2019-04-02, 13:39 | #1 |
Mar 2018
3^{2}×59 Posts |
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? |
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