignore comments without checking most of them. If you really think so, reply the comment and list up the numbers of the issues which you actually checked. Why do you always ignore comments? You might guess that you fully completely understand them, but without replying to them, you could not verify how you got them. If you were so smart that you never misunderstood issues, then it would not be problematic. The fact is that you always made trivial mistakes. Then it is reasonable to reply to feedbacks.
> but still have to figure out how the last equality relates to the first one
Then how could you be confident that your statement was correct? Without solving all issues, you cannot complete the proof. You always check portions of issues and state "I corrected" without clarifying what issues you actually corrected. As long as there are remaining issues, you have not really understood the problem. List up all issues which you actually solved. Do you force readers who pointed out issues to check what issues are solved without your explanation? It is waste of time. Since your "correction" is frequently wrong, you chould clarify what issues you "intend" to complete to solve.
p-adic 21:45, March 11, 2020 (UTC)

I updated my user page with what I was really trying to say in my December through March arguments. Allam(2^^n mod 10^6 for n >= 8) (talk) 02:21, June 10, 2020 (UTC)

See my latest blog post about patterns in last digits, which I haven't yet finished but it covers this matter. Allam(2^^n mod 10^6 for n >= 8) (talk) 00:22, October 2, 2020 (UTC)

Although I have not checked it, does it contain any new point of view which is not included in my explanation using Chinese remainder theorem? (If not, then you can just immitate what I wrote as if you knew it from the beginning although the one who fully explained how to compute last digits in a precise manner is I but not you.)

## On first and last digits

As I wrote above, you are adding your own results to many articles without external sources including a reproducible description on how you computed them. Your own site is not an evidence of your own results, because you have no actual written method to generate them. Please notice that even if you are certain that your results are completely correct, what you are doing is the same as adding random digits to articles for others because you lack descriptions of reproducible methods. If you do not add valid external sources, then I will start to delete all your contributions.

p-adic 03:14, December 27, 2019 (UTC)

I calculated the first digits of 3^3^100 by simply taking the log10 of 3 and multiplying by 3^100, and then taking 10 to the power of the fractional part, and the last digits by using simple modular exponentiation techniques. Which external sources are valid, exactly? Do I just have to add a calculation method to my own site's article, or do I have to search the Web for another source (which there may not be in some cases)? Allam948736 (talk) 03:31, December 27, 2019 (UTC)

Thank you for adding the explanations for $$2^{2^{100}}$$. It is sufficient for me. Also, your site is valid for me, as long as you explicitly declare how you computed (in the article or your site). On the other hand, for example, something like "the last digits of tetration can be computed by modular exponentiation" is a bad explanation, because they are not easily computed in that way for higher torwers of exponentiations unlike lower towers such as $$2^{2^{100}}$$ and $$3^{3^{100}}$$. Could you tell me how you computed the last digits of tetration? As I commented above, your description here is not sufficient for others.
p-adic 04:12, December 27, 2019 (UTC)

@p-bot

This is not really related to the topic at hand, but I think you're being a tad too insistent

Many artciles for numbers contain approximations in various other notations. Most of these approximations are not cited anywhere and there is nothing on the page that justifies the approximation.

Do you think all those approximations should be deleted because of "no source"?

Username5243 (talk) 23:49, December 30, 2019 (UTC)

Usually I do not care about such approximations, because if there is a wrong information, then we can ask the original authors the reasons. But if nobody can explain them in a reasonable way, then we should delete them. I am not requiring a full proof, but just requiring an explanation.
Allam's results are such ones. I asked Allam the precise methods so many times, but somewhy he or she hesitated to fix an answer. Moreover, Allam tried to shift goalposts when I pointed out incorrect logic in his or her "explanation". Then isn't it good for us to doubt that Allam executed a meaningful algorithm?
Also, I added counterexamples for the wrong methods in the original articles on Graham's number and tetration, which Allam believed to be correct. What should I do in addition? Are you requiring me to write a proof that Allam's secret wrong method occasionally outputs correct answers?
p-adic 00:45, December 31, 2019 (UTC)
It was fine for the first digits of exponential factorials (where the last digits were recently removed). Also, why were the first and last digits of triangoogol and triangrolplex removed? Those were included in the source given (which is my site's article where I defined the googolisms). For the last time, the first digits are always computed using logarithms, and the last digits using modular arithmetic.  And no, I didn't "hesitate", because why would I keep a method for calculating numbers of all things secret?  Also, those "counterexamples" themselves contain incorrect logic. First of all, the recursive modular exponentiation method in base 10 fails for just one digit (ex. 2^...6 = ...4 if ...6 is preceded by an even digit, but 2^...6 = ...6 if the first ...6 is preceded by an odd digit), but works for more digits (ex. 2^...36 = ...36 regardless of the digits before ...36 in the exponent). Second, "3^g64 - g64 is divisible by 10^x for any x" implies that we thought the recursive modular exponentiation method (that is, take a^b mod 10^d, then substitute that result for b, then repeat) would continue to work forever (which it doesn't, but in the case of Graham's number the number of digits you would have to go to get to that point is itself extremely large). Allam948736 (talk) 01:10, December 31, 2019 (UTC)
> triangoogol and triangrolplex
It is because the source was referred to at the first sentence. If the computations are also mentioned in the sources, i.e. your website, then it is my mistake. I will revert the two articles, and add the references also at the end of those computations.
> For the last time, the first digits are always computed using logarithms, and the last digits using modular arithmetic.
As I wrote above, "moduar exponentiation" or something like that is too ambiguous. For example, the wrong description in the article is also a modular exponentiation. Therefore I am requiring to add precise explanations of the methods which you used.
> And no, I didn't "hesitate", because why would I keep a method for calculating numbers of all things secret?
Then why haven't you ever written down explicit methods? I said that "modular exponentiation" is not a full reproducible description, but you have never described further details. Please add precise methods to each article.
> Also, those "counterexamples" themselves contain incorrect logic.
As I have already pointed out, your statement that my logic is incorrect is incorrect. Should I repeat the explanation? I showed a counterexample for the method in the original article, which has no restriction on the length of digits as it says "The last d digits of {}^y x in base b is defined by the following recursive formula:". I am not talking about your updated statement after you read my counterexamples.
Also, even if you persist to state that it works for more digits d, you are wrong. Say, consider (x,y,b) = (10^d,2,b). I guess that you will still update your statement by saying "No, I am always considering the case where x is coprime with b" or something like that.
> Second, "3^g64 - g64 is divisible by 10^x for any x" implies that we thought the recursive modular exponentiation method
You might think so. It does not mean that others made the same mistake.
p-adic 01:34, December 31, 2019 (UTC)

