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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
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OFFSET
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1,2
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COMMENTS
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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.
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LINKS
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EXAMPLE
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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)
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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
...
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MATHEMATICA
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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] ]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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