cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-7 of 7 results.

A339929 a(n+1) = a(n-1-a(n)^2) + 1, starting with a(1) = a(2) = 0.

Original entry on oeis.org

0, 0, 1, 1, 1, 2, 1, 2, 2, 2, 2, 3, 1, 3, 2, 3, 3, 2, 2, 4, 2, 4, 2, 3, 4, 3, 4, 3, 3, 3, 5, 2, 5, 2, 4, 3, 4, 5, 4, 5, 4, 4, 5, 4, 5, 3, 4, 4, 6, 4, 6, 4, 5, 5, 4, 6, 3, 5, 3, 7, 3, 7, 4, 4, 5, 5, 6, 4, 7, 3, 8, 3, 8, 3, 5, 7, 4, 8, 2, 4, 5, 5, 7, 6, 5, 4, 8, 5, 8, 4, 9, 3, 6, 7, 5
Offset: 1

Views

Author

Rok Cestnik, Dec 23 2020

Keywords

Comments

To obtain the next term, square the current term and add 1, then count back this number and add 1.
The sequence cannot repeat. Proof: Assume a finite period. Label an arbitrary term in the period x. Because of the back-referencing definition it follows that x-1 has to be in the period, and by the same argument so does x-2 and x-3, x-4, ... until 0. But it is not possible to obtain new 0s since each new term is larger than one already existing term.
Every positive integer appears in the sequence.
First occurrence of n: 1, 3, 6, 12, 20, 31, 49, 60, 71, 91, 129, 163, 214, 265, 303, 354, 516, 594, 792, 915, ...
The sequence appears to grow with the cube root of n, which is expected since f(x) = (3*x)^(1/3) satisfies the definition for large x, i.e., lim_{x->oo} f(x+1)-(f(x-1-f(x)^2)+1) = 0.
The width of the distribution of terms within a range (n^2,n^2+n) appears to be constant for large n and can be defined as: lim_{n->oo} ( 1/n*Sum_{k=n..2n} ( Max_{i=k^2..k^2+k} a(i) - Min_{i=k^2..k^2+k} a(i) ) ) and evaluates to 12.98... (for n^2 = 5*10^8).
Each term is determined by referencing an earlier term. Following the chain of these references we can uniquely trace back every term to one of the two initial zeros, e.g., a(60) -> a(49) -> a(31) -> a(20) -> a(14) -> a(11) -> a(5) -> a(2). These progressions form two trees, with the zeros being at their roots (for visualization see the Links section). The average branching factor B (number of links divided by the number of non-leaf nodes) is numerically evaluated to B = 1.51803... (for n = 10^8). Considering the asymptotic behavior of the sequence a(n) ~ (3*n)^(1/3), we can conclude that the number of links within a range (n,m) is asymptotically equal to m-n and therefore B is the inverse of the proportion of non-leaf terms (B = 1.518... then implies that roughly 34% of all terms never get referenced).
An extended definition can be considered that only requires one initial value: a(0) = -1, and the next terms are obtained via a(n+1) = a(n-1-a(n)*|a(n)|)+1.

Examples

			a(3) = a(2-1-a(2)^2)+1 = a(1)+1 = 1.
a(4) = a(3-1-a(3)^2)+1 = a(1)+1 = 1.
a(5) = a(4-1-a(4)^2)+1 = a(2)+1 = 1.
a(6) = a(5-1-a(5)^2)+1 = a(3)+1 = 2.

Links

Crossrefs

Analogous sequences: A339930, A339931, A339932.
Cf. A330772, A005206, A002516, A062039.

Programs

  • C
    #include
#include
int main(void){
    int N = 1000;
    int *a = (int*)malloc(N*sizeof(int));
    a[0] = 0;
    a[1] = 0;
    for(int n = 1; n < N-1; ++n){
        a[n+1] = a[n-1-a[n]*a[n]]+1;
    }
    free(a);
    return 0;
}
  • Python
    a = [0,0]
for n in range(1,1000):
    a.append(a[n-1-a[n]**2]+1)

Formula

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

A339930 a(n+1) = a(n-2-a(n)^2) + 1, starting with a(1) = a(2) = a(3) = 0.

