20220810, 20:32  #1 
Dec 2008
you know...around...
2·443 Posts 
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 followup 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 kalmost primes  beside the predominant term log(log(p))^k? 
20220810, 21:10  #2 
Jan 2021
California
1045_{8} Posts 

20220810, 21:49  #3 
"Jeppe"
Jan 2016
Denmark
2^{2}×47 Posts 
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

20220810, 22:00  #4  
Dec 2008
you know...around...
886_{10} Posts 
Quote:
Quote:
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". 

20220811, 12:22  #5 
Feb 2017
Nowhere
2^{4}·3·7·19 Posts 
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 baseb "automorphs." They are not relatively prime to the base b. If x is a kdigit 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 kdigit automorph, x^n and x will have the same last k digits. So if k > 1, n > 1 is odd, x is a kdigit 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. 
20220811, 21:21  #6 
Dec 2008
you know...around...
2×443 Posts 
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^k2 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... 
20220820, 09:47  #7 
Dec 2008
you know...around...
376_{16} Posts 
Slightly offtrack
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? 10adic automorph prime thingies? 
20221006, 17:47  #8 
Dec 2008
you know...around...
2·443 Posts 
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?

20221009, 17:21  #9 
Dec 2008
you know...around...
2×443 Posts 
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. 
20221012, 19:54  #10 
Dec 2008
you know...around...
2·443 Posts 
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(n^{c})" 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 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. 200+ digits? No? 
20221015, 16:59  #11  
Dec 2008
you know...around...
1101110110_{2} Posts 
Quote:
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 \(s_{10}(n) \gt s_{10}(n^5)\). 

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