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.

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

Author

David James Sycamore, Mar 15 2023

Keywords

• nonn

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.

Examples

			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.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..10000
• Michael De Vlieger, Log log scatterplot of a(n), n = 1..2^20, showing primes in red.

Crossrefs

Cf. A007947, A361133.

Programs

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 *)