OFFSET
1,15
COMMENTS
a(n) is the number of r in row n of A381801 that are such that rad(r) does not divide n.
For prime p, a(p) = 0 since r < n are coprime to p and k such that rad(k) | p are powers of p with p^0 congruent to 1 (mod p) and p^m congruent to 0 (mod p) for m > 0.
For proper prime power p^m, m > 1, a(p^m) = 0 since k such that rad(k) | p are powers p^j, j > 1, such that p^j mod p^m = p^(j mod m), divisors d of p^m and thus rad(d) | p^m.
LINKS
Michael De Vlieger, Table of n, a(n) for n = 1..10000
EXAMPLE
Let S(n) = row n of A381801 and R(n) = row n of A162306, with n in R(n) instead written as n mod n = 0.
Define quality Q between natural numbers k and n to be rad(k) does not divide n.
a(10) = 1 since S(10) = {0,1,2,4,5,6,8} only contains r = 6 with quality Q.
a(15) = 3 since S(15) = {0,1,3,5,6,9,10,12} and R(15) = {0,1,3,5,9} = {6,10,12}.
a(18) = 2 since S(18) = {0,1,2,3,4,6,8,9,10,12,14,16} and R(18) = {1,2,3,4,6,8,9,12,16,18} = {10,14}.
a(20) = 1 since S(20) = {0,1,2,4,5,8,10,12,16} and R(20) = {0,1,2,4,5,8,10,16} = {12}, etc.
MATHEMATICA
f[x_] := Block[{c, ff, m, r, p, s, w},
c[_] := True; ff = FactorInteger[x][[All, 1]]; w = Length[ff];
s = {1};
Do[Set[p[i], ff[[i]]], {i, w}];
Do[Set[s, Union@ Flatten@ Join[s, #[[-1, 1]]]] &@ Reap@
Do[m = s[[j]];
While[Sow@ Set[r, Mod[m*p[i], x]];
c[r],
c[r] = False;
m *= p[i]],
{j, Length[s]}],
{i, w}]; s ];
rad[x_] := Times @@ FactorInteger[x][[All, 1]];
{0}~Join~Table[Length@ Complement[f[n], {0}~Join~Select[Range[n - 1], Divisible[#, rad[#]] &]], {n, 2, 120}]
CROSSREFS
KEYWORD
nonn,look
AUTHOR
Michael De Vlieger, Mar 14 2025
STATUS
approved
