A281130 -id:A281130 - cp's OEIS frontend

A291598 a(n) = log_2(A281130(n)).

0, 0, 1, 0, 1, 1, 2, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 3, 4, 4, 5, 2, 3, 4, 2, 3, 3, 4, 2, 3, 2, 3, 4, 4, 5, 4, 5, 3, 4, 3, 4, 2, 3, 5, 2, 3, 3, 4, 3, 4, 4, 5, 4, 5, 2, 3, 4, 4, 5, 3, 4, 2, 3, 2, 3, 4, 4, 5, 4, 5, 3, 4, 3, 4, 4, 5, 5, 6

Offset: 1

Rok Cestnik, Aug 27 2017

nonn

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

Cf. A281130, A281131.

#include
#include
#include
int main(void){
   int N = 100;
   int *a = (int*)malloc((N+1)*sizeof(int));
   printf("1 0\n2 0\n");
   a[1] = 0;
   a[2] = 0;
   for(int i = 2; i < N; ++i){
      if(a[i-1] < a[i]) a[i+1] = a[i-(int)(pow(2,a[i]))];
      else a[i+1] = a[i]+1;
      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[Log2@ a@ # &, 105] (* Michael De Vlieger, Aug 29 2017 *)

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

A330772 a(n) = 1 for n<1; for n >= 0, a(n+1) = 2*a(n-a(n)).

Original entry on oeis.org

2, 2, 2, 4, 2, 4, 4, 4, 8, 4, 8, 4, 8, 4, 8, 8, 8, 16, 4, 16, 8, 16, 8, 16, 8, 16, 8, 8, 32, 2, 16, 16, 16, 16, 32, 4, 32, 4, 32, 8, 32, 16, 32, 16, 16, 64, 2, 32, 16, 32, 32, 8, 32, 16, 8, 4, 16, 64, 2, 32, 16, 32, 4, 4, 64, 4, 64, 4, 8, 32, 8, 8, 8, 128

Offset: 1

Rok Cestnik, Dec 30 2019

nonn

From the current term count back the same number of terms and double it to obtain the next term. Because a(n) can exceed n, negative indexes are also occasionally referenced.

			a(1) = 2*a(0-a(0)) = 2*a(-1) = 2.
a(2) = 2*a(1-a(1)) = 2*a(-1) = 2.
a(3) = 2*a(2-a(2)) = 2*a(0) = 2.
a(4) = 2*a(3-a(3)) = 2*a(1) = 4.

Cf. A281130, A070867, A004001.

a = [2]
for n in range(1000):
    if(a[n] > n):
        a.append(2)
    else:
        a.append(2*a[n-a[n]])

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;
}

A364197 a(n+1) = a(|n-a(n)^2|) + 1, a(0) = 0.

Original entry on oeis.org

0, 1, 1, 2, 2, 1, 3, 3, 2, 3, 1, 4, 2, 3, 3, 2, 5, 4, 2, 4, 3, 5, 3, 4, 4, 3, 6, 2, 5, 3, 4, 4, 3, 5, 3, 4, 5, 5, 3, 4, 5, 3, 4, 7, 4, 6, 4, 5, 4, 4, 6, 4, 5, 3, 5, 4, 5, 5, 4, 5, 4, 5, 6, 7, 4, 5, 6, 5, 5, 8, 2, 7, 4, 6, 6, 4, 6, 6, 4, 7, 5, 5, 6, 5, 5, 6, 5, 6, 5, 8, 4, 7, 5, 6, 6, 5, 3, 7, 5, 7

Offset: 0

Rok Cestnik, Jul 13 2023

nonn

			a(1) = a(|0-a(0)^2|)+1 = a(|0-0|)+1 = a(0)+1 = 1.
a(2) = a(|1-a(1)^2|)+1 = a(|1-1|)+1 = a(0)+1 = 1.
a(3) = a(|2-a(2)^2|)+1 = a(|2-1|)+1 = a(1)+1 = 2.
a(4) = a(|3-a(3)^2|)+1 = a(|3-4|)+1 = a(1)+1 = 2.
a(5) = a(|4-a(4)^2|)+1 = a(|4-4|)+1 = a(0)+1 = 1.

