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A355432
a(n) = number of k < n such that rad(k) = rad(n) and k does not divide n, where rad(k) = A007947(k).
15
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 2, 0, 0, 0, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 4, 0, 2, 0, 2, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1
OFFSET
1,48
COMMENTS
a(n) = 0 for prime powers and squarefree numbers.
LINKS
Michael De Vlieger, Plot (k, n) at (x, -y), k = 1..n, n = 1..54, showing k in A126706 in dark blue, n in A360768 in dark red, and for n and nondivisor k such that rad(k) = rad(n), we highlight in large black dots. This sequence counts the number of black dots in row n.
Michael De Vlieger, Condensation of the above plot, showing k = 1..n and only n in A360768 and n <= 54.
FORMULA
a(n) > 0 for n in A360768.
a(n) < A243822(n) < A010846(n).
a(n) = A008479(n) - A005361(n). - Amiram Eldar, Oct 25 2024
EXAMPLE
a(1) = 18, since 18/6 >= 3. We note that rad(12) = rad(18) = 6, yet 12 does not divide 18.
a(2) = 24, since 24/6 >= 3. rad(18) = rad(24) = 6 and 24 mod 18 = 6.
a(3) = 36, since 36/6 >= 3. rad(24) = rad(36) = 6 and 36 mod 24 = 12.
a(6) = 54, since 54/6 >= 3. m in {12, 24, 36, 48} are such that rad(m) = rad(54) = 6, but none divides 54, etc.
MATHEMATICA
rad[n_] := rad[n] = Times @@ FactorInteger[n][[All, 1]]; Table[Which[PrimePowerQ[n], 0, SquareFreeQ[n], 0, True, r = rad[n]; Count[Select[Range[n], Nor[PrimePowerQ[#], SquareFreeQ[#]] &], _?(And[rad[#] == r, Mod[n, #] != 0] &)]], {n, 120}]
PROG
(PARI) rad(n) = factorback(factorint(n)[, 1]); \\ A007947
a(n) = my(rn=rad(n)); sum(k=1, n-1, if (n % k, rad(k)==rn)); \\ Michel Marcus, Feb 23 2023
KEYWORD
nonn,changed
AUTHOR
Michael De Vlieger, Feb 22 2023
STATUS
approved