· Portal  Help Search Members Calendar 
Welcome Guest ( Log In  Register )  Resend Validation Email 
Join the millions that use us for their forum communities. Create your own forum today.  Welcome to Dozensonline. We hope you enjoy your visit. You're currently viewing our forum as a guest. This means you are limited to certain areas of the board and there are some features you can't use. If you join our community, you'll be able to access memberonly sections, and use many memberonly features such as customizing your profile, and sending personal messages. Registration is simple, fast, and completely free. (You will be asked to confirm your email address before we sign you on.) Join our community! If you're already a member please log in to your account to access all of our features: 
Pages: (2) [1] 2 ( Go to first unread post ) 
Double sharp 
Posted: Oct 13 2015, 09:37 AM


Dozens Disciple Group: Members Posts: 1,401 Member No.: 1,150 Joined: 19September 15 
I've been curious about comparing these two values, so here's a list. It gives every integer up to 60, and then the runnerup HCNs and primorials up to 840. I wanted it to show every integer up to 360, but what a task to do manually. I'd also have loved to get a column for φ(n)/d(n), but the alignment is confusing me to no end. So, please treat this as a call for help.


wendy.krieger 
Posted: Oct 13 2015, 09:58 AM

Dozens Demigod Group: Members Posts: 2,432 Member No.: 655 Joined: 11July 12 
I take it that this is Euler Totient divided by number of divisors. Should be able to bash a rexx script to do this.
You can use a text editor to get the alignment right. I use metapad, but it depends on what form of DOS you use. 
Double sharp 
Posted: Oct 13 2015, 12:14 PM


Dozens Disciple Group: Members Posts: 1,401 Member No.: 1,150 Joined: 19September 15 
Yup, it is Euler totient over number of divisors. I would also love to have Euler totient over number of regular digits. You're right, I didn't think of that...next time I'll be composing this sort of thing in a text editor, using a monospaced font instead. 

Double sharp 
Posted: Oct 13 2015, 01:24 PM


Dozens Disciple Group: Members Posts: 1,401 Member No.: 1,150 Joined: 19September 15 
Up to 240:
The numbers that give a value of φ(n)/d(n) less than or equal to 2 are {1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 24, 28, 30, 36, 40, 42, 48, 60, 72, 84, 90, 120}. If we move the limit down to 1.5 (3/2), it becomes {1, 2, 3, 4, 6, 8, 10, 12, 14, 18, 20, 24, 30, 36, 42, 60}. If we move it down further to 1.333... (4/3), we get {1, 2, 3, 4, 6, 8, 10, 12, 18, 20, 24, 30, 36, 60}. An interesting sequence! Given how {18, 20} are comparable in scale, this may be the actual limit (with a slight totative dominance) for whether a base mostly resists us or helps us. If we move it down to 1, we get {1, 2, 3, 4, 6, 8, 10, 12, 18, 24, 30} and no others. I am not convinced completely about this sequence, because {20, 60, (120)} have appeared in pretty advanced societies, and therefore seem to be workable. The 4/3 limit seems better, though I'm not sure if {120} is really as bad as this valuation paints it as (although I think everything else beyond {60} is surely unusable). 

Kodegadulo 
Posted: Oct 13 2015, 01:31 PM


Obsessive poster Group: Moderators Posts: 4,189 Member No.: 606 Joined: 10September 11 
Here you go:
Do a Quote to see the BBNcode/HTML. I used a little macro recording in notepad++ to format one table row line and then just repeated it (rerecording it when I got to 3digit numbers). You could certainly doctor any script to output the same codes. 

icarus 
Posted: Oct 13 2015, 02:48 PM


Dozens Demigod Group: Admin Posts: 1,913 Member No.: 50 Joined: 11April 06 
This comparison always has fascinated me.
Check this out: OEIS A020490. I added code there this morning: Select[Range@ 1000000, EulerPhi@ # <= DivisorSigma[0, #] &] , and once I can track it down I'll add a reference I know about. It's a very interesting sequence. 