Original entry on oeis.org

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

Views

Author

Rok Cestnik, Dec 23 2020

Keywords

Comments

To obtain the next term, square the current term and add 2, then count back this number and add 1.
The sequence cannot repeat. Proof: Assume a finite period. Label an arbitrary term in the period x. Because of the back-referencing definition it follows that x-1 has to be in the period, and by the same argument so does x-2 and x-3, x-4,... until 0. But it is not possible to obtain new 0s since each new term is larger than one already existing term.
Every positive integer appears in the sequence.
First occurrence of n: 1, 4, 8, 15, 22, 34, 50, 69, 108, 171, 210, 277, 376, 464, 567, 670, 775, 993,...
The sequence appears to grow with the cube root of n, which is expected since f(x) = (3*x)^(1/3) satisfies the definition for large x, i.e. lim_{x->oo} f(x+1)-(f(x-2-f(x)^2)+1) = 0.
The width of the distribution of terms within a range (n^2,n^2+n) appears to be constant for large n and can be defined as: lim_{n->oo} ( 1/n*Sum_{k=n..2n} ( Max_{i=k^2..k^2+k} a(i) - Min_{i=k^2..k^2+k} a(i) ) ) and evaluates to 7.41... (for n^2 = 5*10^8).

Examples

			a(4) = a(3-2-a(3)^2)+1 = a(1)+1 = 1.
a(5) = a(4-2-a(4)^2)+1 = a(1)+1 = 1.
a(6) = a(5-2-a(5)^2)+1 = a(2)+1 = 1.
a(7) = a(6-2-a(6)^2)+1 = a(3)+1 = 1.
a(8) = a(7-2-a(7)^2)+1 = a(4)+1 = 2.

Crossrefs

Analogous sequences: A339929, A339931, A339932.
Cf. A330772, A005206, A002516, A062039.

Programs

  • C
    #include
#include
int main(void){
    int N = 1000;
    int *a = (int*)malloc(N*sizeof(int));
    a[0] = 0;
    a[1] = 0;
    a[2] = 0;
    for(int n = 2; n < N-1; ++n){
        a[n+1] = a[n-2-a[n]*a[n]]+1;
    }
    free(a);
    return 0;
}
  • Python
    a = [0, 0, 0]
for n in range(2, 1000):
    a.append(a[n-2-a[n]**2]+1)

Formula

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

A339931 a(n+1) = a(n-3-a(n)^2) + 1, starting with a(1) = a(2) = a(3) = a(4) = 0.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 1, 2, 2, 2, 2, 2, 2, 3, 2, 3, 2, 3, 3, 2, 3, 3, 3, 3, 3, 3, 4, 3, 4, 3, 4, 3, 3, 4, 3, 4, 3, 4, 4, 3, 4, 4, 4, 4, 4, 4, 5, 4, 5, 4, 5, 4, 4, 5, 4, 5
Offset: 1

Views

Author

Rok Cestnik, Dec 23 2020

Keywords

Comments

To obtain the next term, square the current term and add 3, then count back this number and add 1.
The sequence cannot repeat. Proof: Assume a finite period. Label an arbitrary term in the period x. Because of the back-referencing definition it follows that x-1 has to be in the period, and by the same argument so does x-2 and x-3, x-4,... until 0. But it is not possible to obtain new 0s since each new term is larger than one already existing term.
Every positive integer appears in the sequence.
First occurrence of n: 1, 5, 10, 18, 31, 51, 71, 100, 140, 221, 283, 351, 487, 612, 737, 885,...
The sequence appears to grow with the cube root of n, which is expected since f(x) = (3*x)^(1/3) satisfies the definition for large x, i.e. lim_{x->oo} f(x+1)-(f(x-3-f(x)^2)+1) = 0.
The width of the distribution of terms within a range (n^2,n^2+n) appears to be constant for large n and can be defined as: lim_{n->oo} ( 1/n*Sum_{k=n..2n} ( Max_{i=k^2..k^2+k} a(i) - Min_{i=k^2..k^2+k} a(i) ) ) and evaluates to 7.40... (for n^2 = 5*10^8).

Examples

			a(5) = a(4-3-a(4)^2)+1 = a(1)+1 = 1.
a(6) = a(5-3-a(5)^2)+1 = a(1)+1 = 1.
a(7) = a(6-3-a(6)^2)+1 = a(2)+1 = 1.
a(8) = a(7-3-a(7)^2)+1 = a(3)+1 = 1.
a(9) = a(8-3-a(8)^2)+1 = a(4)+1 = 1.
a(10) = a(9-3-a(9)^2)+1 = a(5)+1 = 2.

Crossrefs

Analogous sequences: A339929, A339930, A339932.
Cf. A330772, A005206, A002516, A062039.

