A361629 For n <= 2, a(n) = n. Thereafter let p be the greatest prime which divides the least number of terms in U = {a(n-2), a(n-1)}, then a(n) is the smallest multiple of p that is not yet in the sequence.

1, 2, 4, 6, 3, 8, 9, 12, 10, 5, 14, 7, 16, 21, 28, 15, 35, 42, 20, 49, 56, 18, 63, 70, 25, 77, 11, 84, 22, 33, 24, 44, 55, 30, 66, 88, 27, 99, 110, 40, 121, 132, 36, 143, 13, 154, 26, 39, 45, 52, 65, 50, 78, 91, 98, 104, 117, 48, 130, 156, 60, 169, 182, 105, 195, 208, 75, 221, 17, 234, 34, 51, 54, 68, 85, 80, 102, 119, 112, 136, 153, 57, 19

Offset: 1

Author

David James Sycamore, Mar 18 2023

Keywords

• nonn

Comments

In other words, if R = A007947(a(n-2)*a(n-1)), and S (= 1 or 2) is the smallest number of terms in U divisible by any p|R, then a(n) is the least novel multiple of the greatest p|R which divides S terms in U.

When for some m > 1, a(n-1) = m*p, is the first occurrence of a multiple of prime p, then a(n) = p. The sequence after a(1) = 1 can be represented as an irregular table in which the n-th row T(n,k) starts with p = prime(n), and ends with the first subsequent term divisible by q = prime(n+1). The next row begins with q, and so on. The range of k is not always known; see Formula and Example.

Some primes (2,7,11,13,47,79,113,127,...) exhibit fully developed initial trajectories T(n,k) dominated by multiples of p, until p*q is reached, whereas for others the trajectory ends with a term m*q where m < p.

The scatterplot resembles a fine-toothed comb (as in A361133, A361534), in which each "tooth" represents the initial trajectory of a prime. It is not known whether infinitely many primes exhibit fully developed initial trajectories.

Conjectured to be a permutation of the positive integers with primes in order.

Examples

			a(1,2) = 1,2 and 2 is the smallest p|R = 2 which divides the least number (1) of terms in U = {1,2}. Therefore a(3) = 4, is the least multiple of 2 which has not already occurred.
a(2,3) = 2,4 and since rad(2) = rad(4) = 2, S = 2 then a(4) = 6, the least multiple of 2 which has not occurred earlier.
a(3,4) = 4,6 so S = 1 since 3|6 but not 2, so a(5) = 3, since 3 has not appeared previously.
The sequence after a(1) can be shown as an irregular table of finite subsequences where each row T(n) starts with prime(n) and ends with the first subsequent multiple of prime(n+1). The table includes the fully developed trajectories of 2,7,11,13 with 3,15,24 terms respectively, each having final term in A006094. T(4) starts with 7, ends with 77 and has 3*(prime(5)-1)/2 = 15 terms.
Table T(n) for n = 1..10:
  2,4,6
  3,8,9,12,10
  5,14
  7,16,21,28,15,35,42,20,49,56,18,63,70,25,77
  11,84,22,33,24,44,55,30,66,88,27,99,110,40,121,132,36,143
  13,154,26,39,45,52,65,50,78,91,98,104,117,48,130,156,60,169,182,105,195,208,75,221
  17,234,34,51,54,68,85,80,102,119,112,136,153,57
  19,69,
  23,72,46,92,138,81,115,161,126,184,207,87
  29,90,58,116,174,93
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..16384
• Michael De Vlieger, Log log scatterplot of a(n), n = 1..2^20.
• Michael De Vlieger, Log log scatterplot of a(n), n = 1..2^12, showing primes in red, composite prime powers in gold, squarefree composites in green, and numbers neither prime power nor squarefree in blue, accentuating those last-mentioned whose prime power factors have multiplicities exceeding 1 in light blue.

Crossrefs

Cf. A006094, A007947, A361133, A361534.

Programs

Mathematica

Block[{a, c, g, h, i, j, k, p, q, u, w, nn},
nn = 120; c[] = False; q[] = 1;
a[1] = i = 1; a[2] = j = 2; c[1] = c[2] = True; w = {1, 2}; u = 3;
Do[If[Equal @@ w, p = Times @@ w[[1]],
  g = Reverse@ Union@ Flatten[w];
  p = First[
     Join @@ Map[ReverseSort,
       SplitBy[SortBy[
         Table[{p, Count[{i, j}, _?(Divisible[#, p] &)]}, {p, g}],
         Last], Last]] ][[1]] ];
  k = q[p]; While[c[k*p], k++]; k *= p; While[c[p*q[p]], q[p]++]
  Set[{a[n], c[k], w, i, j},
      {k, True, Append[w[[2 ;; -1]], FactorInteger[k][[All, 1]]], j, k}];
  If[k == u, While[c[u], u++]], {n, 3, nn}];
Array[a, nn] ] (* Michael De Vlieger, Mar 18 2023 *)

Formula

The number of terms T(n,k) is <= 3*(prime(k+1) - 1)/2, with equality when the trajectory is fully developed. The number of terms in shorter trajectories is not readily known. Therefore the irregular table (see Example) is expressed as T(n), rather than T(n,k).