## Higher Order Veblen Defintion

I commented an idea for a definition on your "Ordinal Questions" blog page C7X (talk) 14:50, March 6, 2020 (UTC)

$$\varphi_0(0) = 1$$

$$\varphi_0(\alpha) = \omega^\alpha$$

$$\varphi_\gamma(\alpha, \beta)$$ = the limit of $$\varphi_\gamma(\alpha-1, \varphi_\gamma(\alpha, \beta-1)+1$$, $$\varphi_\gamma(\alpha-1, \varphi_\gamma(\alpha-1, \varphi_\gamma(\alpha, \beta-1)+1)$$, $$\varphi_\gamma(\alpha-1, \varphi_\gamma(\alpha-1, \varphi_\gamma(\alpha-1, \varphi_\gamma(\alpha, \beta-1)+1))$$...for successor ordinals $$\alpha$$ and $$\beta$$, or $$\sup\{\varphi_1(\alpha, \delta\}$$ for all $$\delta \lt \beta$$ for limit $$\beta$$

$$\varphi_\gamma(\alpha, 0)$$ = $$\sup\{\varphi_\gamma(\delta, 0)\}$$ for all $$\delta \lt \alpha$$ for limit $$\alpha$$, or the first fixed point of $$\varphi_\gamma(\alpha-1, \beta)$$ for successor $$\alpha$$

$$\varphi_\gamma(\alpha, \beta+1)$$ = $$\sup\{\varphi_\gamma(\delta, \varphi_\gamma(\alpha, \beta)+1)\}$$ for all $$\delta \lt \alpha$$ for limit $$\alpha$$

$$\varphi_\delta(\alpha, \beta+1, \gamma+1)$$ = the limit of $$\varphi_\delta(\alpha, \beta, \varphi_\delta(\alpha, \beta+1, \gamma)+1), (\varphi_\delta(\alpha, \beta, \varphi_\delta(\alpha, \beta, \varphi(\alpha, \beta+1, \gamma)+1)), \dots$$

$$\varphi_{\alpha+1}(\beta)$$ = $$\varphi_\alpha(^1_\beta)$$

$$\varphi_\alpha(0)$$ = $$\sup\{\varphi_\delta(0)\}$$ for all $$\delta \lt \alpha$$ iff $$\alpha$$ is a limit ordinal

$$\varphi_\beta(\alpha+1)$$ = the limit of $$\varphi_\delta(\varphi_\beta(\alpha)+1)$$ for all $$\delta \lt \beta$$ for limit $$\beta$$

$$\varphi_\alpha(\gamma+1, 0, 0)$$ = the first fixed point of $$\beta \mapsto \varphi_\alpha(\gamma, \beta, 0)$$

$$\varphi_\alpha(\gamma+1, 0, 0, 0)$$ = the first fixed point of $$\beta \mapsto \varphi_\alpha(\gamma, \beta, 0, 0)$$

$$\varphi_\alpha(\beta, \gamma, \delta+1, 0)$$ = the first fixed point of $$\zeta \mapsto \varphi_\alpha(\beta, \gamma, \delta, \zeta)$$

$$\varphi_\alpha(^1_{\gamma+1})$$ = the first fixed point of $$\beta \mapsto \varphi_\alpha(^\beta_\gamma)$$ for countable ordinals $$\alpha$$ and $$\gamma$$

$$\varphi_\alpha(^1_\beta)$$ = $$\sup\{\varphi_\alpha(^1_{\delta})\}$$ for all $$\delta \lt \beta$$ for limit $$\beta$$

$$\varphi_\alpha(0)$$ = $$\sup\{\varphi_\delta(0)\}$$ for all $$\delta \lt \alpha$$ if and only if $$\alpha$$ is a limit ordinal

$$\varphi_\alpha(\beta+1)$$ = $$\sup\{\varphi_\delta(\varphi_\alpha(\beta)+1)\}$$ iff $$\alpha$$ is a limit ordinal

$$\varphi_\alpha(\beta+1_{\gamma+1}, 0_\gamma, ..., 0, 0)$$ = the first fixed point of $$\delta \mapsto \varphi_\alpha(\beta_{\gamma+1}, \delta_\gamma, ..., 0, 0)$$ (here the subscript denotes the position of the arguments)

$$\varphi_\delta(\alpha, \beta, \dots\dots, \zeta+1, \eta+1)$$ = the limit of $$\varphi_\delta(\alpha, \beta, \dots\dots, \zeta, (\varphi_\delta(\alpha, \beta, \dots\dots, \zeta+1, \eta)+1)), \varphi_\delta(\alpha, \beta, \dots\dots, \zeta, \varphi_\delta(\alpha, \beta, \dots\dots, \zeta, \varphi_\delta(\alpha, \beta, \dots\dots, \zeta+1, \eta)+1)))\dots$$ (an arbitrary number of arguments)

The first fixed point of $$\alpha \mapsto \varphi_\alpha(0)$$ using these functions is probably $$\psi(\Omega_\omega)$$.

