20220810, 20:32  #1 
Dec 2008
you know...around...
2^{4}·53 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
11×47 Posts 

20220810, 21:49  #3 
"Jeppe"
Jan 2016
Denmark
2^{3}×23 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...
2^{4}×53 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
14101_{8} 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^{4}×53 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...
350_{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^{4}·53 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^{4}·53 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^{4}×53 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...
848_{10} 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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