login
A361534
Let h,i,j be the latest 3 terms in the sequence, starting with a(1)=1, a(2)=2, a(3)=3. Let R = rad(h*i*j), where rad is A007947, and let S be the smallest number of terms in U = {h,i,j} which are divisible by any prime p dividing R. Then, a(n) is the least novel multiple of the greatest such prime p.
3
1, 2, 3, 6, 9, 4, 12, 15, 5, 8, 18, 10, 20, 21, 7, 25, 30, 14, 28, 35, 40, 42, 24, 49, 56, 27, 33, 11, 22, 36, 39, 13, 16, 45, 26, 52, 50, 55, 65, 78, 44, 66, 91, 104, 77, 88, 117, 130, 99, 110, 143, 156, 60, 121, 169, 182, 132, 154, 195, 208, 165, 176, 221, 17, 234, 48, 34, 51, 68, 54, 85, 70, 102
OFFSET
1,2
COMMENTS
In the first 2^20 terms there are only 3 occasions of adjacent terms with same rad, (e.g., a(12,13) = 10, 20); only two occasions of rad(h) = rad(j) (e.g., a(3,4,5) = 3,6,9) and there is no occasion of rad(h) = rad(i) = rad(j) (possible iff S = 3), so it's reasonable to assume that it never happens. In any case the definition would ensure continuation of the sequence since a(n) would be the least novel multiple of the greatest prime divisor of R.
The sequence is conjectured to be a permutation of the positive integers with the primes appearing in their natural order.
The scatterplot resembles a "fine-toothed comb" similar to that of A361133.
LINKS
Michael De Vlieger, Log log scatterplot of a(n), n = 1..2^20, showing primes in red.
EXAMPLE
a(1,2,3) = 1,2,3 and every prime divisor in R divides only one term (S = 1). Since 3 is the greatest such prime, a(4) = 6, the least novel multiple of 3.
a(2,3,4) = 2,3,6 and S = 2 since both 2 and 3 divide two terms in U and no prime divides more than that. Since 3 is the greatest such prime a(5) must be 9.
a(3,4,5) = 3,6,9 and S = 1 because 2|6 but no other term in {3,6,9}, therefore a(6) = 4, the least novel multiple of 2.
MATHEMATICA
nn = 120; c[_] = False; q[_] = 1;
Array[Set[{a[#], c[#]}, {#, True}] &, 3];
Set[{h, i, j}, {a[1], a[2], a[3]}]; u = 4;
w = Array[FactorInteger[#][[All, 1]] &, 3];
Do[g = Reverse@ Union@ Flatten[w];
p = First[
Join @@ Map[ReverseSort,
SplitBy[SortBy[
Table[{p, Count[{h, i, j}, _?(Divisible[#, p] &)]}, {p, g}],
Last], Last]]][[1]] ;
(k = q[#]; While[c[k #], k++]; k *= #; While[c[# q[#]], q[#]++]) &[p];
Set[{a[n], c[k], w, h, i, j}, {k, True,
Append[w[[2 ;; -1]], FactorInteger[k][[All, 1]]], i, j, k}];
If[k == u, While[c[u], u++]], {n, 4, nn}];
Array[a, nn] (* Michael De Vlieger, Mar 15 2023 *)
CROSSREFS
Sequence in context: A191397 A358764 A361133 * A359857 A087494 A328843
KEYWORD
nonn
AUTHOR
STATUS
approved