Whenever I use $$n$$, it indicates a natural number (that is, $$\lt \omega$$). The domain of the function is the countable ordinals, meaning that all variables in the definition represent countable ordinals. In general, each argument whose position is a successor ordinal enumerates fixed points of the argument immediately to the right, just as in the regular Veblen function. $$\varphi_0$$ is just the regular Veblen function. Note that these are not to be confused with the notation sometimes used as a shorthand for the 2-argument Veblen function. Note that (countably) infinitely many 0s can be added at the beginning without affecting the value of the expression.

Allam(2^^n mod 10^6 for n >= 8) (talk) 15:19, March 6, 2020 (UTC)

Continuing the discussion from Talk:Bachmann-Howard ordinal:  No, I didn't try to use $$\omega-1$$ and $$\Omega-1$$ in my definition. If there is an $$\alpha-1$$ in my definition, then I am assuming $$\alpha$$ is a successor ordinal. Allam(2^^n mod 10^6 for n >= 8) (talk) 23:35, March 6, 2020 (UTC)
Quantify all variables which you use by clarifying something like "for any successor ordinal α" or something like that. Then you will soon understand that after the restriction above, φ_1(ω,ω) and φ_1(Ω,Ω) are ill-defined because there is no rule to characterise them.
• Quantify all variables which you want to use.
• Clarify the domain of all functions which you want to define.
• Write down the precise definition applicable to any case.
Do you understand?
p-adic 23:41, March 6, 2020 (UTC)

I have updated my definition to include what to do if an argument is a limit ordinal, and clarification for the expressions with $$\alpha-1$$. Allam(2^^n mod 10^6 for n >= 8) (talk) 02:04, March 7, 2020 (UTC)
Read my comment above.
p-adic 03:35, March 7, 2020 (UTC)
I have further updated my definition to include the case of the 2-argument function where the second argument is a limit ordinal, so now phi_1(w, w) should be the limit of phi_1(w, 0), phi_1(w, 1), phi_1(w, 2), phi_1(w, 3), ... Allam(2^^n mod 10^6 for n >= 8) (talk) 14:53, March 9, 2020 (UTC)
Read my comment above. Write down the precise definition applicable to any case. All of α[n] for a limit α, φ_α(2,0)[n], and so on are ill-defined.
p-adic 15:02, March 9, 2020 (UTC)
I've generalized the definition, removing the examples that used $$\varphi_1$$ and replacing the definition of $$\varphi_\omega$$ with a general definition applicable to any limit ordinal. Allam(2^^n mod 10^6 for n >= 8) (talk) 22:31, March 9, 2020 (UTC)

Now that I have fixed all the errors in my definition (such as the direct invocation of fundamental sequences), what does $$\varphi_4(0) = \varphi_3(1) = \varphi_2(1, 0)$$ evaluate to? $$\varphi_\omega(0)$$? $$\varphi_\Omega(0)$$ (here the $$\Omega$$ indicates the first fixed point) Allam(2^^n mod 10^6 for n >= 8) (talk) 00:27, March 12, 2020 (UTC)

No. It is just ill-defined. Why do you always ignore the issues which have already been pointed out? It does not have a rule applicable to any cases. (By the way, is 0 a limit ordinal in your convention?)
p-adic 01:42, March 12, 2020 (UTC)
If you had pointed out exactly which lines in the definition have errors, I would have probably fixed them by now. Allam(2^^n mod 10^6 for n >= 8) (talk) 02:32, March 12, 2020 (UTC)
If you had replied to all feedbacks for you from others, then you would have improved yourself. Why could you say such a thing by ignoring my question above on 0?
> In general, each argument enumerates fixed points of the argument immediately to the right, just as in the regular Veblen function.
Please write down what you intend in mathematical formulae. Since you were not even confident on the values of Veblen function, it is unreasonable for others to guess what you really intend. This is too ambiguous. If I point out an error, then you can say "As I said, this is defined similarly to Veblen. Therefore your correction is wrong." or something like that. Please fix the precise definition.
p-adic 03:13, March 12, 2020 (UTC)
I added a definition for an arbitrary number of arguments. Also, 0 is neither a limit or successor ordinal. Allam(2^^n mod 10^6 for n >= 8) (talk) 14:48, March 12, 2020 (UTC)
Then it is ill-defined because there is no rule applicable to φ_0(0). Also, delete multiply-defined cases. Also, "an arbitrary number of arguments" is ambiguous. Is it a natural number greater than or equal to 4? Or are you considering transfinite numbers?
p-adic 15:22, March 12, 2020 (UTC)

phi_0(0) is just $$\varphi(0)$$ in the regular Veblen function which is just 1. Allam(2^^n mod 10^6 for n >= 8) (talk) 15:29, March 12, 2020 (UTC)

