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%I #48 May 09 2020 00:45:48
%S 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,
%T 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,
%U 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
%N a(n) = A000720(A006530(A334468(n))).
%C Indices of the greatest prime factor of A334468(n).
%C 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).
%C 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.
%e 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.
%e We list row n of A217438 below, starting with n aligned in columns:
%e 1 2 3
%e 2 3
%e 3 4 5
%e 4 5 6 7
%e 5 6 7
%e 6 7
%e 7 8 9 10 11
%e 8 9 10 11
%e 9 10 11
%e 10 11 12 13 14
%e 11 12 13 14 15
%e 12 13 14 15
%e 13 14 15
%e 14 15
%e ...
%e 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, ...}.
%e Least indices of prime(k) in a(n):
%e i p(i) n a(n)
%e ---------------------
%e 1 2 1 4
%e 2 3 2 6
%e 3 5 5 15
%e 4 7 18 63
%e 5 11 59 308
%e 6 13 49 234
%e 7 17 68 374
%e 8 19 84 475
%e 9 23 292 2392
%e 10 29 401 3625
%e 11 31 518 4991
%e 12 37 791 8547
%e ...
%t 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] ]
%Y Cf. A000720, A006530, A217287, A217438, A334468.
%K nonn
%O 1,2
%A _Michael De Vlieger_, May 05 2020