mersenneforum.org  

Go Back   mersenneforum.org > Math Stuff > Other Mathematical Topics

Reply
 
Thread Tools
Old 2022-08-10, 20:32   #1
mart_r
 
mart_r's Avatar
 
Dec 2008
you know...around...

83210 Posts
Default Sum of digits of x and powers of x

Given a base b >= 2 and an odd integer exponent n >= 3, consider the difference (or ratio) between the sum of the last d digits (in base b) of x and the sum of the last d digits of x^n, where x mod b <> 0.
How quickly/efficiently can one find the minimum (or maximum) of these quantities for all possible values of x (0 < x < b^d)? Are there any "interesting" solutions or ways of solving?
(This is a generalization of the ideas that lead to e.g. A122484 or A138597. A very old question of mine, that I haven't asked anyone yet, is "is there an x such that the sum of digits of x^5 is less than or equal to the sum of digits of x?")


In case this sounds too complicated to start thinking about, here's already a follow-up question on another subject:
We know that the sum of the reciprocals of primes <= p is ~ log(log(p))+M with M=0.2614972...; how do the asymptotics look like for the sum of the reciprocals of semiprimes, or k-almost primes - beside the predominant term log(log(p))^k?
mart_r is offline   Reply With Quote
Old 2022-08-10, 21:10   #2
slandrum
 
Jan 2021
California

1E316 Posts
Default

Quote:
Originally Posted by mart_r View Post
A very old question of mine, that I haven't asked anyone yet, is "is there an x such that the sum of digits of x^5 is less than or equal to the sum of digits of x?")
I assume you are not looking for trivial answers like x = 0, or x = 10^y
slandrum is offline   Reply With Quote
Old 2022-08-10, 21:49   #3
JeppeSN
 
JeppeSN's Avatar
 
"Jeppe"
Jan 2016
Denmark

B616 Posts
Default

You will be interested in the thread Integers n for which the digit sum of n exceeds the digit sum of n^5 in which I once wrote comments. /JeppeSN
JeppeSN is offline   Reply With Quote
Old 2022-08-10, 22:00   #4
mart_r
 
mart_r's Avatar
 
Dec 2008
you know...around...

34016 Posts
Default

Quote:
Originally Posted by slandrum View Post
I assume you are not looking for trivial answers like x = 0, or x = 10^y
Nope. I just phrased the question as simply as possible.

Quote:
Originally Posted by JeppeSN View Post
You will be interested in the thread Integers n for which the digit sum of n exceeds the digit sum of n^5 in which I once wrote comments. /JeppeSN
Interesting indeed! Thanks
I was in fact typing the question into mathstackexchange but didn't get this particular result... maybe also because I used "x" instead of "n".
mart_r is offline   Reply With Quote
Old 2022-08-11, 12:22   #5
Dr Sardonicus
 
Dr Sardonicus's Avatar
 
Feb 2017
Nowhere

137158 Posts
Default

I'm afraid I'm pretty much drawing a blank on this one. However, I did notice a couple of minor things.

If the base b = p, a prime number, or p > 2 is prime and b = p^e or 2*p^e, the multiplicative group of integers mod p is cyclic.

If gcd(x, b) = 1 the multiplicative order of x^n will depend to some extent on gcd(n, eulerphi(b)).

If the base b has at least two prime factors, there will be "automorphs" to the base b. If b = A*B with A > 1, B > 1, and gcd(A,B) = 1 then for any positive integer k, the Chinese Remainder Theorem insures that there is a simultaneous solution to the congruences

x == 0 (mod A^k), x == 1 (mod B^k). Such x are called base-b "automorphs." They are not relatively prime to the base b.

If x is a k-digit automorph, we have x^2 == x (mod b^k), and x ≠ 0, 1 (mod b). Note that b^k + 1 - x satisfies x == 1 (mod A^k) and x == 0 (mod B^k), so automorphs occur in pairs.

To the base b = 2*5, there are two automorphs, ...6259918212890625 and ...3740081787109376.

