OFFSET
1,2
COMMENTS
After 12427560 terms the sequence is periodic with period 2985984. The largest periodic term is 158.
The largest term is 159 which appears only 3 times: a(4942686), a(7756992) and a(9818928).
If banning for n terms the sequence just repeats [1,2,1,3] (A364447). If banning for n^3 terms the sequence continues growing forever (A364449).
The first 41 terms are shared with A028920.
LINKS
Rok Cestnik, Table of n, a(n) for n = 1..10000
FORMULA
a(n) = a(n-2985984) for n >= 15413545.
EXAMPLE
a(n) ban 1 2 3 4 5 6 7 ...
1 | | | | | | |
2 x | | | | | |
1 | x | | | | |
3 x x | | | | |
1 | x x | | | |
4 x x x | | | |
1 | | x x | | |
2 x | x x | | |
1 | x x x | | |
5 x x x x | | |
1 | x x x x | |
6 x x x x x | |
1 | | x x x x |
2 x | | x x x |
1 | x | x x x |
3 x x | x x x |
1 | x x x x x |
7 x x x x x x |
.
.
.
PROG
(Python)
a = []
ban = [0 for n in range(160)]
for i in range(1000):
can = ban.index(0, 1)
ban = [max(b-1, 0) for b in ban]
a.append(can)
ban[can] = can**2
(C)
#include<stdlib.h>
int main(void){
int N = 1000;
int *a = (int*)malloc((N+1)*sizeof(int));
int *ban = (int*)malloc(160*sizeof(int));
for(int n = 0; n < N; ++n){
for(int can = 1; can < 160; ++can){
if(ban[can] == 0){
a[n] = can;
ban[can] = can*can+1;
break;
}
}
for(int can = 1; can < 160; ++can) if(ban[can]) --ban[can];
}
free(a); free(ban);
return 0;
}
CROSSREFS
KEYWORD
nonn,look
AUTHOR
Rok Cestnik, Jul 25 2023
STATUS
approved