Programs

  • C
    #include
#include
int main(void){
    int N = 1000;
    int *a = (int*)malloc(N*sizeof(int));
    a[0] = 0;
    a[1] = 0;
    a[2] = 0;
    a[3] = 0;
    for(int n = 3; n < N-1; ++n){
        a[n+1] = a[n-3-a[n]*a[n]]+1;
    }
    free(a);
    return 0;
}
  • Python
    a = [0, 0, 0, 0]
for n in range(3, 1000):
    a.append(a[n-3-a[n]**2]+1)

Formula

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

A339932 a(n+1) = a(n-4-a(n)^2) + 1, starting with a(1) = a(2) = a(3) = a(4) = a(5) = 0.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 2, 1, 2, 2, 2, 2, 2, 2, 2, 3, 2, 3, 2, 3, 3, 2, 3, 3, 3, 3, 3, 3, 3, 4, 3, 4, 3, 4, 3, 3, 4, 3, 4, 3, 4, 4, 3, 5, 3, 5, 3, 5, 3, 4, 5, 3, 5, 4, 5
Offset: 1

Views

Author

Rok Cestnik, Dec 23 2020

Keywords

Comments

To obtain the next term, square the current term and add 4, then count back this number and add 1.
The sequence cannot repeat. Proof: Assume a finite period. Label an arbitrary term in the period x. Because of the back-referencing definition it follows that x-1 has to be in the period, and by the same argument so does x-2 and x-3, x-4,... until 0. But it is not possible to obtain new 0s since each new term is larger than one already existing term.
Every positive integer appears in the sequence.
First occurrence of n: 1, 6, 12, 21, 35, 49, 70, 100, 130, 171, 212, 266, 320, 406, 564, 669, 849,...
The sequence appears to grow with the cube root of n, which is expected since f(x) = (3*x)^(1/3) satisfies the definition for large x, i.e. lim_{x->oo} f(x+1)-(f(x-4-f(x)^2)+1) = 0.
The width of the distribution of terms within a range (n^2,n^2+n) appears to be constant for large n and can be defined as: lim_{n->oo} ( 1/n*Sum_{k=n..2n} ( Max_{i=k^2..k^2+k} a(i) - Min_{i=k^2..k^2+k} a(i) ) ) and evaluates to 8.10... (for n^2 = 5*10^8).

Examples

			a(6) = a(5-4-a(5)^2)+1 = a(1)+1 = 1.
a(7) = a(6-4-a(6)^2)+1 = a(1)+1 = 1.
a(8) = a(7-4-a(7)^2)+1 = a(2)+1 = 1.
a(9) = a(8-4-a(8)^2)+1 = a(3)+1 = 1.
a(10) = a(9-4-a(9)^2)+1 = a(4)+1 = 1.
a(11) = a(10-4-a(10)^2)+1 = a(5)+1 = 1.
a(12) = a(11-4-a(11)^2)+1 = a(6)+1 = 2.

Crossrefs

Analogous sequences: A339929, A339930, A339931.
Cf. A330772, A005206, A002516, A062039.

Programs

  • C
    #include
#include
int main(void){
    int N = 1000;
    int *a = (int*)malloc(N*sizeof(int));
    a[0] = 0;
    a[1] = 0;
    a[2] = 0;
    a[3] = 0;
    a[4] = 0;
    for(int n = 4; n < N-1; ++n){
        a[n+1] = a[n-4-a[n]*a[n]]+1;
    }
    free(a);
    return 0;
}
  • Python
    a = [0, 0, 0, 0, 0]
for n in range(4, 1000):
    a.append(a[n-4-a[n]**2]+1)

Formula

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

A340134 a(n+1) = a(n-2*a(n)) + 1, starting with a(1) = a(2) = 0.

Original entry on oeis.org

0, 0, 1, 1, 1, 2, 1, 2, 2, 2, 3, 2, 3, 2, 3, 3, 3, 4, 3, 4, 3, 4, 3, 4, 4, 4, 5, 4, 5, 4, 5, 4, 5, 4, 5, 5, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 6, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 7, 7, 8, 7, 8, 7, 8, 7, 8, 7, 8, 7, 8, 7, 8, 7, 8, 8, 8, 9, 8, 9, 8, 9
Offset: 1

Views

Author

Rok Cestnik, Dec 29 2020

Keywords

Examples

			a(3) = a(2-2*a(2))+1 = a(2)+1 = 1.
a(4) = a(3-2*a(3))+1 = a(1)+1 = 1.
a(5) = a(4-2*a(4))+1 = a(2)+1 = 1.
a(6) = a(5-2*a(5))+1 = a(3)+1 = 2.

