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%I #45 Feb 28 2023 13:13:11
%S 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,
%T 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,
%U 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
%N a(n) = number of k < n such that rad(k) = rad(n) and k does not divide n, where rad(k) = A007947(k).
%C a(n) = 0 for prime powers and squarefree numbers.
%H Michael De Vlieger, <a href="/A355432/b355432.txt">Table of n, a(n) for n = 1..16384</a>
%H Michael De Vlieger, <a href="/A355432/a355432_1.png">Plot (k, n) at (x, -y)</a>, 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.
%H Michael De Vlieger, <a href="/A355432/a355432_2.png">Condensation of the above plot</a>, showing k = 1..n and only n in A360768 and n <= 54.
%F a(n) > 0 for n in A360768.
%F a(n) < A243822(n) < A010846(n).
%e a(1) = 18, since 18/6 >= 3. We note that rad(12) = rad(18) = 6, yet 12 does not divide 18.
%e a(2) = 24, since 24/6 >= 3. rad(18) = rad(24) = 6 and 24 mod 18 = 6.
%e a(3) = 36, since 36/6 >= 3. rad(24) = rad(36) = 6 and 36 mod 24 = 12.
%e 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.
%t 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}]
%o (PARI) rad(n) = factorback(factorint(n)[, 1]); \\ A007947
%o a(n) = my(rn=rad(n)); sum(k=1, n-1, if (n % k, rad(k)==rn)); \\ _Michel Marcus_, Feb 23 2023
%Y Cf. A007947, A010846, A013929, A020639, A024619, A027750, A126706, A162306, A243822, A272618, A360589, A360768.
%K nonn
%O 1,48
%A _Michael De Vlieger_, Feb 22 2023
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