Now if n is odd, x^n - x is divisible by x^2 - x, so if x is a k-digit automorph, x^n and x will have the same last k digits. So if k > 1, n > 1 is odd, x is a k-digit automorph, and d is the number of digits in x^n, then d > k, so x^n will have a larger digit sum than x.
Dr Sardonicus is offline   Reply With Quote
Old 2022-08-11, 21:21   #6
mart_r
 
mart_r's Avatar
 
Dec 2008
you know...around...

15008 Posts
Default

Ah yes, the automorphs
I dug out some calculations of mine from 2010 - man, it's already been twelve years! -, where I found that if b is the product of exactly k distinct primes or prime powers, then there will be 2^k-2 automorphs in base b. I have a nice printout with the last 100 digits of automorphs of each base b <= 36. For example, for b = 30, there are six automorphs: ...LB2J6 / ...OH13A / ...E1Q7F / ...FS3MG / ...5CSQL / ...8IRAP.
I'm sure there are also extensive lists on the web, but it's late and I'm too lazy to dig them up...
mart_r is offline   Reply With Quote
Old 2022-08-20, 09:47   #7
mart_r
 
mart_r's Avatar
 
Dec 2008
you know...around...

26×13 Posts
Default Slightly off-track

All this talk about automorphs made me kinda hungry and I tried to find primes in the sequences A214882 and A214883.

In A214882, a(n) is prime for n = {3, 5, 7, 9, 11, 12, 14, 23, 41, 69, 78, 223, 264, 293, 294, 312, 314, 316, 555, 624, 917, 1750, 2183, 2266, 4858, ...}
In A214883, a(n) is prime for n = {1, 2, 3, 5, 6, 27, 191, 206, 220, 3499, ...}
(tested up to n=5000 in both cases)

What could we call them? Autoprimes? Primorphs? 10-adic automorph prime thingies?
mart_r is offline   Reply With Quote
Old 2022-10-06, 17:47   #8
mart_r
 
mart_r's Avatar
 
Dec 2008
you know...around...

26×13 Posts
Default Happy to be curious still

Just out of curiosity: would it impress anyone to have 100+ digit integers not ending in zero whose digital sum is larger than the digital sum of their cubes?
mart_r is offline   Reply With Quote
Old 2022-10-09, 17:21   #9
mart_r
 
mart_r's Avatar
 
Dec 2008
you know...around...

26·13 Posts
Default

Has even any work been done toward proving or disproving the infinitude of integers <> 0 mod b such that the digit sum in base b is larger than the digit sum of their cube?
I was working my way through some figures that suggest these numbers might be finite (at least for base 10; contrary to what I used to believe for a long time).

BTW, I noticed that there is a user on math.stackexchange named "Martin R" - that is not me, just in case someone is curious.
mart_r is offline   Reply With Quote
Old 2022-10-12, 19:54   #10
mart_r
 
mart_r's Avatar
 
Dec 2008
you know...around...

26·13 Posts
Default The internet is sooo overrated

Looking through various works concerning the sum of digits of different functions (a rather recent one being "The sum of digits of floor(nc)" by J. Morgenbesser: see here), the one I'm looking for still seems elusive. Probably I'm overlooking something. The papers all seem concerned with the distribution of digits rather than certain extreme cases:
For \(s_b(n)\) denoting the sum of digits for a given base \(b \geq 2\) and positive integer n, let \(\Delta_b(d) := \max(n<b^d, n \mod b \neq 0: s_b(n)-(s_b(n^3 \mod b^d)))\).
For \(d \to \infty\), \(\Delta_b(d)\) appears to exhibit a linear asymptotic behaviour: \(\Delta_b(d) \sim d \cdot v(b)\) for some variable \(v\) depending on \(b\):
Code:
v(2) ~ 0.78
v(3) ~ 1.57
v(4) ~ 2.4
v(5) ~ 3.3
v(6) ~ 4.1
v(7) ~ 4.9
v(8) ~ 5.7
v(9) ~ 6.6
v(10) ~ 7.57
v(20) ~ 16.7
v(30) ~ 25.8
v(40) ~ 35.4
v(100) ~ 92.5
e.g. \(\Delta_2(100) = 75\) (close to \(100 \cdot 0.78\)), where, for \(n=633825300114114700575210732923\), \(s_2(n)=92\) and \(s_2(n^3 \mod 2^{100})=17\).
Has the behaviour of \(v_b\) been looked at by anyone? The values are slightly different but similar for higher odd powers other than 3, from what I've calculated so far. Confusion, lost, 1 ea.

