A318281 a(n+1) = a(n-a(n)) if a(n-1) != a(n), otherwise a(n+1) = a(n) + 3; a(1) = a(2) = a(3) = a(4) = 1.

1, 1, 1, 1, 4, 1, 4, 1, 4, 4, 7, 1, 7, 1, 7, 1, 7, 4, 1, 4, 1, 4, 4, 7, 7, 10, 1, 10, 4, 7, 4, 1, 4, 4, 7, 10, 10, 13, 7, 1, 7, 4, 13, 7, 10, 7, 7, 10, 13, 10, 1, 10, 4, 13, 7, 10, 7, 10, 10, 13, 7, 13, 13, 16, 10, 7, 10

Offset: 1

Author

Rok Cestnik, Aug 23 2018

Keywords

• nonn

Comments

Sequences with an analogous condition a(n+1) = a(n) + s for s != 3 tend towards repetitive patterns:

for even values of s this is obvious, e.g.:

s = 2: 1,1,1,3,1,3,1,3,... (1,3 repeats)

s = 4: 1,1,1,1,1,5,1,5,1,5,1,5,... (1,5 repeats)

for odd values of s it has been checked up to s <= 19:

s = 1: 1,1,2,1,2,2,3,1,3,2,1,2,2,3,1,3,... (2,1,2,2,3,1,3 repeats)

s = 5: settles to a repetitive pattern of 192 terms

s = 7: settles to a repetitive pattern of 25 terms

s = 9: settles to a repetitive pattern of 31 terms

s = 11: settles to a repetitive pattern of 37 terms

s = 13: settles to a repetitive pattern of 43 terms

s = 15: settles to a repetitive pattern of 49 terms

s = 17: settles to a repetitive pattern of 55 terms

s = 19: settles to a repetitive pattern of 61 terms

It appears that for further values of s such sequences settle to a repetitive pattern of 4 + 3*s terms.

Examples

			a(5) = a(4) + 3 = 4, because a(3) == a(4).
a(6) = a(5-a(5)) = a(1) = 1, because a(4) != a(5).
		

Links

• Rok Cestnik, Table of n, a(n) for n = 1..9999

Crossrefs

See A318282 for (a(n)-1)/3.

Cf. A281130, A291598, A004001, A005085, A005229.

Programs

C

#include
#include
#include
int main(void){
  int N = 100; //number of terms
  int *a = (int*)malloc((N+1)*sizeof(int));
  printf("1 1\n2 1\n3 1\n4 1\n");
  a[1] = 1;
  a[2] = 1;
  a[3] = 1;
  a[4] = 1;
  for(int i = 4; i < N; ++i){
    if(a[i-1] != a[i]) a[i+1] = a[i-a[i]];
    else a[i+1] = a[i]+3;
    printf("%d %d\n", i+1, a[i+1]);
  }
  free(a);
  return 0;
}