No. There is no rule to deduce it. Also, why do you ignore two other comments on multiply-defined cases and "an arbitrary number of arguments?
p-adic 15:44, March 12, 2020 (UTC)
The "arbitrary number of arguments" rule applies to natural numbers >= 4 and transfinitely many arguments. Allam(2^^n mod 10^6 for n >= 8) (talk) 15:49, March 12, 2020 (UTC)
Then φ_1(1,1,1) ill-defined. There are many other ill-defined values. I am not interested in pointing out all ill-defined values. Could you find all of them? If you cannot, then it just means that you do not understand what you wrote. Also, what does "transfinitely many arguments" in this context precisely mean when you deal with your original function different from Veblen's function? Of course, it is ambiguous because there are many classes of maps dealt with as "transfinitely many arguments", e.g. the class of all maps from On to On. Clarify the definition of "transfinitely many arguments" in terms of a map.
p-adic 22:50, March 12, 2020 (UTC)
phi_1(1, 1, 1) should now be well-defined, as I added a definition for 3 arguments. Allam(2^^n mod 10^6 for n >= 8) (talk) 23:03, March 12, 2020 (UTC)
As I wrote above, it is still ill-defined. Also, why dou you ignore the question about "transfiitly many arguments"? Also, your rule does not ensure that φ_0(x) is the same as Veblen's φ_0(x), i.e. ω^x unless x = 0.
p-adic 23:21, March 12, 2020 (UTC)
I added a rule to ensure that phi_0 is the same as the regular Veblen function (as it should be). Allam(2^^n mod 10^6 for n >= 8) (talk) 00:36, March 13, 2020 (UTC)
Why do you always consider only a little porsion of issues? Why do you continue to ignore feedbacks?
p-adic 01:29, March 13, 2020 (UTC)

No. What feedbacks am I ignoring? After you pointed out that the fundamental sequences are ill-defined, I removed all direct invocations of fundamental sequences from my definition. After you pointed out that there was no rule that ensures that $$\varphi_0$$ is the same as the regular Veblen function as intended, I added one. Could you give me an example of a value that is still ill-defined so that I can address it? Allam(2^^n mod 10^6 for n >= 8) (talk) 01:34, March 13, 2020 (UTC)

Your "No" is almost always wrong. Read back what you wrote before stating "No". Say, here is a list of what you are still ignoring:
1. Delete all multiple-definitions. You have redundant descriptions of rules applicable to a common case.
2. Specify the precise definition of transfinitely many arguments in your context.
3. Read back carefully what you wrote in order to find errors, i.e. cases to which no rule is applicable.
You always conclude "fixed" without trying hard to find errors. Should I point out all errors by spending time although you spend little effort (actually at least less than a day) to do it by yourself? Should I do it for the one who dishonestly ignores many of feedbacks? If you were honestly paying attensions to feedbacks from others, then I might spend more time. But you are always ignoring feedbacks, and frequently talk as if you had noticed all errors by yourself.
p-adic 01:47, March 13, 2020 (UTC)

I tried to write my definition for transfinitely many arguments, but I just couldn't get the LaTeX to work. How do I write two-row expressions with more than one column? Allam(2^^n mod 10^6 for n >= 8) (talk) 02:46, March 13, 2020 (UTC)

You can use "array" environment. For example, see this blogpost.
p-adic 03:45, March 13, 2020 (UTC)

I removed the multiple-definitions, and added a definition for transfinitely many arguments (the line with the subscript gammas). Allam(2^^n mod 10^6 for n >= 8) (talk) 03:03, March 14, 2020 (UTC)

> I removed the multiple-definitions
No. Remove all multiple-definitions. If two rules are applicable to a common case, (even if the result is believed to be the same value) restrict one of the two in order to clarify what you intend. Usually, you do not have to care about them as long as the actual values are the same, but since you have troubles in expressing what you intend precisely, avoid all multiple-definitions. All does not mean one although you often use all instead of one. (And do not bother others by saying "if you had pointed out all errors, ..." or something like that without showing sufficient effort.)
> Note that (countably) infinitely many 0s can be added at the beginning without affecting the value of the expression.
It does not follow from the written rules. Write down all precise rules. Adding explanations which is not based on the precise definition is not good, because you can always say "It is well-defined as well as Veblenn's function" without specifying the precise rules.
> and added a definition for transfinitely many arguments (the line with the subscript gammas).
Read my comment above. I asked you to clarify the definition of the notion of "transfinitely many arguments" in your ambiguous context unlike the original Veblen function.
p-adic 05:52, March 14, 2020 (UTC)

## Veblen Extension Ideas

My definition may work up to $$\varphi_\omega$$. The ideas below are ill-defined.

After $$\varphi_\omega(0)$$, $$\varphi_{\varphi_\omega(0)}(0)$$, etc., we have:

• $$\varphi_{(1,0)}(0)$$ is the 1st fixed point $$\alpha=\varphi_\alpha(0)$$
• $$\varphi_{(1,0)}(1)$$ is the 2nd fixed point $$\alpha=\varphi_\alpha(0)$$
• $$\varphi_{(1,0)}(1,0)$$ is the 1st fixed point $$\alpha=\varphi_{(1,0)}(\alpha)$$
• $$\varphi_{(1,0)+1}(0)$$ is the 1st fixed point $$\alpha=\varphi_{(1,0)}\begin{pmatrix}1 \\ \alpha\end{pmatrix}$$
• $$\varphi_{(1,0)+2}(0)$$ is the 1st fixed point $$\alpha=\varphi_{(1,0)+1}\begin{pmatrix}1 \\ \alpha\end{pmatrix}$$
• $$\varphi_{(1,0)+\omega}(0)$$
• $$\varphi_{(1,0)+\varphi_{(1,0)}(0)}(0)$$
• $$\varphi_{(1,0)+\varphi_{(1,0)+\varphi_{(1,0)}(0)}(0)}(0)$$
• $$\varphi_{(1,0)*2}(0)$$ is the 1st fixed point $$\alpha=\varphi_{(1,0)+\alpha}(0)$$
• $$\varphi_{(1,0)*2+1}(0)$$ is the 1st fixed point $$\alpha=\varphi_{(1,0)*2}\begin{pmatrix}1 \\ \alpha\end{pmatrix}$$
• $$\varphi_{(1,0)*3}(0)$$ is the 1st fixed point $$\alpha=\varphi_{(1,0)*2+\alpha}(0)$$
• $$\varphi_{(1,0)*\omega}(0)$$
• $$\varphi_{(1,0)*\varphi_{(1,0)*\omega}(0)}(0)$$
• $$\varphi_{(1,0)*\varphi_{(1,0)*\varphi_{(1,0)*\omega}(0)}(0)}(0)$$
• $$\varphi_{(1,0)^2}(0)$$ is the 1st fixed point $$\alpha=\varphi_{(1,0)*\alpha}(0)$$