Michael De Vlieger, Table of n, a(n) for n = 0..10000
Rok Cestnik, Term-referencing tree for 1000 terms.

Cf. A364198 (indices of record highs).
Cf. A339929, A340224.
Cf. A281130, A330772, A340134, A002516, A003056, A004001, A005185.

a[0] = 0; a[n_] := a[n] = a[Abs[n - 1 - a[n - 1]^2]] + 1; Array[a, 100, 0] (* Amiram Eldar, Jul 13 2023 *)

a = [0]; [a.append(a[abs(n-a[n]**2)]+1) for n in range(100)]

a(n) ~ (3n)^(1/3) (conjectured).

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;
}

A330631 a(n+1) = a(n-a(n))+1 if a(n) > a(n-1), otherwise a(n+1) = 2*a(n); a(1) = a(2) = 1.

Original entry on oeis.org

1, 1, 2, 2, 4, 2, 4, 3, 6, 3, 6, 5, 10, 3, 6, 7, 7, 14, 3, 6, 4, 8, 4, 8, 8, 16, 4, 8, 7, 14, 8, 16, 8, 16, 15, 30, 3, 6, 17, 9, 18, 5, 10, 9, 18, 5, 10, 4, 8, 19, 9, 18, 17, 34, 7, 14, 6, 12, 6, 12, 5, 10, 19, 10, 20, 19, 38, 8, 16, 18, 19, 19, 38, 16, 32

Offset: 1

Rok Cestnik, Dec 21 2019

nonn

a(n) < n (with exception a(1) = 1). Proof: Suppose a(s) = s+x, x >= 0, is the first occurrence of a(n) >= n. From here we branch into two possibilities: Possibility #1: a(s-2) < a(s-1), from which it follows that a(s) = a(s-1-a(s-1))+1 and therefore a(s-1-a(s-1)) = s-1+x is an earlier example of a(n) >= n. Possibility #2: a(s-2) >= a(s-1) and the terms can be expressed as a(s-1) = (s+x)/2 and a(s-2) = (s+x)/2+y with y >= 0. From this it follows that a(s-2-((s+x)/2+y))+1 = (s+x)/2, which when simplified reveals that a((s+x)/2-2-x-y) = (s+x)/2-1 is an earlier example of a(n) >= n. Both possibilities lead to a contradiction of the first statement, therefore we conclude that there is no occurrence of a(n) >= n (with exception a(1) = 1).

Some numbers never seem to appear in the sequence; the smallest of these are 328, 329, 331, 332, 333, 445, 667, 668, 669, ...

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

See A281130 for a similar sequence.

Cf. A004001, A005229, A055748, A318281.

Nest[Append[#, If[Less @@ Take[#, -2], #[[Length@ # - #[[-1]] ]] + 1, 2 #[[-1]] ]] &, {1, 1}, 73] (* Michael De Vlieger, Dec 23 2019 *)

a = [1,1]
for n in range(1, 1000):
    if(a[n] > a[n-1]):
        a.append(a[n-a[n]]+1)
    else:
        a.append(2*a[n])

cp's OEIS Frontend

A291598 a(n) = log_2(A281130(n)).

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A281131 First appearance of 2^n in A281130.

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Extensions

A330772 a(n) = 1 for n<1; for n >= 0, a(n+1) = 2*a(n-a(n)).

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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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A364197 a(n+1) = a(|n-a(n)^2|) + 1, a(0) = 0.

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

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A330631 a(n+1) = a(n-a(n))+1 if a(n) > a(n-1), otherwise a(n+1) = 2*a(n); a(1) = a(2) = 1.

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