A359727 - OEIS

%I #23 Jan 28 2023 13:44:56

%S 2,4,7,8,13,14,17,23,24,28,30,39,40,43,46,49,50,53,59,65,66,70,72,75,

%T 76,81,86,88,92,96,98,107,108,114,117,118,123,127,134,140,143,144,149,

%U 150,153,156,159,160,163,166,175,176,179,182,185,191,195

%N 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.

%C 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.

%H <a href="/index/Si#sieve">Index entries for sequences generated by sieves</a>

%e 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.

%t bp[limit_] := (*Find all the Beattific primes up to limit*)

%t   Module[{r = (1 + Sqrt[5])/2, sieve = ConstantArray[1, limit]},

%t    Do[If[sieve[[n]] == 1,

%t      sieve[[Table[Floor[k n r], {k, (limit + 1)/(n r)}]]] = 0;

%t      sieve[[Table[Floor[k n r r], {k, (limit + 1)/(n r r)}]]] = 0],

%t     {n, 2, limit}];

%t    Rest@Flatten@Position[sieve, 1]];

%o (PARI) h(n)=(n+sqrtint(5*n^2))\2

%o 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

%Y Cf. A001622, A104457, A000201, A001950.

%K nonn

%O 1,1

%A _James Propp_, Jan 11 2023