Then you can use $$(1,0)^3$$, $$(1,0)^\omega$$, $$(1,0)^{(1,0)}$$, $$\varepsilon_{(1,0)+1}$$, $$\varphi(1,0,(1,0)+1)=\Gamma_{(1,0)+1}$$, $$\varphi_1((1,0)+1)$$, $$\varphi_{(1,0)}((1,0)+1)$$, $$\varphi_{\varphi_{(1,0)}((1,0)+1)}((1,0)+1)$$, etc. until we reach $$(1,1)$$, which is the 1st fixed point $$\alpha=\varphi_\alpha((1,0)+1)$$. Maybe (1,2), (1,3), (1,$$\omega$$), etc. can be defined similarly, but I don't knwo how to extend this to (2,0). —Preceding unsigned comment added by C7X (talkcontribs)

You could use (1, (1, 0)), (1, (1, (1, 0))), (1, (1, (1, (1, 0))))... to get to (2, 0). Allam(2^^n mod 10^6 for n >= 8) (talk) 16:26, March 9, 2020 (UTC)
That's good; also (1,0,0)=(((...,0),0),0), (1,0,0,0)=(((...,0,0),0,0),0,0), $$\begin{pmatrix}1 \\ \omega\end{pmatrix}$$ is the limit of $$(1,\underbrace{\cdots,0,0,0}_n)$$ for finite n, etc. Then the limit is the 1st fixed point $$\alpha=\begin{pmatrix}1 \\ \alpha\end{pmatrix}$$ —Preceding unsigned comment added by C7X (talkcontribs)
@C7X
No way. How can ill-defined functions be good? It is bad to show one, who does not understand the ill-definedness, an ill-defined work without showing the issue. Could you see that he has never listened to feedbacks on the ill-definedness? It might be unbelievable, but he does not even understand what to do in order to define a function, unlike you. Spoiling a beginner in thay way causes an awful problem such as the infamous incident that beginners without understanding ordinals started to use UNOCF. Since he is not ready for distinguishing well-defined functions from imaginary poems, please be very careful. Otherwise, he will restart to create fake articles as he did so many times.
p-adic 22:10, March 9, 2020 (UTC)

No, I know how to define a function (with ordinary numbers at least, see the reptend function that I invented last fall). It's just that I don't know very much about ordinals beyond $$\vartheta(\Omega^\Omega)$$, and I'm trying to extend the phi function further. Allam(2^^n mod 10^6 for n >= 8) (talk) 22:40, March 9, 2020 (UTC)

You are wrong. You are not even understanding how to define simple functions, because you never follow the besic rules below:
• Quantify all variables which you want to use.
• Clarify the domain of all functions which you want to define.
• Write down the precise definition applicable to any case.
Up to $$\vartheta(\Omega^\Omega)$$? No way. You are just overestimating yourself. I am 95% certain that you are currently unable to define a function at that level because you always ignore the basic rules, which are required for well-defined functions. I guess that you are not even understanding the precise definition of $$\vartheta(\Omega^\Omega)$$. (No, calling it by another known name is not showing a definition and calculation.)
p-adic 22:54, March 9, 2020 (UTC)

It becomes more difficult to define the function as each expression based on ordered pairs gets more complex. Here is an attempt of a definition up to (but not including) $$\varphi_{(1,0)*2}$$ when the < relation is defined:

Before defining the function, define a set of ordered tuples $$P$$ (some of the ordered tuples are labeled with a +):

• For all $$\alpha\in\text{Ord}$$, $$(\omega^\alpha)\in P$$
• For all $$p_0,p_1,\cdots,p_n\in P$$ for $$n\in\mathbb{N}$$, $$(p_0,p_1,\cdots,p_n)_+\in P$$
• For $$\alpha\in\text{Ord}-\{0\}$$ and $$\beta\in\text{Ord}$$, $$(\alpha,\beta)\in P$$ (however, there is no (1,(1,0)) )

An ordered tuple $$p\in P$$ is in the set of standard ordered tuples $$SP$$ if it satisfies any of these conditions:

• $$p$$ is of the form $$(\alpha)$$ with $$\alpha\in\text{Ord}$$
• $$p$$ is of the form $$(p_0,p_1,\cdots,p_n)_+$$ with $$p_0,p_1,\cdots,p_n\in SP$$ and $$p_n\le\cdots p_1\le p_0$$ (I will define $$<$$ later)
• $$p$$ is of the form $$(\alpha,\beta)$$ with $$\alpha\in\text{Ord}-\{0\}$$ and $$\beta\in\text{Ord}$$

Define a relation < over $$P$$:

• $$(0)<q\iff q\neq (0)$$
• If $$p$$ is of the form $$(\alpha)$$ with $$\alpha\in\text{Ord}$$, then:
• If $$q$$ is of the form $$(\beta)$$ with $$\beta\in\text{Ord}$$, then:
• $$p<q\iff\alpha<\beta$$
• Else $$q$$ is not of the form $$(\beta)$$ with $$\beta\in\text{Ord}$$. Then:
• If $$q$$ is of the form $$(p_0,p_1,\cdots,p_n)_+$$ with $$p_0,p_1,\cdots,p_n\in P$$, then:
• Else $$q$$ is not of the form $$(p_0,p_1,\cdots,p_n)_+$$ with $$p_0,p_1,\cdots,p_n\in P$$. Then:
• If $$q=(1,0)$$, then:
• Else $$q\neq (1,0)$$. Then:
• Else $$p$$ is not of the form $$(\alpha)$$ with $$\alpha\in\text{Ord}$$. Then:
• I will define this later

$$\varphi_p$$ is defined when its argument is a 2-row array of ordinals and when $$p\in P$$.

$$\varphi_{(0)}\begin{pmatrix}\alpha \\ 0\end{pmatrix}=\omega^\alpha$$

For $$p\in P$$, $$\varphi_{(p,1)_+}\begin{pmatrix}\alpha \\ 0\end{pmatrix}=\varphi_{n}\begin{pmatrix}1 \\ \alpha\end{pmatrix}$$

For $$\alpha\in\text{Ord}$$, $$\varphi_{(\alpha)}(\text{ (Any array) })=\sup(\{\varphi_{\beta}(\text{ (Array) }):\beta<\alpha\})$$

$$\varphi_p\begin{pmatrix}\cdots & \alpha+1 & \gamma \\ \cdots & \beta+1 & 0\end{pmatrix}$$ is the $$\gamma$$th ordinal in the set $$\{\delta:\delta=\varphi_p\begin{pmatrix}\cdots & \alpha & \delta \\ \cdots & \beta+1 & \beta\end{pmatrix}\}$$

If $$\alpha$$ is a limit ordinal, then $$\varphi_p\begin{pmatrix}\cdots & \alpha & \gamma \\ \cdots & \beta+1 & 0\end{pmatrix}$$ is the $$\gamma$$th ordinal in the set $$\bigcap_{\varepsilon<\alpha}\{\delta:\delta=\varphi_p\begin{pmatrix}\cdots & \varepsilon & \delta \\ \cdots & \beta+1 & \beta\end{pmatrix}\}$$

If $$\beta$$ is a limit ordinal, then $$\varphi_p\begin{pmatrix}\cdots & \alpha+1 & \gamma \\ \cdots & \beta & 0\end{pmatrix}$$ is the $$\gamma$$th ordinal in the set $$\bigcap_{\varepsilon<\beta}\{\delta:\delta=\varphi_p\begin{pmatrix}\cdots & \alpha & \delta \\ \cdots & \beta & \varepsilon\end{pmatrix}\}$$

If $$\alpha$$ and $$\beta$$ are limit ordinals, then $$\varphi_p\begin{pmatrix}\cdots & \alpha & \gamma \\ \cdots & \beta & 0\end{pmatrix}$$ is the $$\gamma$$th ordinal in the set $$\bigcap_{\varepsilon<\alpha, \eta<\beta}\{\delta:\delta=\varphi_p\begin{pmatrix}\cdots & \varepsilon & \delta \\ \cdots & \beta & \eta\end{pmatrix}\}$$

If $$p=(1,0)$$, then $$\varphi_p(0)=\sup(\{\alpha_n\})$$ given $$\alpha_0=0$$ and $$\alpha_{n+1}=\varphi_{\alpha_n}(0)$$

@Allam948736 The reason that my previous comment was ill-defined is because it was a list of values with no rules to determine intermediate values. It would be like defining f(x) as "the enumerating function of {1,2,4,8,16,...}" without specifying a rule. If the set refers to 2^x, then f(5)=32, but if the set refers to "the maximum possible number of ways that x circles can divide the plane", then f(5)=31. It's impossible to tell which one just given the example, because the values match for x<5. C7X (talk) 23:11, March 9, 2020 (UTC)

No, I understand that $$\vartheta(\Omega^\Omega)$$ is where you'd get if you kept "iterating the number of arguments in the Veblen phi function" forever, or the fixed point of $$\vartheta(\Omega^\alpha)$$. I also updated my definition seen above to always give general-case definitions instead of specific cases. Allam(2^^n mod 10^6 for n >= 8) (talk) 00:29, March 10, 2020 (UTC)

Dear... You are just confident because you do not understand what you do not know. Say, remember how you talked about "the second fixed point of $$\alpha \mapsto \vartheta(\Omega^\alpha)$$". If you knew the definition of $$\vartheta$$, then you would not talk in that way... Hello? Could you tell me what "the second fixed point of $$\alpha \mapsto \vartheta(\Omega^\alpha)$$" means? Please show us the existence by a precise argument. I am certain that you are not even able to show that the first fixed point of $$\alpha \mapsto \vartheta(\Omega^\alpha)$$ is $$\vartheta(\Omega^\Omega)$$.
Also, read my comment above. You have not checked all of the three. Should I enumerate them so that you can remember?
1. Quantify all variables which you want to use.
2. Clarify the domain of all functions which you want to define.
3. Write down the precise definition applicable to any case.
Then could you answer which rule you checked?
p-adic 00:59, March 10, 2020 (UTC)

I mentioned a second fixed point of $$\alpha \mapsto \vartheta(\Omega^\alpha)$$ because I thought all such ordinal functions had infinitely many fixed points like $$\omega^\alpha$$ or $$\varepsilon_\alpha$$ do. I now know that there is no such "second fixed point". I have now checked numbers 1 and 3, but not 2. The domain of the function is (at least) all countable ordinals. Allam(2^^n mod 10^6 for n >= 8) (talk) 01:19, March 10, 2020 (UTC)