Double sharp 
Posted: Oct 14 2015, 01:44 PM


Dozens Disciple Group: Members Posts: 1,401 Member No.: 1,150 Joined: 19September 15 
Yes, it's the same sequence. However, given the use of {20} by the Mayans and {60} by the Sumerians, I think the limit might be just a little higher than 1. To admit these two bases, we need to raise the limit to at most a 3:4 ratio between divisors and totatives, which adds {20, 36, 60} to the list. I'm somewhat convinced by this as {60} does appear to be very useful, {20} and {18} are comparable in scale (having the same number of divisors and opaque totatives), and removing the transparent nonunitary totatives from {36}'s count (5, 7, and 35) brings it to totativedivisor parity. OTOH, {120} has φ(120)/d(120) = 2. So I am not quite sure if going up to {120} is worth it, because all you gain from {60} are singledigit eighths (a significant plus if you use 6:10 and 12:10; not very significant if you use pure sexagesimal or centovigesimal), and better neighbour relationships (but they only ameliorate four totatives out of thirtytwo, so that the ratio remaining is still 1.75). Perhaps regular digits are significant too. Instinctively it feels wrong to me to count semidivisors equally with divisors, but I don't know what a fair valuation would be, so I have to try it: Senary (6)  5 regular, 2 totatives: ratio is 2.5 Octal (8)  4 regular, 4 totatives: ratio is 1 Decimal (10)  4 regular, 4 totatives: ratio is 1 Duodecimal (12)  8 regular, 4 totatives: ratio is 2 Tetradecimal (14)  6 regular, 6 totatives: ratio is 1 Hexadecimal (16)  5 regular, 8 totatives: ratio is 0.625 Octodecimal (18)  10 regular, 6 totatives: ratio is 1.666... Vigesimal (20)  8 regular, 8 totatives: ratio is 1 Tetravigesimal (24)  11 regular, 8 totatives: ratio is 1.375 Octovigesimal (28)  8 regular, 12 totatives: ratio is 0.666... Trigesimal (30)  18 regular, 8 totatives: ratio is 2.25 Hexatrigesimal (36)  14 regular, 12 totatives: ratio is 1.166... Duoquadragesimal (42)  19 regular, 12 totatives: ratio is 1.583... Octoquadragesimal (48)  15 regular, 16 totatives: ratio is 0.9375 Sexagesimal (60)  26 regular, 16 totatives: ratio is 1.625 Septuagesimal (70)  18 regular, 24 totatives: ratio is 0.75 Duoseptuagesimal (72)  18 regular, 24 totatives: ratio is 0.75 Octogesimal (80)  14 regular, 32 totatives: ratio is 0.4375 Tetroctogesimal (84)  28 regular, 24 totatives: ratio is 1.166... Nonogesimal (90)  32 regular, 24 totatives: ratio is 1.333... Hexanonogesimal (96)  20 regular, 32 totatives: ratio is 0.625 Centesimal (100)  15 regular, 40 totatives: ratio is 0.375 Centoctonary (108)  21 regular, 36 totatives: ratio is 0.583... Centoduodecimal (112)  14 regular, 48 totatives: ratio is 0.2916... Centovigesimal (120)  36 regular, 32 totatives: ratio is 1.125... Centotetraquadragesimal (144)  23 regular, 48 totatives: ratio is 0.47916... Duocentodecimal (210)  50 regular, 48 totatives: ratio is 1.0416... Duocentoquadragesimal (240)  51 regular, 64 totatives: ratio is 0.796875 Trecentosexagesimal (360)  61 regular, 96 totatives: ratio is 0.635416... Septingentovigesimal (720)  76 regular, 192 totatives: ratio is 0.39583... (I'd have loved to include 2310 and 2520, but couldn't find counts of regulars.) The 1andabove club in this list (ignoring the really small bases below 5 that get in simply for having very few totatives) is {6, 8, 10, 12, 14, 18, 20, 24, 30, 36, 42, 60, 84, 90, 120, 210}. But I think that this equal valuation is biased in favour of primorials, and does not consider that regulars like 512 in {720} are of no practical help whatsoever (they're too "rich", dividing too high a power of the base). So I don't really know of a measure that seems to reflect {120} as an "island of stability" in the sea of instability beyond 36, as I thought it would be. ({60} is assuredly an island.) Perhaps the added totative resistance is really not worth it. Is this because this measurement is geared towards pure radices instead of things like {12:10}? 

