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A357910 The natural numbers ordered lexicographically by their prime factorization, with prime factors written in decreasing order (see comments).

%I A357910 #17 Jun 20 2025 20:22:44
%S A357910 1,2,4,3,6,8,9,12,5,10,15,30,16,27,18,25,20,45,60,7,14,21,42,35,70,
%T A357910 105,210,32,81,24,125,40,75,90,49,28,63,84,175,140,315,420,11,22,33,
%U A357910 66,55,110,165,330,77,154,231,462,385,770,1155,2310,64,243,36,625,50
%N A357910 The natural numbers ordered lexicographically by their prime factorization, with prime factors written in decreasing order (see comments).
%C A357910 Offset to match A019565, to which this sequence is analogous.
%C A357910 Let c(x) be the cardinality of x^e in a(1..n-1). Set multiplier m(x) = 1, increasing whereupon m(x)*x appears in the sequence.
%C A357910 Define function f(x) to be such that prime x implies x^c(x), otherwise x implies m(x)*x such that rad(m(x)) | rad(x).
%C A357910 Let b(n) be the expression of n through powers 2^e, thus, b(5) = "101" = 2^0 + 2^2. Then Heinz number h(n) = Product { prime(e+1) }, thus, h(n) = 10 = prime(0+1) * prime(2+1).
%C A357910 Define S(n) = { k : rad(k) = k, gpf(k) = prime(n) } such that the ordering of elements is according to h(k). Thus, S(4) = {7, 14, 21, 42, 35, 70, 105, 210} since, writing 1 when divisible by prime(j) in the j-th place read from the right, these are {1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111}, and interpreted as binary numbers these are {8, 9, 10, 11, 12, 13, 14, 15}. S(4) appears as row 4 of A019565 when interpreted as an irregular triangle.
%C A357910 We may generate T(n) by mapping f across S(1..n-1) and appending S(n), which is equivalent to mapping f across S(1..n). Note that we ignore the empty product, the sole element of S(0) = {1}, for n > 0.
%C A357910 Let R(k) = { r : rad(r) | rad(k) }. Hence, R(2) = A000079, R(6) = A003586, and R(10) = A003592. Let K = A019565(k). Column k contains K*R(K). Therefore, T(n,1) = A000079, T(n,3) = { 6*A003586 }, T(n,5) = { 10 * A003592 } and generally, for k = 2^e and p = prime(e+1), we have T(n, 2^e) = { p * p^m }.
%C A357910 R(k) is the tensor product of { p^e : e >= 0 } for all p | k and is therefore infinite for k > 1. A019565 is a permutation of squarefree numbers A005117, an infinite sequence. Therefore this sequence is infinite and a permutation of natural numbers.
%H A357910 Michael De Vlieger, <a href="/A357910/b357910.txt">Table of n, a(n) for n = 0..16384</a>
%H A357910 Michael De Vlieger, <a href="/A357910/a357910.png">Log log scatterplot of a(n)</a> n = 1..10^5.
%H A357910 Michael De Vlieger, <a href="/A357910/a357910_1.png">Log log scatterplot of a(n)</a>, n = 1..503, showing primes in red, composite prime powers in gold, squarefree composites in green, primorials in dark blue, and other numbers in light blue.
%H A357910 Michael De Vlieger, <a href="/A357910/a357910_2.png">Plot prime(k)^e(k) | a(n) at (x,y) = (n,k)</a> for n = 0..1024, showing multiplicity e(k) with a color function where black = 1, red = 2, ..., magenta = 9.
%e A357910 T(0) = S(0) = {1},
%e A357910 T(1) = S(1) = {2},
%e A357910 T(2) = {4, 3, 6}, since, mapping f across S(1..2) = {2, 3, 6};
%e A357910   2 is prime and c(2) = 2, hence 2^c(2) = 4, but
%e A357910   3 and 6 are new to this sequence, therefore {4, 3, 6}.
%e A357910 T(3) = {8, 9, 12, 5, 10, 15, 30}, since S(1..2) = {2, 3, 6};
%e A357910   2 is prime and c(2) = 3, hence 2^c(2) = 2^3 = 8,
%e A357910   3 is prime and c(3) = 2, hence 3^c(3) = 3^2 = 9,
%e A357910   6 is composite and m(6) = 2, 2 | 6, hence 2*6 = 12,
%e A357910   and we append S(3).
%e A357910 T(4) = {16, 27, 18, 25, 20, 45, 60, 7, 14, 21, 42, 35, 70, 105, 210},
%e A357910   since S(1..3) = {2, 3, 6, 5, 10, 15, 30};
%e A357910   2 is prime and c(2) = 4, hence 2^c(2) = 2^4 = 16,
%e A357910   3 is prime and c(3) = 3, hence 3^c(3) = 3^3 = 27,
%e A357910   6 is composite and m(6) = 3, 3 | 6, hence 3*6 = 18,
%e A357910   5 is prime and c(5) = 2, hence 5^c(5) = 5^2 = 25,
%e A357910   10 is composite and m(10) = 2, 2 | 10, hence 2*10 = 20,
%e A357910   15 is composite and m(15) = 2, but 2 does not divide 15,
%e A357910     therefore we increment m(15) = 3,
%e A357910     3 | 15, hence 3*15 = 45.
%e A357910   30 is composite and m(30) = 2, 2 | 30, hence 2*30 = 60,
%e A357910   and we append S(3), etc. Let f(x) = phi(x)/x.
%e A357910 First terms of the first rows appear as follows:
%e A357910   n\k    1      2    3       4    5     6    7       8
%e A357910   ----------------------------------------------------
%e A357910    1     2
%e A357910    2     4      3    6
%e A357910    3     8      9   12       5   10    15   30
%e A357910    4    16     27   18      25   20    45   60       7  ...
%e A357910    5    32     81   24     125   40    75   90      49  ...
%e A357910    6    64    243   36     625   50   135  120     343  ...
%e A357910    7   128    729   48    3125   80   225  150    2401  ...
%e A357910    8   256   2187   54   15625  100   375  180   16807  ...
%e A357910    9   512   6561   72   78125  160   405  240  117649  ...
%e A357910   10  1024  19683   96  390625  200   675  270  823543  ...
%t A357910 nn = 16; rad[n_] := rad[n] = Times @@ FactorInteger[n][[All, 1]]; q[_] = 1; r = t[0] = {1}; c[1] = True; Do[Set[s, Join[r, Prime[n]*r]]; Set[t[n], Map[If[PrimeQ[#], Set[m, #^q[#]]; q[#]++; m, Set[m, q[#] #]; q[#]++; While[! Divisible[#, rad[q[#]]], q[#]++]; m] &, Rest[r]]]; r = s, {n, nn}]; {{1}}~Join~Rest@ Array[t, nn] // Flatten
%Y A357910 Cf. A000040, A000079, A002110, A003586, A003592, A007947, A019565, A182944.
%K A357910 nonn,tabf
%O A357910 0,2
%A A357910 _Michael De Vlieger_, Jan 23 2023