So now you understand that you are not understanding $$\vartheta$$ well. Also, you are probabry unable to show that $$\vartheta(\Omega^\Omega)$$ is the first fixed point in that case. It means that you are not really understanding it. You just know the fact that it is the first fixed point, because you heard it from others.
You have not checked 3. Say, what are the definitions of PTO(ZFC)[3], ω_3^{CK}[4], φ_3(0,0), φ_1(2,0,0), and so on? Maybe you are still believing that listing finitely many small values characterises the whole functions. As I taught you so many times, you need full definitions for all notions which you use and which others have not defined yet applicable to any cases.
p-adic 01:40, March 10, 2020 (UTC)

phi_1(2, 0, 0) is the first fixed point of $$\alpha \mapsto \varphi_1(1, \alpha, 0)$$. There is no phi_3(0, 0), instead it's phi_3(1, 0) and it's the first fixed point of $$\alpha \mapsto \varphi_3(\alpha)$$ (phi_3(0) is equal to phi_2(1)). As for those first two ordinals, we don't know their fundamental sequences, and PTO(ZFC)[3] and w_3^CK[4] would depend on how you would actually define the fundamental sequences of these ordinals. If we don't really know the fundamental sequence of the ordinal well, that can't make an ordinal function ill-defined. It just makes the fundamental sequence of the ordinal ill-defined. I am no longer listing smaller values, and have made sure that my definition includes only variables (which indicate a general-case definition), not specific values.

>Also, you are probably unable to show that $$\vartheta(\Omega^\Omega)$$ is the first fixed point in that case.

No, I understand that by definition, the $$\Omega$$ indicates a fixed point. This is true of all OCFs. Allam(2^^n mod 10^6 for n >= 8) (talk) 02:38, March 10, 2020 (UTC)

> phi_1(2, 0, 0) is the first fixed point of ...
No. It does not follow from your "definition".
> There is no phi_3(0, 0), instead it's phi_3(1, 0)
It contradicts your descriotion "The domain of the function is the countable ordinals, meaning that all variables in the definition represent countable ordinals."
> If we don't really know the fundamental sequence of the ordinal well, that can't make an ordinal function ill-defined. It just makes the fundamental sequence of the ordinal ill-defined.
No. You directy referred to the fundamental sequences. If you "define" a function using undefined and unquantified notions, then it is ill-defined,
> No, I understand that by definition,
I am certain that you do not. How could you verify something about all OCFs even though you do not really understand single elementary OCF or the meaning of all? The statement itself is meaningless, because the notion of "all OCFs" in your sense is ambiguous. If I show you a counterexample such as ψ_0(φ_Ω(1)) with respect to Buchholz's OCF, then you will shift your goalpost by saying something like "no, Buchholz's OCF is not an OCF" as you always do. The fact that you do not know a counterexample does not mean that you are correct. Are you happy to overestimate yourself on what you know few about? Such an attitude just prevents you from improving yourself.
p-adic 03:03, March 10, 2020 (UTC)

When I said that $$\Omega$$ always indicates a fixed point, I meant inside the parentheses, as in $$\psi(\Omega^{\Omega^\Omega})$$. Actually, phi_3(0, 0) does exist, but is the same as phi_3(0).

>No. You directly referred to the fundamental sequences ... then it is ill-defined.

No, it just makes the function with arguments with ill-defined fundamental sequences ill-defined. If the fundamental sequence is well-defined, for example $$\varphi_2(\omega^2, 0)$$, then it is well-defined.  Allam(2^^n mod 10^6 for n >= 8) (talk) 03:20, March 10, 2020 (UTC)

> When I said that $$\Omega$$ always indicates a fixed point, I meant inside the parentheses,
Unfortunately, the shifting goalpost is not successful. Both of Ω's in φ_Ω(0) and Ω^Ω^Ω are in the parenses of ψ. Are you happy if I write another counter example, e.g. ψ_0(ψ_1(ψ_2(Ω)))? Then will you say that Ω should be right inside the parenses, although Ω^Ω is just an abbreviation of the value of a two-ary function given by exponential?
> No, it just makes the function with arguments with ill-defined fundamental sequences ill-defined.
Wow! Are you believing that arguments, i.e., countable limit ordinals are equipped with fundamental sequences? Then you are not ready for understanding limit ordinals. An ordinal is a sort of a well-founded set, and itself does not have an information of a fundamental sequence. Go ahead. Continuing to answer what you know few about shows how poorly you misunderstood them. It is helpful for you, because you can fix your ovrestimation of yourself.
p-adic 03:30, March 10, 2020 (UTC)

While there isn't a single fundamental sequence attached to every countable limit ordinal, there are agreed-upon fundamental sequences for all limit ordinals $$\le \varepsilon_0$$. Allam(2^^n mod 10^6 for n >= 8) (talk) 03:35, March 10, 2020 (UTC)

It irrelevant to your function. You wrote that the domain of your variadic function is all countable ordinals. Then φ_3(0,0) should be defined because 0 is a countable ordinal, and φ_3(ω_3^{CK}) should be defined because ω_3^{CK} is a countable ordinal. Or are you starting to shift goalpost by stating 0 and ordinals greater than ε_0 are not countable ordinals in your convention?
p-adic 03:41, March 10, 2020 (UTC)
No, I only said that all ordinals below $$\varepsilon_0$$ have agreed-upon fundamental sequences. Obviously there are countable ordinals larger than $$\varepsilon_0$$, such as $$\zeta_0$$ and $$\Gamma_0$$, but their fundamental sequences are usually not agreed on.  (There are some ordinals beyond $$\varepsilon_0$$ with agreed-upon fundamental sequences, however). I already said that phi_3(0, 0) is defined, but is equal to phi_3(0) = phi_2(1) = phi_1(1, 0) = LVO. If the fundamental sequences of the arguments are unknown (as in $$\omega_3^CK$$), then the expression is ill-defined. Otherwise, it is well-defined. Allam(2^^n mod 10^6 for n >= 8) (talk) 14:05, March 10, 2020 (UTC)
You finally start to mix mathematical sstatements, e.g. the well-definedness, and intuition-based non-mathematical sentences, e.g. "someone knows a fundamental sequence"... It is waste of time for you to talk about ordinals, which you do not know at all.
p-adic 14:26, March 10, 2020 (UTC)

