login
A359727
Beattific 'primes': numbers n > 1 not equal to floor(k*m*phi) or floor(k*m*phi^2) for any smaller element k in this sequence and any positive integer m.
0
2, 4, 7, 8, 13, 14, 17, 23, 24, 28, 30, 39, 40, 43, 46, 49, 50, 53, 59, 65, 66, 70, 72, 75, 76, 81, 86, 88, 92, 96, 98, 107, 108, 114, 117, 118, 123, 127, 134, 140, 143, 144, 149, 150, 153, 156, 159, 160, 163, 166, 175, 176, 179, 182, 185, 191, 195
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