Quote:
Originally Posted by mart_r View Post
Just out of curiosity: would it impress anyone to have 100+ digit integers not ending in zero whose digital sum is larger than the digital sum of their cubes?
200+ digits? No?
mart_r is offline   Reply With Quote
Old 2022-10-15, 16:59   #11
mart_r
 
mart_r's Avatar
 
Dec 2008
you know...around...

26·13 Posts
Default

Quote:
Originally Posted by mart_r View Post
Has even any work been done toward proving or disproving the infinitude of integers <> 0 mod b such that the digit sum in base b is larger than the digit sum of their cube?
I was working my way through some figures that suggest these numbers might be finite (at least for base 10; contrary to what I used to believe for a long time).
Brain malfunction, sorry. Those integers should be infinite in number. If we find an example where \(\Delta = 7.57 d\) (in fact, \(\Delta = 5 d\) would suffice), it would take about \(10^{0.054 d}\) trials to find an example, if my logic doesn't fool me (again).
In contrast, for powers of 5, it would take \(\gt 10^{2.8 d}\) trials per \(10^d\), thus almost surely there is no integer where \(s_{10}(n) \gt s_{10}(n^5)\).
mart_r is offline   Reply With Quote
Reply

Thread Tools


Similar Threads
Thread Thread Starter Forum Replies Last Post
Raising groups of digits to powers, equalling itself Boltzmann brain Miscellaneous Math 2 2020-02-09 18:35
What powers your farm/cpu? Uncwilly Lounge 15 2010-03-31 07:13
Prime Powers plandon Math 7 2009-06-30 21:29
2^x using powers of e? nibble4bits Math 31 2007-12-11 12:56
Powers, and more powers Numbers Puzzles 3 2005-07-13 04:42

All times are UTC. The time now is 11:40.


Wed Nov 30 11:40:54 UTC 2022 up 104 days, 9:09, 0 users, load averages: 1.28, 1.08, 1.04

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2022, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.

≠ ± ∓ ÷ × · − √ ‰ ⊗ ⊕ ⊖ ⊘ ⊙ ≤ ≥ ≦ ≧ ≨ ≩ ≺ ≻ ≼ ≽ ⊏ ⊐ ⊑ ⊒ ² ³ °
∠ ∟ ° ≅ ~ ‖ ⟂ ⫛
≡ ≜ ≈ ∝ ∞ ≪ ≫ ⌊⌋ ⌈⌉ ∘ ∏ ∐ ∑ ∧ ∨ ∩ ∪ ⨀ ⊕ ⊗ 𝖕 𝖖 𝖗 ⊲ ⊳
∅ ∖ ∁ ↦ ↣ ∩ ∪ ⊆ ⊂ ⊄ ⊊ ⊇ ⊃ ⊅ ⊋ ⊖ ∈ ∉ ∋ ∌ ℕ ℤ ℚ ℝ ℂ ℵ ℶ ℷ ℸ 𝓟
¬ ∨ ∧ ⊕ → ← ⇒ ⇐ ⇔ ∀ ∃ ∄ ∴ ∵ ⊤ ⊥ ⊢ ⊨ ⫤ ⊣ … ⋯ ⋮ ⋰ ⋱
∫ ∬ ∭ ∮ ∯ ∰ ∇ ∆ δ ∂ ℱ ℒ ℓ
𝛢𝛼 𝛣𝛽 𝛤𝛾 𝛥𝛿 𝛦𝜀𝜖 𝛧𝜁 𝛨𝜂 𝛩𝜃𝜗 𝛪𝜄 𝛫𝜅 𝛬𝜆 𝛭𝜇 𝛮𝜈 𝛯𝜉 𝛰𝜊 𝛱𝜋 𝛲𝜌 𝛴𝜎𝜍 𝛵𝜏 𝛶𝜐 𝛷𝜙𝜑 𝛸𝜒 𝛹𝜓 𝛺𝜔