OFFSET
0,2
COMMENTS
Offset to match A019565, to which this sequence is analogous.
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.
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).
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).
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.
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.
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 }.
LINKS
Michael De Vlieger, Table of n, a(n) for n = 0..16384
Michael De Vlieger, Log log scatterplot of a(n) n = 1..10^5.
Michael De Vlieger, Log log scatterplot of a(n), 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.
Michael De Vlieger, Plot prime(k)^e(k) | a(n) at (x,y) = (n,k) for n = 0..1024, showing multiplicity e(k) with a color function where black = 1, red = 2, ..., magenta = 9.
EXAMPLE
T(0) = S(0) = {1},
T(1) = S(1) = {2},
T(2) = {4, 3, 6}, since, mapping f across S(1..2) = {2, 3, 6};
2 is prime and c(2) = 2, hence 2^c(2) = 4, but
3 and 6 are new to this sequence, therefore {4, 3, 6}.
T(3) = {8, 9, 12, 5, 10, 15, 30}, since S(1..2) = {2, 3, 6};
2 is prime and c(2) = 3, hence 2^c(2) = 2^3 = 8,
3 is prime and c(3) = 2, hence 3^c(3) = 3^2 = 9,
6 is composite and m(6) = 2, 2 | 6, hence 2*6 = 12,
and we append S(3).
T(4) = {16, 27, 18, 25, 20, 45, 60, 7, 14, 21, 42, 35, 70, 105, 210},
since S(1..3) = {2, 3, 6, 5, 10, 15, 30};
2 is prime and c(2) = 4, hence 2^c(2) = 2^4 = 16,
3 is prime and c(3) = 3, hence 3^c(3) = 3^3 = 27,
6 is composite and m(6) = 3, 3 | 6, hence 3*6 = 18,
5 is prime and c(5) = 2, hence 5^c(5) = 5^2 = 25,
10 is composite and m(10) = 2, 2 | 10, hence 2*10 = 20,
15 is composite and m(15) = 2, but 2 does not divide 15,
therefore we increment m(15) = 3,
3 | 15, hence 3*15 = 45.
30 is composite and m(30) = 2, 2 | 30, hence 2*30 = 60,
and we append S(3), etc. Let f(x) = phi(x)/x.
First terms of the first rows appear as follows:
n\k 1 2 3 4 5 6 7 8
----------------------------------------------------
1 2
2 4 3 6
3 8 9 12 5 10 15 30
4 16 27 18 25 20 45 60 7 ...
5 32 81 24 125 40 75 90 49 ...
6 64 243 36 625 50 135 120 343 ...
7 128 729 48 3125 80 225 150 2401 ...
8 256 2187 54 15625 100 375 180 16807 ...
9 512 6561 72 78125 160 405 240 117649 ...
10 1024 19683 96 390625 200 675 270 823543 ...
MATHEMATICA
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}], n]; {{1}}~Join~Rest@ Array[t, nn]] // Flatten
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Michael De Vlieger, Jan 23 2023
STATUS
approved