 mersenneforum.org Sum of digits of x and powers of x
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 Register FAQ Search Today's Posts Mark Forums Read 2022-08-10, 20:32 #1 mart_r   Dec 2008 you know...around... 32916 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 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?   2022-08-10, 21:10   #2
slandrum

Jan 2021
California

46010 Posts Quote:
 Originally Posted by mart_r 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   2022-08-10, 21:49 #3 JeppeSN   "Jeppe" Jan 2016 Denmark 2×7×13 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   2022-08-10, 22:00   #4
mart_r

Dec 2008
you know...around...

809 Posts Quote:
 Originally Posted by slandrum 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 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".   2022-08-11, 12:22 #5 Dr Sardonicus   Feb 2017 Nowhere 599710 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 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.   2022-08-11, 21:21 #6 mart_r   Dec 2008 you know...around... 809 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^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...   2022-08-20, 09:47 #7 mart_r   Dec 2008 you know...around... 809 Posts 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?  Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post Boltzmann brain Miscellaneous Math 2 2020-02-09 18:35 Uncwilly Lounge 15 2010-03-31 07:13 plandon Math 7 2009-06-30 21:29 nibble4bits Math 31 2007-12-11 12:56 Numbers Puzzles 3 2005-07-13 04:42

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