A333518 a(n) = A000720(A006530(A334468(n))).

1, 2, 1, 2, 3, 1, 2, 2, 2, 3, 1, 2, 3, 3, 2, 2, 3, 4, 1, 4, 2, 3, 3, 2, 3, 2, 3, 4, 2, 3, 3, 1, 3, 4, 2, 3, 3, 2, 4, 4, 3, 4, 2, 3, 4, 2, 4, 3, 6, 3, 2, 3, 1, 3, 4, 2, 4, 3, 5, 6, 4, 3, 2, 4, 4, 4, 3, 7, 3, 4, 2, 4, 3, 3, 6, 4, 6, 2, 4, 4, 3, 5, 6, 8, 7, 3, 2, 5, 3, 4, 1, 4, 5, 3, 5, 4, 4, 2, 4, 5, 3, 5, 6, 3, 4

Offset: 1

Author

Michael De Vlieger, May 05 2020

Keywords

• nonn

Comments

Indices of the greatest prime factor of A334468(n).

Consider A334468, a list of numbers m = n+j such that j > 0 is also the smallest number such that n+j has no prime factor > j for some n and j = A217287(n).

Since prime q always contributes a novel prime divisor (i.e., q itself) to the set of distinct primes that divide at least 1 number i the range n + i (1 <= i <= j), the numbers m in A334468 are composite, and given the above, m is a product of relatively small prime factors.

Examples

			Start with n = 1, the empty product. Incrementing n and storing the distinct prime factors each time, we encounter 2, which does not divide any previous number n. Therefore we proceed to n = 3, which is prime and its distinct prime divisor again does not divide any previous number. Finally, at 4, we have the distinct prime divisor 2, since 2 divides the product of the previous range {1, 2, 3}, we end the chain. Therefore 4 is the first term of this sequence.
We list row n of A217438 below, starting with n aligned in columns:
1  2  3
   2  3
      3  4  5
         4  5  6  7
            5  6  7
               6  7
                  7  8  9  10  11
                     8  9  10  11
                        9  10  11
                           10  11  12  13  14
                               11  12  13  14  15
                                   12  13  14  15
                                       13  14  15
                                           14  15
                                               ...
Adding 1 to the last numbers seen in all the rows, we generate the sequence A334468: {4, 6, 8, 12, 15, 16, ...}. Of these, we have greatest prime factors {2, 3, 2, 3, 5, 2, ...} with indices {1, 2, 1, 2, 3, 1, ...}.
Least indices of prime(k) in a(n):
   i  p(i)    n    a(n)
  ---------------------
   1    2     1      4
   2    3     2      6
   3    5     5     15
   4    7    18     63
   5   11    59    308
   6   13    49    234
   7   17    68    374
   8   19    84    475
   9   23   292   2392
  10   29   401   3625
  11   31   518   4991
  12   37   791   8547
  ...
		

Crossrefs

Cf. A000720, A006530, A217287, A217438, A334468.

Programs

Mathematica

Block[{nn = 2^10, r}, r = Array[If[# == 1, 0, Total[2^(PrimePi /@ FactorInteger[#][[All, 1]] - 1)]] &, nn]; Map[PrimePi@ FactorInteger[#][[-1, 1]] &, #] &@ Union@ Array[Block[{k = # + 1, s = r[[#]]}, While[UnsameQ[s, Set[s, BitOr[s, r[[k]] ] ] ], k++]; k] &, nn - Ceiling@ Sqrt@ nn] ]