In general, the function is ill-defined because the results for certain arguments depend on the definition of fundamental sequences, but expressions such as the one I gave in a comment above where the fundamental sequences are agreed-upon are well-defined. Allam(2^^n mod 10^6 for n >= 8) (talk) 15:43, March 10, 2020 (UTC)

Don't you know the definition of the domain? You clarified that the domain is the set of all countable ordinals. It means that any countable ordinal lies in the domain. It is irrelevant to your intuition-based description on the exostence of an agreed-upon fundamental sequence. Say, ω_3^{CK} is a countable ordinal, and hence the value φ_1(ω_3^{CK}) should be well-defined even if there is no agreed-upon fundamental sequence. Well, I understand that you are not willing to honestly learn more.
p-adic 16:02, March 10, 2020 (UTC)

@Allam I advice you to join Discord (if you're not there). It might be helpful for learning things. — Best regards, Triakula 16:40, March 10, 2020 (UTC) No. What I'm saying is that these would be well-defined if a universally agreed-upon definition of fundamental sequences were fixed. For instance, if we fix $$\omega_1^{CK}[3] = \omega$$, then $$\varphi_1(\omega_2^{CK}+\omega_1^{CK}, 0)[3] = \varphi_1(\omega_2^{CK}+\omega, 0)$$. Allam(2^^n mod 10^6 for n >= 8) (talk) 22:49, March 10, 2020 (UTC)

No, you are wrong. You are not allowed to use undefined unquantified object other than the input in the mathematical definition of a function. When you define a function whose domain is the set of all countable ordinals, you need to specify the definition of all unquantified fundamental sequences used in the definition. It is irrelevant to the non-mathematical sentence such as "agreed-upon" and "unknown". How many times should I repeat the same thing?
p-adic 22:55, March 10, 2020 (UTC)
I will update my definition to not directly invoke fundamental sequences. Allam(2^^n mod 10^6 for n >= 8) (talk) 03:03, March 11, 2020 (UTC)
It is good. But before that, it is better to understand why you were wrong. Otherwise, you will repeat the falilure, as you always do. Here is a list to avoid the same errors in the definitin of a function:
1. You need to clarify the domain of the function.
2. You need to define all objects appearing in the definition unless they are inputs, already defined objects, quantified objects, or the function itself.
3. You need to write down the precise rule to characterise the value of the function applicable to any inputs.
For example, if the domain is the set of all countable ordinals, the written rule should be applicable to any countable ordinal such as PTO(ZFC) or ω_3^{CK}. It is irrelevant to the non-mathematical fact that we do not have aagreed-upon fundamental sequences or any other good informations on them. In the definition, you are not allowed to use undefined unquantified objects such as undefined unquantified fundamental sequences or intuition-based predicates.
p-adic 03:16, March 11, 2020 (UTC)

## Hey

I don't know if you saw my comment on your blog post regarding the binomial hierarchy, but I hope my analysis was helpful. QuasarBooster (talk) 15:56, March 16, 2020 (UTC)

Yes, I did find the analysis helpful. In fact, I have also found that the values of the binomial hierarchy can be lower-bounded by the corresponding values in SGH (and is identical for an argument of 2). This means that $$F_{\varepsilon_0}(3) \gt 7625597484987$$, but I have no idea how to find the actual value. Allam(2^^n mod 10^6 for n >= 8) (talk) 16:53, March 16, 2020 (UTC)

Glad I could help! Indeed, it would be very hard to fully evaluate $$F_{\varepsilon_0}(3)$$ based on how messy its recursion rules are. Of course, the other growth hierarchies have the same problem too. But once the numbers get big enough, having exact precision isn't really even necessary. Using my upper bound, I can tell you that $$F_{\varepsilon_0}(3)<3^{7625597484987}$$. QuasarBooster (talk) 21:54, March 16, 2020 (UTC)

I found the exact value! Yesterday I figured that $$F_{\varepsilon_0}(3)$$ is definitely less than trn(7625597484987), meaning it is less than $$3^{54}$$. The exact value is trn(7625597484987) - 3*trn(2541865828328) - 18*trn(84728609442) - 216*trn(282429536480) - 324*trn(94143178826) - 1944*trn(31381059608) - 23328*trn(10460353202) - 34992*trn(3486784400) - 209952*trn(1162261466) - 3464208*trn(387420488) - 944784*trn(129140162) - 5668704*trn(43046720) - 68024448*trn(14348906) - 102036672*trn(4782968) - 612220032*trn(1594322) - 7346640384*trn(531440) - 11019960576*trn(177146) - 66119763456*trn(59048) - 1090976097024*trn(19682) - 297538935552*trn(6560) - 1785233613312*trn(2186) - 21422803359744*trn(728) - 32134205039616*trn(242) - 192805230237696*trn(80) - 2313662762852352*trn(26) - 3470494144278528*trn(8) - 20822964865671168*trn(2) or 306,030,478,267,498,284. Allam(2^^n mod 10^6 for n >= 8) (talk) 16:49, September 18, 2020 (UTC)

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