Double sharp 
Posted: Oct 19 2015, 01:00 PM


Dozens Disciple Group: Members Posts: 1,401 Member No.: 1,150 Joined: 19September 15 
One thing that I find particularly interesting about this list is that the only bases here that have a gap in their prime factorization are {3, 10}. If we add {20, 36, 60}, in effect raising the maximum acceptable totative:divisor ratio to 4/3, then the list becomes {3, 10, 20}. {3} is probably in there due to what I call "edge effects": in the very low bases, there are so few digits that you can get oddities like all but one digit being regular ({4, 6}), no nontrivial totatives and divisors ({2}), and primes with no opaque digits ({2, 3, 5}). This strange behaviour seems to stop at {7}, at which a new humanscale regime appears to start and continue to {15} (excepting {13}). So, if we discount {3} for being too small for today's society, and odd at that, it does appear than {20}, and {10} even more so, are really exceptional bases, that can still compete in spite of their less efficient prime decomposition. 

Double sharp 
Posted: Oct 19 2015, 04:06 PM


Dozens Disciple Group: Members Posts: 1,401 Member No.: 1,150 Joined: 19September 15 
The sequence A072938 shows the HCNs which are half of the next HCN: {1, 2, 6, 12, 60, 360, 2520} and no others. (Here's a proof in German.) The first five appear in our sequence with totative:divisor ratio set to 4:3, but the last two are huge, at {360, 2520}, only to add the prime powers {8, 9, 7}. {360} concentrates extra weight on previously unrepresented powers of already existing primes, for no benefit in reducing the totient ratio, so that the totatives end up swamping the divisors. And {2520} does add another prime factor 7, but is even more swamped with totatives due to its huge size. Now I think I see why {120} does so well despite having a large totative:divisor ratio! It handles 8 as a divisor and 7 as a keen omegatotative, only lacking the 9 contributed by {360}! It's acting like the little brother of the absent {2520}, only failing to handle ninths in one place! And in exchange it handles the next prime {11} gracefully too, as well as the second prime after that, {17}! (The missing prime {13} is not even maximal, being handled by the cubeomega!) The wonderful sevenths seem to be a good tradeoff for the slightly worse ninths when comparing {120} against {360}! The drawback of this is that while I am very sure now that {120} truly does rank in usefulness equally with the numbers with at most 4/3 as many totatives as divisors  {1, 2, 3, 4, 6, 8, 10, 12, 18, 20, 24, 30, 36, 60}  I do not really have a legitimate mathematical formula that gives this valuation of {120} yet. All I have to support my case here is how {120} handles sevenths and eighths pretty well at an incredible twentyfirst of the scale of {2520}, the first SHCN to handle them both as divisors. It really seems to be one of the last islands of usefulness, if not the last, thanks to it being the only low ({5040} and below) SHCN that has only composite neighbours. As Icarus said, comparing the most useful small SHCN bases {2, 6, 12, 60, 120}, with a brief look at decimal (emphasis mine):


Double sharp 
Posted: Oct 20 2015, 01:56 PM

Dozens Disciple Group: Members Posts: 1,401 Member No.: 1,150 Joined: 19September 15 
Another totativerelated comparison I've recently been thinking about is the incidence of opaque composite totatives, like sexagesimal 49. After having done some work in sexagesimal with pen and paper, I find myself very often noting 49 as a prime, before realizing that it isn't. (Even though it isn't, it is a sexagesimal totative and so primes can end in "49", such as "1'49", decimal 109.)
The bases that lack opaque composite totatives are {210, 12, 15, 18, 24, 30}. I'm not surprised by the low range {210} as there simply are so few digits (hmm, maybe {210} is the "extended" low range while {26} is the main low range?) that even inefficient prime decompositions like {5} and {7} still work thanks to the very composite neighbours. After that, it's only the multiples of six, which work so well to govern the primes until {24}, just below 5^{2} (the square of the first unrepresented senary prime), at which 5 steps into factorization for {30} and no more bases. I am really amazed by pentadecimal's performance here, and while I'm not sure what the best base overall is, I am now almost totally convinced that the best odd base is {15}. An added bonus of the midscale bases {18, 24, 30} here is that not only do they lack composite totatives, they also have φ(n) = d(n) (the main subject of this thread), and their abbreviated multiplication tables have only regular product lines (as their first totatives  5, 5, and 7 respectively  are above their square roots). I must say that my opinion of trigesimal has now raised significantly! I've got to experiment with it to see if I think the loss of a clean quarter over sexagesimal is worth its smaller size, totativedivisor parity, and lack of composite totatives. I'm also impressed by tetravigesimal, as it has 5 as a base24 alpha Wieferich prime. Octodecimal I am less impressed by, as it does not seem to offer important added advantages over duodecimal that would offset its size: but I will investigate it further nonetheless. 
wendy.krieger 
Posted: Oct 21 2015, 10:08 AM

