A380672 Triangle read by rows where row n lists divisors d | n such that rad(d) != rad(n), where rad = A007947.
1, 1, 1, 1, 1, 2, 3, 1, 1, 1, 1, 2, 5, 1, 1, 2, 3, 4, 1, 1, 2, 7, 1, 3, 5, 1, 1, 1, 2, 3, 9, 1, 1, 2, 4, 5, 1, 3, 7, 1, 2, 11, 1, 1, 2, 3, 4, 8, 1, 1, 2, 13, 1, 1, 2, 4, 7, 1, 1, 2, 3, 5, 6, 10, 15, 1, 1, 1, 3, 11, 1, 2, 17, 1, 5, 7, 1, 2, 3, 4, 9, 1, 1, 2, 19
Offset: 2
Examples
D(6) = {1, 2, 3, 6}; of these, {1, 2, 3} are such that rad(d) != rad(6).
D(10) = {1, 2, 5, 10}; of these, {1, 2, 5} are such that rad(d) != rad(10).
D(12) = {1, 2, 3, 4, 6, 12}; of these, {1, 2, 3, 4} are such that rad(d) != rad(12).
D(36) = {1, 2, 3, 4, 6, 9, 12, 18, 36}; of these, {1, 2, 3, 4, 9} are such that rad(d) != rad(36), etc.
Table begins:
n: row n
---------------
2: 1;
3: 1;
4: 1;
5: 1;
6: 1, 2, 3;
7: 1;
8: 1;
9: 1;
10: 1, 2, 5;
11: 1;
12: 1, 2, 3, 4;
13: 1;
14: 1, 2, 7;
15: 1, 3, 5;
...
Links
- Michael De Vlieger, Table of n, a(n) for n = 2..11751 (rows n = 2..2000, flattened).
Programs
-
Mathematica
rad[x_] := rad[x] = Times @@ FactorInteger[x][[All, 1]]; Table[r = rad[n]; Select[Divisors[n], rad[#] != r &], {n, 2, 40}] // Flatten
Formula
Row 1 is empty since rad(1) | rad(1).
The first term of row n is 1 for all n > 1.
n is not in row n since rad(n) = rad(n).
Length of row n = A183093(n) = tau(n) - tau(n/rad(n)).
Let S(n) = row n of A284318 and let D(n) = row n of A027750. Then row n of this sequence is D(n) \ S(n).
For prime p and m > 0, row p^m = {1}, since d | p^m, d > 1, are such that rad(d) = p.
For squarefree composite n, row n = D(n) \ {n} with length 2^(omega(k)-1).
Comments