A291598 -id:A291598 - cp's OEIS frontend

A281130 a(n+1) = a(n-a(n)) if a(n-1) < a(n), otherwise a(n+1) = 2*a(n); a(1) = a(2) = 1.

1, 1, 2, 1, 2, 2, 4, 2, 4, 2, 4, 4, 8, 2, 4, 4, 8, 4, 8, 4, 8, 8, 16, 4, 8, 8, 16, 4, 8, 8, 16, 4, 8, 8, 16, 8, 16, 8, 16, 16, 32, 4, 8, 16, 4, 8, 8, 16, 4, 8, 4, 8, 16, 16, 32, 16, 32, 8, 16, 8, 16, 4, 8, 32, 4, 8, 8, 16, 8, 16, 16, 32, 16, 32, 4, 8, 16, 16, 32, 8

Offset: 1

Rok Cestnik, Jan 15 2017

nonn

The sequence is well defined, namely a(n) < n (with the exception of a(1)). Proof: Suppose a(s) = s+m, with m >= 0, is the first occurrence of a(n) >= n. It follows that a(s-1) = a(s-2) = (s+m)/2 and from there it follows that a(s-2-(s+m)/2) = (s+m)/2, but s-2-(s+m)/2 < (s+m)/2 which is a contradiction to the first statement. - Rok Cestnik, Aug 29 2017
From Robert G. Wilson v, Jan 16 2017: (Start)
a(n) = 2^k, k >= 0.
a(n)=1 for n: 1, 2, 4;
a(n)=2 for n: 3, 5, 6, 8, 10, 14;
a(n)=4 for n: 7, 9, 11, 12, 15, 16, 18, 20, 24, 28, 32, 42, 45, 49, 51, 62, 65, 75, 82, 84, 101, 108, 110, 118, 127, 175, 240;
First occurrence of 2^k (A281131): 1, 3, 7, 13, 23, 41, 98, 223, 437, 699, 1213, 2624, 4674, 11163, 21300, 40858, 73977, etc.;
Last occurrence of 2^k: 4, 14, 240, 1314, 10565, 35893, 62417, 638149, 2030926, etc.;
Number of occurrences of 2^k: 3, 6, 27, 77, 167, 330, 706, 1756, 3811, etc.
(End)

			a(3) = 2*a(2) = 2 because a(1) !< a(2).
a(4) = a(3-a(3)) = 1 because a(2) < a(3).

Rok Cestnik, Table of n, a(n) for n = 1..10000
Rok Cestnik, Self-referencing visualization

Cf. A281131, A291598.

#include
#include
int main(void){
   int N = 1000;
   int *a = (int*)malloc((N+1)*sizeof(int));
   printf("1 1\n2 1\n");
   a[1] = 1;
   a[2] = 1;
   for(int i = 2; i < N; ++i){
      if(a[i-1] < a[i]) a[i+1] = a[i-a[i]];
      else a[i+1] = 2*a[i];
      printf("%d %d\n", i+1, a[i+1]);
   }
   return 0;
}

a[n_] := a[n] = If[a[n -2] < a[n -1], a[n -1 -a[n -1]], 2 a[n -1]]; a[1] = a[2] = 1; Array[a, 100]

A281131 First appearance of 2^n in A281130.

Original entry on oeis.org

1, 3, 7, 13, 23, 41, 98, 223, 437, 699, 1213, 2624, 4674, 11163, 21300, 40858, 73977, 148591, 297394, 567076, 1100738, 2243474, 4340628, 8726122, 17397270, 34701556, 68372147, 136254352, 271069771, 546613630, 1088921640, 2163138108, 4334318825

Offset: 0

Rok Cestnik, Jan 15 2017

Conjecture: a(n) ~ 2^n.

			a(1) = 3 because A281130(3) = 2^1 and A281130(i) != 2^1 for i < 3.
a(2) = 7 because A281130(7) = 2^2 and A281130(i) != 2^2 for i < 7.

Rok Cestnik, C program (utilizing hard drive memory)

Cf. A281130, A291598.

#include
#include
int main(void){
   int N = 1000000000;
   int *a = (int*)malloc((N+1)*sizeof(int));
   int max = 1;
   int maxindex = 1;
   a[1] = 1;
   a[2] = 1;
   for(int i = 2; i < N; ++i){
      if(a[i-1] < a[i]) a[i+1] = a[i-a[i]];
      else a[i+1] = 2*a[i];
      if(a[i+1] > max){
         max = a[i+1];
         printf("%d %d\n", maxindex, i+1);
         maxindex++;
      }
   }
   return 0;
}

a[n_] := a[n] = If[a[n - 2] < a[n - 1], a[n - 1 - a[n - 1]], 2 a[n - 1]]; a[1] = a[2] = 1; Function[w, Function[e, First /@ Lookup[w, 2^e]]@ Range[0, Log2@ Max@ Keys@ w]]@ PositionIndex@ Array[a, 10^7] (* Michael De Vlieger, Jan 21 2017, Version 10. *)

a(31)-a(32) from Rok Cestnik, Aug 27 2017

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.

Original entry on oeis.org

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

Rok Cestnik, Aug 23 2018

nonn

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.

			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).

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

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

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

#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;
}

A318282 a(n) = (A318281(n) - 1)/3.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 1, 0, 1, 1, 2, 0, 2, 0, 2, 0, 2, 1, 0, 1, 0, 1, 1, 2, 2, 3, 0, 3, 1, 2, 1, 0, 1, 1, 2, 3, 3, 4, 2, 0, 2, 1, 4, 2, 3, 2, 2, 3, 4, 3, 0, 3, 1, 4, 2, 3, 2, 3, 3, 4, 2, 4, 4, 5, 3, 2, 3, 2, 2, 3, 4, 3, 4, 4, 5

Offset: 1

Rok Cestnik, Aug 23 2018

nonn

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

Cf. A318281, A281130, A291598, A004001, A005185, A005229.

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

cp's OEIS Frontend

A281130 a(n+1) = a(n-a(n)) if a(n-1) < a(n), otherwise a(n+1) = 2*a(n); a(1) = a(2) = 1.

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A281131 First appearance of 2^n in A281130.

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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.

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A318282 a(n) = (A318281(n) - 1)/3.

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