|
| |
|
|
A365000
|
|
a(n) = n for n <= 2; for n > 2, rad(a(m)) != rad(a(n)), m < n, implies a(n+1) is the least novel k such that rad(k) | a(n), rad(a(m)) = rad(a(n)), m < n, implies a(n+1) = q*a(n), where q is the least novel prime such that gcd(q, a(n)) = 1 and rad(a(m)) != rad(q*a(n)), m < n.
|
|
1
|
|
|
|
1, 2, 4, 12, 3, 9, 45, 5, 25, 50, 8, 56, 7, 49, 147, 21, 42, 6, 30, 10, 70, 14, 154, 11, 121, 242, 16, 208, 13, 169, 507, 27, 297, 33, 66, 18, 234, 24, 408, 17, 289, 578, 32, 608, 19, 361, 1083, 57, 114, 36, 828, 23, 529, 1058, 46, 230, 20, 220, 22, 286, 26, 130
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,2
|
|
|
COMMENTS
|
In other words, starting with 1 and 2, thereafter, if a(n) is such that rad(a(m)) != rad(a(n)), m < n, then a(n+1) is the smallest k != a(m) such that rad(k) | a(n).
If there is a term a(m) such that rad(a(m)) = rad(a(n)), then a(n+1) is the smallest k != a(m) obtainable from a(n) by multiplying a(n) by a nondivisor prime q, and the result must be such that there is no prior term a(m) with rad(a(m)) = rad(a(n)).
Primes appear in order as a consequence of first condition.
If a(n) = p, then a(n+1) = p^2 for prime p.
The second condition implies a(m) != q*a(n), m < n.
Conjectures:
1. Sequence is a permutation of natural numbers.
2. Primorials outside of {1, 2, 30} enter late. a(18) = 6, a(19) = 30, a(20476) = 210, a(20477) = 2310. Thereafter many long-missing k such that 30 | k and 210 | k enter the sequence.
|
|
|
LINKS
|
Michael De Vlieger, Log log scatterplot of a(n), n = 1..100000, with a color code showing primes in red, other prime powers in gold, squarefree composites in green, and numbers neither squarefree nor prime powers in blue, accentuating terms that are powerful but not prime powers appear in magenta.
Michael De Vlieger, Log log scatterplot of a(n), n = 1..1024, with the same color function as above, except accentuating powerful numbers that are not prime powers in light blue.
|
|
|
EXAMPLE
|
a(3) = 4 since rad(a(2)) = rad(2) = 2, and rad(1) = 1, and the smallest missing number k such that rad(k) | 2 is 4.
a(4) = 12 since rad(a(3)) = rad(4) = 2, rad(2) = 2, and thus since q = 3 is the smallest prime that does not divide a(3) = 4, a(4) = q * a(3) = 3 * 4 = 12.
a(5) = 3 since rad(a(4)) = rad(12) = 6, and no a(m), m < 4 has rad(a(m)) = 6. Thus we select the smallest number missing number k such that rad(k) | 12, so k = 3.
a(6) = 9 since a(5) is squarefree and thus rad(a(5)) = 3; no a(m), m < 5, has rad(a(m)) = 3. We select the smallest missing number k such that rad(k) | 3, and that is 9.
a(7) = 45 since a number a(m), m < 6, with rad(a(m)) = 3 has appeared. Therefore we find the smallest prime q that does not divide 9, but also is such that no a(m), m < n, rad(a(m)) | q*a(6) appears in the sequence. Since numbers a(m) with rad(a(m)) = 6 already appear, we are forced to q = 5, hence a(7) = 5 * a(6) = 5 * 9 = 45, etc.
|
|
|
MATHEMATICA
|
nn = 120; c[_] := False; r[_] := False;
rad[x_] := rad[x] = Times @@ FactorInteger[x][[All, 1]];
Array[Set[{a[#], c[#]}, {#, True}] &, 2];
j = 2; r[1] = True; u = 3;
Do[
If[r[#],
k = 2;
While[Or[! CoprimeQ[#, k], r[rad[k j]]], k = NextPrime[k]];
k *= j,
If[PrimeQ[#],
k = Floor@ Log[#, u]; While[c[#^k], k++]; k = #^k,
k = u; While[Or[c[k], ! Divisible[#, rad[k]]], k++]]] &[
rad[j]];
Set[{a[n], c[k], r[rad[j]], j}, {k, True, True, k}];
If[k == u, While[c[u], u++]], {n, 3, nn}];
Array[a, nn]
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