Dozens Demigod Group: Members Posts: 2,432 Member No.: 655 Joined: 11July 12 
16 and 18 are unique in having a perimeter equal to the area (4,4) and (3,6), these being regular tilings, but that's the same formula.
2,4,6,10,12, and 18 are the only bases to have prime decades, such as decimal 101, 103, 107, 109, and dozenal 81, 85, 87, 8E. The coprimes of the form 7,9,2,X in base 18 are also primes (this is the second such run). Base 30 has prime decades if one supposes a fully symmetric digits (14 to +15), there are eight primes within 15 of 1230. 
icarus 
Posted: Oct 21 2015, 09:56 PM

Dozens Demigod Group: Admin Posts: 1,913 Member No.: 50 Joined: 11April 06 
@Double Sharp,
I feel your pain, regarding the sexagesimal totative 7^2. It is a "pseudoprime" that way, because our "sieve" (sexagesimal) passes it as perhaps prime because it is coprime to 60 and no other composite digit is coprime to 60. {{1, 7, 11, 13}, {17, 19, 23, 29}, {31, 37, 41, 43}, {47, [49], 53, 59}} seems so complete. I used to consider it a major thing that a base had a composite totative but all bases greater than 30 have at least one. Then I considered the basecomplements of the totatives {1, 59}, {7, 53}, etc., that there must be something up with 11. In base 360 the lineups seem to indicate that composite totatives have composite complements  until they don't. 
Double sharp 
Posted: Oct 22 2015, 07:45 AM


Dozens Disciple Group: Members Posts: 1,401 Member No.: 1,150 Joined: 19September 15 
In base 360 the pair {143, 217} are complements and both are totatives to 360. However, both are composite: 143 = 11 * 13 and 217 = 7 * 31. This is the only such pair in base 360. The only bases where all opaque totatives are primes are {210, 12, 15, 18, 24, 30}, as previously mentioned. If we restrict this further to all totatives (even if they are transparent), the only bases with only prime or unitary totatives are {2, 3, 4, 6, 8, 12, 18, 24, 30}. I feel this is a little unfair to decimal as it is disqualified by its omega, which is surely transparent; thus I'd add it in parentheses, giving {2, 3, 4, 6, 8, (10), 12, 18, 24, 30}. This counts omega as transparent but not neighbourfactors. 

Double sharp 
Posted: Feb 4 2017, 04:34 PM

Dozens Disciple Group: Members Posts: 1,401 Member No.: 1,150 Joined: 19September 15 
Come to think of it, to compute resistance, maybe we should really consider φ(n)  2 instead of φ(n). The reason is that 1 never gives any resistance at all, and that the procedure for omega products is the same in any base and never requires much memorisation beyond addition. (Decrement the other multiplicand by one to give the first digit, and then pick the second digit so that they sum to give the omega again.)
By that metric, the following bases have at least as much leverage as resistance: {(1), 2, 3, 4, 5, 6, 8, 10, 12, 14, 18, 20, 24, 30} which seems to accord better with how 14 and 20 seem about equally functional to 10 and 18 respectively. If we looked at φ(n)  3, we would also admit to the club {9, 16, 36}; φ(n)  4 admits also {7, 15, 42, 60}. The latter seems to accord best with my experience: 11 and 13 sneak into the humanscale range not so much because they're easy as because they're small enough that the chore of memorising the multiplication table at least has an end in sight. 
Double sharp 
Posted: Feb 4 2017, 05:06 PM

