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%I #44 Jun 13 2022 19:33:53
%S 1,1,2,1,3,2,4,1,5,3,6,2,7,8,4,9,1,10,11,5,12,3,13,14,6,15,2,16,17,7,
%T 18,19,8,20,21,4,22,23,24,9,25,1,26,27,28,10,29,30,31,11,32,33,5,34,
%U 35,36,12,37,3,38,39,40,13,41,42,43,14,44,45,6,46,47,48,15,49,2,50,51,52,53,16,54,55,56,57,17,58,59,7,60,61,62,63,18,64,65,66
%N Removing the first occurrence of 1, 2, 3, ... reproduces the sequence itself. Each run of consecutive removed terms is separated from the next one by a term a(k) <= a(k-1) such that floor(sqrt(a(k))) equals the length of the run.
%C A fractal sequence : If one deletes the first occurrence of 1, 2, 3, ... the original sequence is reproduced.
%C Subsequent runs of consecutive terms which are these first occurrences are separated by a term whose square root yields the length of the preceding run (when rounded down).
%C Motivated by Project Euler problem 535, see LINKS.
%H Martin Møller Skarbiniks Pedersen, <a href="/A265650/b265650.txt">Table of n, a(n) for n = 1..1000</a>
%H Project Euler, <a href="https://projecteuler.net/problem=535">Problem 535: Fractal Sequence</a>
%H Clark Kimberling, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL10/Kimberling3/kimberling602.html">Interspersions and Fractal Sequences Associated with Fractions c^j/d^k</a>, Journal of Integer Sequences, Issue 5, Volume 10 (2007), Article 07.5.1
%F The sequence contains marked numbers and non-marked numbers.
%F The marked numbers are consecutive starting with a(1)=1.
%F Immediately preceding each non-marked number in a(n), there are exactly floor(sqrt(a(n)) [= A000196(a(n))] adjacent marked numbers.
%e The runs of first occurrences of the positive integers are {1}, {2}, {3}, {4}, {5}, {6}, {7, 8}, {9}, {10, 11}, ... each separated from the next one by, respectively, 1, 1, 2, 1, 3, 2, 4, 1, 5, ... where 4 and 5 follow the groups {7, 8} and {10, 11} of length 2 = sqrt(4) = floor(sqrt(5)). - _M. F. Hasler_, Dec 13 2015
%o (C)
%o #include <stdio.h>
%o #include <math.h>
%o #define SIZE 1000
%o unsigned int numbers[SIZE];
%o int main() {
%o unsigned int pointer=0, next=1, circle_count=1, next_circle_number=2, sqrt_non_circle=1;
%o numbers[0]=1; printf("1");
%o while (next<SIZE) {
%o if (circle_count==sqrt_non_circle) {
%o numbers[next]=numbers[pointer]; circle_count=0; pointer++;
%o sqrt_non_circle=sqrt(numbers[pointer]);
%o } else {
%o circle_count++; numbers[next]=next_circle_number;
%o next_circle_number++;
%o }
%o printf(",%u",numbers[next]); next++;
%o }
%o }
%o (PARI) A265650(n, list=0, a=[1], cc=0, nc=1, p=0)={for(i=2, n, a=concat(a, if(0<=cc-=1, nc+=1, cc=sqrtint(a[!!p+p+=1]); a[p]))); list&&return(a); a[n]} \\ Set 2nd optional arg.to 1 to return the whole list. - _M. F. Hasler_, Dec 13 2015
%Y Cf. A000196, A003603, A035513.
%K easy,nonn
%O 1,3
%A _Martin Møller Skarbiniks Pedersen_, Dec 11 2015
%E New name from _M. F. Hasler_, Dec 13 2015
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