A062039 -id:A062039 - cp's OEIS frontend

A193927 Numbers m such that A062039(m+1) < A062039(m).

3, 9, 13, 15, 19, 23, 34, 38, 42, 45, 50, 54, 57, 61, 71, 81, 92, 96, 99, 107, 120, 133, 150, 157, 161, 180, 190, 214, 234, 236, 246, 254, 261, 270, 273, 279, 292, 297, 301, 322, 329, 335, 339, 361, 386, 390, 396, 398, 419, 425, 427, 453, 463, 468, 478, 494

Offset: 1

Reinhard Zumkeller, Aug 10 2011

nonn

a(n) = - A193926(a(n));

complement of A193928, A062039(a(n)+1) < A062039(a(n)).

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

import Data.List (findIndices)
a193927 n = a193927_list !! (n-1)
a193927_list = findIndices (< 0) a193926_list

A193854 Numbers m such that A062039(m) = 1.

Original entry on oeis.org

0, 1, 4, 24, 191, 3096, 3728, 6320, 314102, 9645152, 187895163

Offset: 1

Reinhard Zumkeller, Aug 10 2011

A062039(a(n)) = 1.

import Data.List (elemIndices)
a193854 n = a193854_list !! (n-1)
a193854_list = elemIndices 1 a062039_list

a(10)-a(11) from Donovan Johnson, Aug 11 2011

A193926 First differences of A062039.

Original entry on oeis.org

0, 1, 2, -3, 1, 2, 1, 2, 5, -9, 4, 5, 12, -13, 7, -15, 4, 5, 12, -19, 2, 5, 12, -23, 1, 2, 1, 2, 5, 12, 1, 2, 5, 27, -34, 2, 5, 27, -38, 7, 7, 25, -42, 12, 24, -45, 12, 7, 7, 25, -50, 3, 24, 32, -54, 18, 32, -57, 7, 18, 32, -61, 1, 2, 5, 12, 1, 2, 5, 27, 15

Offset: 0

Reinhard Zumkeller, Aug 10 2011

sign

a(n) = A062039(n+1) - A062039(n);

a(A193927(n)) = - A193927(n).

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000

a193926 n = a193926_list !! n
a193926_list = zipWith (-) (tail a062039_list) a062039_list

A123643 Record values in A062039.

Original entry on oeis.org

1, 2, 4, 5, 7, 12, 24, 25, 27, 32, 59, 60, 69, 74, 86, 111, 126, 135, 139, 163, 222, 226, 365, 371, 372, 478, 540, 609, 795, 917, 965, 989, 1645, 2311, 2639, 2695, 2771, 2794, 3034, 3096, 3773, 4134, 4645, 4987, 5263, 5845, 6468, 8373, 11225, 11471, 12989

Offset: 0

Reinhard Zumkeller, Oct 04 2006

nonn

a(n) = A062039(A123644(n)).

A123644 Where record values occur in A062039.

Original entry on oeis.org

0, 2, 3, 7, 8, 9, 13, 31, 32, 33, 34, 42, 54, 71, 81, 92, 96, 133, 149, 150, 157, 214, 234, 254, 297, 335, 386, 396, 425, 542, 618, 656, 666, 759, 1098, 1294, 1378, 1481, 1487, 1549, 1707, 1771, 2004, 2044, 2303, 2528, 2776, 2852, 3700, 4467, 4580, 5453, 6514

Offset: 0

Reinhard Zumkeller, Oct 04 2006

nonn

A123643(n) = A062039(a(n)).

A193928 Numbers m such that A062039(m) <= A062039(m+1).

Original entry on oeis.org

0, 1, 2, 4, 5, 6, 7, 8, 10, 11, 12, 14, 16, 17, 18, 20, 21, 22, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 35, 36, 37, 39, 40, 41, 43, 44, 46, 47, 48, 49, 51, 52, 53, 55, 56, 58, 59, 60, 62, 63, 64, 65, 66, 67, 68, 69, 70, 72, 73, 74, 75, 76, 77, 78, 79, 80, 82

Offset: 1

Reinhard Zumkeller, Aug 10 2011

nonn

Complement of A193927, A062039(a(n)) <= A062039(a(n)+1).

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

A193854 is a subsequence.

import Data.List (findIndices)
a193928 n = a193928_list !! (n-1)
a193928_list = findIndices (0 <=) a193926_list

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

Rok Cestnik, Dec 23 2020

nonn

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.

			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.

Rok Cestnik, Table of n, a(n) for n = 1..1000
Rok Cestnik, Term-referencing tree for 1000 terms
Rok Cestnik, Program for plotting the term-referencing tree

Analogous sequences: A339930, A339931, A339932.

Cf. A330772, A005206, A002516, A062039.

#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;
}

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

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

Rok Cestnik, Dec 23 2020

nonn

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).

			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.

Analogous sequences: A339929, A339931, A339932.

Cf. A330772, A005206, A002516, A062039.

#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;
}

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

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

Rok Cestnik, Dec 23 2020

nonn

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).

			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.

Analogous sequences: A339929, A339930, A339932.

Cf. A330772, A005206, A002516, A062039.

#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;
}

a = [0, 0, 0, 0]
for n in range(3, 1000):
    a.append(a[n-3-a[n]**2]+1)

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

Rok Cestnik, Dec 23 2020

nonn

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).

			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.

Analogous sequences: A339929, A339930, A339931.

Cf. A330772, A005206, A002516, A062039.

#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;
}

a = [0, 0, 0, 0, 0]
for n in range(4, 1000):
    a.append(a[n-4-a[n]**2]+1)

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

cp's OEIS Frontend

A193927 Numbers m such that A062039(m+1) < A062039(m).

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Haskell

A193854 Numbers m such that A062039(m) = 1.

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Haskell

Extensions

A193926 First differences of A062039.

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Haskell

A123643 Record values in A062039.

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A123644 Where record values occur in A062039.

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A193928 Numbers m such that A062039(m) <= A062039(m+1).

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Haskell

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

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C

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Formula

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

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C

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Formula

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

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C

Python

Formula

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

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