Dozens Disciple Group: Members Posts: 1,401 Member No.: 1,150 Joined: 19September 15 
Which makes me think of another thing: what leverage is this function computing? You see, for all that the multiplication table of base thirty might be highly patterned, it would be insane to try to memorise it. (I confess to having tried to do it. In my defence, I expected to fail, and I did fail.)
And what about rote memorisation? If the number of facts is within the range that memorising all of them is reasonable, then there's not so much difference. The difference seems to only come in the midscale where there is a difference, including the low range {16, 18, 20} which overlaps with the top of the human scale in its methods. So this seems to, more than anything else, simply compute leverage for bases that can use the reciprocal divisor method. Bases 15 and below don't really need this help. If you use reciprocal divisors, the neighbours of half the base are also pretty easy: they are really only one multiplicative step (the halving), so the stipulation that d(n) be at least φ(n)  4 seems to make some sense. I think this is why Icarus has said before that 60 is semipractical in a way that 120 and 360 really aren't. So when this metric seems to praise {24, 30, 36, 42, 60}, we have to think of it in terms of reciprocal divisors. Indeed I find that they are all pretty good for it, but it should be noted that {30, 42} don't get much of the reason why φ(n)  4 makes sense, because they are only singly even, and the bar is a little higher for them: they need φ(n)  2 in the midscale. Thirty scrapes past but fortytwo doesn't. So {24, 30, 36, 60} seem to be recommended. So now all I've done is recreate the midscale list that you get if you say that you want no more than fourthirds as many totatives as divisors. Oh well. It's good to have confirmation, at least. (Yay, 800 posts!) 
Double sharp 
Posted: Nov 19 2017, 12:07 PM

Dozens Disciple Group: Members Posts: 1,401 Member No.: 1,150 Joined: 19September 15 
Also, perhaps we might want to consider counting regulars instead of divisors, in which case we have regular parity or dominance over other digits only at {1, 2, 3, 4, 6, 8, 10, 12, 18, 30}. Which is the same list as the first, except without 24, and is A275581 in the OEIS (submitted by icarus).

wendy.krieger 
Posted: Nov 19 2017, 02:13 PM

Dozens Demigod Group: Members Posts: 2,432 Member No.: 655 Joined: 11July 12 
The number of regulars under a given base, is not really governed by the signature, but by some form
ln^n(B )/(n1)! ln(p_1)ln(p_2) ... ln(p_n) eg ln³(120) / 2 ln(2) ln(3) ln(5). 
Double sharp 
Posted: Nov 19 2017, 02:20 PM


Dozens Disciple Group: Members Posts: 1,401 Member No.: 1,150 Joined: 19September 15 
Which evaluates to about 5.859 if ln^3 is read as ln (120^3), and about 0.183 if ln^3 means ln ln ln (which it probably doesn't), and there are clearly more 5smooth numbers under 120 than that. So what exactly did you mean here? 

wendy.krieger 
Posted: Nov 19 2017, 02:34 PM


Dozens Demigod Group: Members Posts: 2,432 Member No.: 655 Joined: 11July 12 
ln³(x) is (ln(x))^3. It comes to 44.766173. You consider the regulars in logrithmetic space, like 2, 3, 5 axis. Each regular occupies a cell, and you are drawing lines around the regulars. This is code from my regulars script, that looks compares a set of regulars to some b^n, as formulae: mantissa = digit stream regardless of radix, so 2 = 2,0 = 2,0,0 counted as one entity. new regulars mantissa in ripple n, g0 * n  gc total regulars mantissa to ripple n g0 * n²/2 + ga n + gb. It's only evaluated for bases with three prime divisors.


