Go Back > Extra Stuff > Blogorrhea > enzocreti

Thread Tools
Old 2019-04-02, 13:39   #1
Mar 2018

2·5·53 Posts
Default recursive relation

let be c(x) the function that counts the composite numbers less or equal to x.
the number 128 is a record because no composite less than 128 generates more composite numbers with this algorithm:
if x-c(x) is 1 or prime stop. Otherwise
x-c(x) is a composite m and you repeat the algorithm with m.
128-c(128)=32, so repeat the algorhitm 32-c(32)=12 which is composite so repeat with 12...12-c(12)=6 which is composite so repeat 6-c(6)=4 which is composite so repeat and have 4-c(4)=2 which is prime so stop. so starting from 128 you goes to 32 to 12 to 6 to 4 and to 2. 128 goes to 32 to 12 to 6 to 4 to 2. the lenght is 5 steps for reaching a prime. so 128 is a record because all the composites below 128 have a lenght chain <5. with pari gp i found that the records are 4,6,12,32,128,710,.... I Call a(1)=4, a(2)=6, a(3)=12 the terms of the sequence. My question is why the terms of the sequence satisfy the recursive relation a(n+1)=p_(a(n)-1)? where p is the (a(n)-1)-th prime.
Any formal proof of that?
enzocreti is offline   Reply With Quote

Thread Tools

Similar Threads
Thread Thread Starter Forum Replies Last Post
Yafu: relation packing LaurV YAFU 6 2017-08-06 08:33
error: cannot locate relation cardmaker Factoring 16 2017-07-17 12:38
The relation of time taken for NFS and CPU/Hardware didgogns Factoring 2 2017-06-11 18:04
Error reading relation jux YAFU 24 2016-02-13 10:43
Recursive Primes? jinydu Lounge 18 2004-06-23 02:36

All times are UTC. The time now is 08:03.

Thu Dec 9 08:03:56 UTC 2021 up 139 days, 2:32, 0 users, load averages: 1.49, 1.60, 1.45

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2021, 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.