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A339932 a(n+1) = a(n-4-a(n)^2) + 1, starting with a(1) = a(2) = a(3) = a(4) = a(5) = 0. 3
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,12

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).

LINKS

Table of n, a(n) for n=1..60.

FORMULA

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

EXAMPLE

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.

PROG

(Python)

a = [0, 0, 0, 0, 0]

for n in range(4, 1000):

a.append(a[n-4-a[n]**2]+1)

(C)

#include<stdio.h>

#include<stdlib.h>

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;

}

CROSSREFS

Analogous sequences: A339929, A339930, A339931.

Cf. A330772, A005206, A002516, A062039.

Sequence in context: A036485 A331971 A030547 * A254690 A156642 A155124

Adjacent sequences: A339929 A339930 A339931 * A339933 A339934 A339935

KEYWORD

nonn

AUTHOR

Rok Cestnik, Dec 23 2020

STATUS

approved

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Last modified December 2 11:09 EST 2022. Contains 358493 sequences. (Running on oeis4.)