Double sharp 
Posted: Nov 19 2017, 02:48 PM


Dozens Disciple Group: Members Posts: 1,401 Member No.: 1,150 Joined: 19September 15 
Yes, but there are only 36 regular digits in base 120:
Regular Digits g ≤ r
So how does the value of 44.766173 relate to that? 

wendy.krieger 
Posted: Nov 19 2017, 02:52 PM

Dozens Demigod Group: Members Posts: 2,432 Member No.: 655 Joined: 11July 12 
The actual number of regulars between consecutive powers of 120, is
44.766173 n²  12.95 n + 4. This value is to within \pm 2 up to n=30. See, eg this thread. 
icarus 
Posted: Nov 19 2017, 09:29 PM


Dozens Demigod Group: Admin Posts: 1,913 Member No.: 50 Joined: 11April 06 
I did a really massive study of regular numbers. Wendy is approximating them, which is okay. An algorithm that generates or counts regular numbers does function much like the divisor counting function. It is helpful to look at the former first. The divisor counting function d(n) = (e_1 + 1) * (e_2 + 1) * ... (e_k + 1) (i.e., a tensor product), where the standard form prime decomposition is n = p_1^e_1 * p_2^e_2 * ... * p_k^e^k. In this case, the function produces an orthogonal figure similar to an mdimensional parallelepiped, with m = omega(n), i.e., the number of distinct prime divisors of n. (See OEIS 275055). The regular counting function r(n) is similar but not so easily generated this way, because of the number n acting as a bound. The charts Double sharp posted are a good way to look at it. We made m axes, one for each distinct prime divisor, and include all powers of p from 0 up to floor(log_p(n)). We make a matrix as we do for the divisors, except we don't right any numbers r > n, meaning that the table will have an irregular shape similar to a "rightsimplex" in m dimensions. For squarefree semiprimes this will look roughly triangular, with the "hypotenuse" a slightly convex, ragged diagonal edge. (See OEIS 275280). Thus the shape is amenable to some analysis that Wendy suggests. I've written an algorithm that computes regulars this way and it is the very most efficient method. (See OEIS A244052, function f):
which writes a program for each number based on its factorization:
It could be written by a Do loop and compiled and would be greased effin' lightning:
(This is the counter). 

Double sharp 
Posted: Nov 19 2017, 11:51 PM

Dozens Disciple Group: Members Posts: 1,401 Member No.: 1,150 Joined: 19September 15 
Ah, okay, I get it now, though Wendy might certainly have been clearer, especially about this being an approximation.

icarus 
Posted: Nov 20 2017, 03:28 PM

Dozens Demigod Group: Admin Posts: 1,913 Member No.: 50 Joined: 11April 06 
It may be keen, and perhaps there is a way via calculus to compute a "volume", however the bound by magnitude of n requires that the curve that is the "hyperhypotenuse" ( hypertenuse? ; ) ) must include only full "cells." This nondiscrete method will tend to overestimate regulars. For very large numbers this might be acceptable. I have computed the regular counting function A010846 for numbers as great as primorial p_15# (which for me takes a workday) and a solid range of 32,000,000 (I am heading for p_9# and it will take months) in order to generate data for sequences A292867, A293555, etc. and refine the conjectures therein. It would be interesting to have a nondiscrete or analytical estimation method.
Where the divisor counting function has n as a boundary by nature (no number greater than n may divide n) the regular counting function has n as a less"natural" boundary. The series of regular numbers in base n is infinite; we select n not arbitrarily, but it does "cut off" the sequence. The tensor R of regulars is the same for numbers with the same "core" (squarefree root); that of n = {6, 12, 18, 24, 36, ...} is R_6. The regular counting function merely "cuts off" a portion of R_6 at n in different places. In this case, only squarefree numbers (A005117) have unique R. Note the same is true for d(n), since this function cuts off the axes at the last power of p that divides n. R_6 is cut off at 2^1 and 3^1, and R_12 at 2^2 and 3^1, whereas r(6) cuts R_6 at any number in the tensor greater than 6, r(12) at any number in the tensor greater than 12. This shows that r(n) >= d(n) and divisors a subset of the regulars of n. It also implies that the number of semidivisors (i.e., A243822(n)) for numbers n for which omega(n) > 1 eventually swamp d(n) as n increases, and that divisors are a vanishingly small subset of the regulars of such n as n gets large. 
Pages: (2) [1] 2 