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Lexicographically earliest sequence of distinct integers greater than 1 such that a(n) mod a(i)^2 >= a(i) for all i < n.
1

%I #27 Sep 05 2024 17:16:33

%S 2,3,6,7,14,15,22,23,26,30,31,34,35,42,43,58,59,62,66,67,70,71,78,79,

%T 86,87,94,95,106,107,114,115,122,123,130,131,134,138,139,142,143,158,

%U 159,166,167,170,174,175,178,179,186,187,194,195,210,211,214,215,222

%N Lexicographically earliest sequence of distinct integers greater than 1 such that a(n) mod a(i)^2 >= a(i) for all i < n.

%C In the sieve of Eratosthenes, first the even numbers are removed, then the multiples of 3, then multiples of 5. In this sieve first the numbers greater than 2 and modulo 0 or 1 (mod 4) are removed leaving (1) 2, 3, 6, 7, 10, 11, 14, 15. Then the numbers greater than 3 and modulo 0, 1, 2 (mod 9) are removed leaving (1) 2, 3, 6, 7, 14, 15. Then numbers modulo 0, 1, 2, 3, 4, 5 (mod 36) are removed.

%H Robert Israel, <a href="/A335099/b335099.txt">Table of n, a(n) for n = 1..10000</a>

%p N:= 1000: # for terms <= N

%p R:= NULL:

%p Cands:= [$2..N]:

%p while Cands <> [] do

%p r:= Cands[1];

%p R:= R,r;

%p Cands:= select(t -> t mod r^2 >= r, Cands[2..-1]);

%p od:

%p R; # _Robert Israel_, Sep 05 2024

%o (Python3)

%o from math import sqrt

%o length=100

%o s=list(range(2,length))

%o for p in range(int(sqrt(length))):

%o x = s[p]

%o if x==0 : continue

%o for i,e in enumerate(s):

%o if e>x and e%(x*x)<x:

%o s[i]=0 # mark sieved values as zero

%o result =[j for j in s if j!=0] # remove zeros

%o print(result)

%o (PARI) seq(n)={my(a=vector(n), k=1); for(n=1, #a, while(1, k++; my(f=1); for(i=1, n-1, if(k%a[i]^2<a[i], f=0; break)); if(f, a[n]=k; break))); a} \\ _Andrew Howroyd_, Sep 12 2020

%Y Cf. A000960, A000040.

%K nonn,easy

%O 1,1

%A _James Kilfiger_, Sep 12 2020 (suggested by student)

%E Terms a(29) and beyond from _Andrew Howroyd_, Sep 12 2020