Crossrefs

For a(n+1) = a(n-a(n)) + 1, starting with a(1) = 0, see A003056.
Cf. A339929, A330772, A005206.

Programs

  • Mathematica
    Table[Floor[Sqrt[n-1]] + ((-1)^(n+Floor[Sqrt[n]])-1)/2,{n,87}] (* Stefano Spezia, Dec 29 2020 *)
  • Python
    a = [0, 0]
for n in range(1, 1000):
    a.append(a[n-2*a[n]]+1)

Formula

a(n) = floor(sqrt(n-1)) + ((-1)^(n+floor(sqrt(n)))-1)/2.

Extensions

More terms from Stefano Spezia, Dec 29 2020

A340224 a(n+1) = a(n-a(n)*(a(n)-1)/2) + 1, starting with a(1) = 0.

Original entry on oeis.org

0, 1, 2, 2, 3, 2, 4, 1, 2, 2, 3, 2, 4, 5, 3, 3, 5, 5, 2, 6, 4, 4, 4, 6, 3, 5, 4, 5, 6, 6, 4, 4, 6, 6, 3, 5, 6, 5, 6, 7, 3, 6, 5, 7, 5, 4, 8, 3, 6, 7, 7, 7, 5, 6, 7, 7, 4, 8, 7, 6, 6, 5, 8, 4, 9, 7, 6, 8, 8, 4, 5, 7, 8, 6, 8, 9, 8, 7, 5, 9, 8, 6, 7, 6, 9, 7, 10, 7, 7, 9, 7, 5, 7, 8, 8
Offset: 1

Views

Author

Rok Cestnik, Jan 01 2021

Keywords

Comments

To obtain the next term, compute the binomial(a,2) with the current term, then count back its value and add 1.
The sequence cannot repeat. Proof: Assume a finite period. Label an arbitrary term in the period x. Because of the back-referencing definition it follows that x-1 has to be in the period, and by the same argument so does x-2 and x-3, x-4, ... until 0. But it is not possible to obtain new 0s since each new term is larger than one already existing term.
Every positive integer appears in the sequence.
First occurrence of n: 1, 2, 3, 5, 7, 14, 20, 40, 47, 65, 87, 124, 170, 210, 258, 310, 389, 469, 548, 640, 758, 864, ...
The sequence appears to grow with the cube root of n, which is expected since f(x) = (6*x)^(1/3) satisfies the definition for large x, i.e., lim_{x->oo} f(x+1)-(f(x-f(x)*(f(x)-1)/2)+1) = 0.
Each term is determined by referencing an earlier term. Following the chain of these references we can uniquely trace back every term to the initial zero, e.g., a(40) -> a(24) -> a(17) -> a(13) -> a(11) -> a(9) -> a(8) -> a(1). These progressions form a tree, with the zero being at its root (for visualization see the Links section). The average branching factor B (number of links divided by the number of non-leaf nodes) is numerically evaluated to B = 1.380... (for n = 10^8). Considering the asymptotic behavior of the sequence a(n) ~ (6*n)^(1/3), we can expect that the number of links within a range (n,m) is asymptotically equal to m-n and therefore B is the inverse of the proportion of non-leaf terms (B = 1.380... then implies that roughly 27.5% of all terms never get referenced).

Examples

			a(2) = a(1-a(1)*(a(1)-1)/2)+1 = a(1)+1 = 1.
a(3) = a(2-a(2)*(a(2)-1)/2)+1 = a(2)+1 = 2.
a(4) = a(3-a(3)*(a(3)-1)/2)+1 = a(2)+1 = 2.
a(5) = a(4-a(4)*(a(4)-1)/2)+1 = a(3)+1 = 3.

Links

Crossrefs

Related to A339929.
Cf. A330772, A340134, A005206, A002516.

Programs

  • Python
    a = [0]
for n in range(1000):
    a.append(a[int(n-a[n]*(a[n]-1)/2)]+1)

Formula

a(n) ~ (6*n)^(1/3) (conjectured).

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

Views

Author

Rok Cestnik, Jul 13 2023

Keywords

Examples

			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.

Links

Crossrefs

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

Programs

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

Formula

a(n) ~ (3n)^(1/3) (conjectured).
Showing 1-7 of 7 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.