A334468 - OEIS

%I #9 May 05 2020 01:19:04

%S 4,6,8,12,15,16,18,24,27,30,32,36,40,45,48,54,60,63,64,70,72,75,80,81,

%T 90,96,100,105,108,120,125,128,135,140,144,150,160,162,168,175,180,

%U 189,192,200,210,216,224,225,234,240,243,250,256,270,280,288,294,300

%N List of distinct values of n + A217287(n).

%C This sequence is a list of primitive least m > n whose distinct prime factors p are not a subset of those prime factors p found in the range n..(m - 1), i.e., the smallest A217287(n)-smooth number m > n. These numbers serve as "obstructions" that end or break the chains described at A217287.

%C The number (a(n) - 1) can be found in at least one row of A217438. In other words, this sequence includes any number T(n, A217287(n)) + 1 where T(n, k) is the irregular triangle described at A217438.

%H Michael De Vlieger, <a href="/A334468/b334468.txt">Table of n, a(n) for n = 1..10000</a>

%H Michael De Vlieger, <a href="/A334468/a334468.png">Plot (x,y)</a> of x in rows 1 <= y <= 4096 of A217438 in gray, with the single pixel m (in this sequence) that exceeds the largest term of A217438 in red.

%H Michael De Vlieger, <a href="/A334468/a334468.txt">Analysis of prime decompositions of terms in this sequence.</a>

%F a(n) > n + 2 for all n.

%F a(m) = m + 2 for m = 2^k - 2 and k > 1, since m is even and 2^k only has the distinct prime divisor 2. Therefore 2^k for k > 1 is in this sequence.

%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 and considering each value only once, we generate the sequence {4, 6, 8, 12, 15, 16, ...}

%t Block[{nn = 2^9, r}, r = Array[If[# == 1, 0, Total[2^(PrimePi /@ FactorInteger[#][[All, 1]] - 1)]] &, nn + Ceiling@ Sqrt@ nn]; Union@ Array[Block[{k = # + 1, s = r[[#]]}, While[UnsameQ[s, Set[s, BitOr[s, r[[k]] ] ] ], k++]; k] &, nn] ]

%Y Cf. A000918, A217287, A217438, A334469.

%K nonn

%O 1,1

%A _Michael De Vlieger_, May 02 2020