A252868 Squarefree version of A252867.

1, 2, 3, 10, 21, 5, 6, 35, 22, 15, 14, 33, 7, 30, 77, 26, 105, 13, 42, 65, 66, 91, 11, 70, 143, 210, 187, 39, 55, 78, 385, 34, 165, 182, 51, 110, 273, 85, 154, 195, 119, 330, 17, 231, 170, 429, 238, 715, 102, 455, 374, 1365, 38, 1155, 442, 57, 770, 663, 95, 462, 1105, 114, 1001, 255

Offset: 1

Author

Franklin T. Adams-Watters, Dec 23 2014

Keywords

• nonn

Comments

a(n) = n if n <= 3, otherwise the first squarefree number not occurring earlier having at least one common factor with a(n-2), but none with a(n-1). The squarefree numbers are ordered by their occurrence in A019565.

These represent the same sets of integers as A252867 does, but using the factorization of squarefree numbers for the representation.

This is a permutation of the squarefree numbers. [I believe this is at present only a conjecture. - N. J. A. Sloane, Jan 10 2015]

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..10000
• David L. Applegate, Hans Havermann, Bob Selcoe, Vladimir Shevelev, N. J. A. Sloane, and Reinhard Zumkeller, The Yellowstone Permutation, arXiv preprint arXiv:1501.01669, 2015 and J. Int. Seq. 18 (2015) 15.6.7.

Crossrefs

Cf. A019565, A048793, A098550, A252865, A252867.

Programs

Mathematica

(* b = A019565, c = A252867 *)
b[n_] := Times @@ Prime[Flatten[Position[#, 1]]]&[Reverse[IntegerDigits[n, 2]]];
c[n_] := c[n] = If[n<3, n, For[k=3, True, k++, If[FreeQ[Array[c, n-1], k], If[BitAnd[k, c[n-2]] >= 1 && BitAnd[k, c[n-1]] == 0, Return[k]]]]];
a[n_] := b[c[n-1]];
Array[a, 64] (* Jean-François Alcover, Oct 03 2018 *)

PARI

invecn(v, k, x)=for(i=1, k, if(v[i]==x, return(i))); 0
squarefree(n)=local(r=1,i=1);while(n>0,if(n%2,r*=prime(i));i++;n\=2);r
alist(n)=local(v=vector(n, i, i-1), x); for(k=4, n, x=3; while(invecn(v, k-1, x)||!bitand(v[k-2], x)||bitand(v[k-1], x), x++); v[k]=x); vector(n,i,squarefree(v[i]))

Python

from operator import mul
from functools import reduce
from sympy import prime
def A019565(n):
    return reduce(mul,(prime(i+1) for i,v in enumerate(bin(n)[:1:-1]) if v == '1')) if n > 0 else 1
A252868_list, l1, l2, s, b = [1,2,3], 2, 1, 3, set()
for _ in range(10**4):
    i = s
    while True:
        if not (i in b or i & l1) and i & l2:
            A252868_list.append(A019565(i))
            l2, l1 = l1, i
            b.add(i)
            while s in b:
                b.remove(s)
                s += 1
            break
        i += 1 # Chai Wah Wu, Dec 25 2014

Formula

a(n) = A019565(A252867(n-1)).