OFFSET
1,1
COMMENTS
Given r = (1+sqrt(5))/2 and s = r^2, we sieve the set {2,3,4,...}, where each time we discover a new "prime" p, we sieve out the numbers floor(p*r), floor(2p*r), floor(3p*r), ... and floor(p*s), floor(2p*s), floor(3p*s), ... It appears that significantly more than half the terms are even.
EXAMPLE
The Beattific prime 2 causes us to sieve out 3, 6, 9, ... and 5, 10, ...; then the next Beattific prime, 4, doesn't cause us to throw out anything new; then the next Beattific prime is 7.
MATHEMATICA
bp[limit_] := (*Find all the Beattific primes up to limit*)
Module[{r = (1 + Sqrt[5])/2, sieve = ConstantArray[1, limit]},
Do[If[sieve[[n]] == 1,
sieve[[Table[Floor[k n r], {k, (limit + 1)/(n r)}]]] = 0;
sieve[[Table[Floor[k n r r], {k, (limit + 1)/(n r r)}]]] = 0],
{n, 2, limit}];
Rest@Flatten@Position[sieve, 1]];
PROG
(PARI) h(n)=(n+sqrtint(5*n^2))\2
list(lim)=my(v=vectorsmall(lim\=1, i, 1), u=List()); for(n=2, #v, if(v[n]==0, next); listput(u, n); forstep(k=n, h(lim+1)-lim-1, n, v[h(k)]=0); forstep(k=n, 2*lim+1-h(lim+1), n, v[h(k)+k]=0)); Vec(u) \\ Charles R Greathouse IV, Jan 25 2023
CROSSREFS
KEYWORD
nonn
AUTHOR
James Propp, Jan 11 2023
STATUS
approved