Michael De Vlieger, <a href="/A372322/b372322_1.txt">Table of n, a(n) for n = 1..10000</a>

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a(n) = A010846(A372111(n)).

1, 2, 5, 6, 5, 11, 18, 8, 16, 22, 5, 28, 13, 33, 23, 38, 11, 26, 12, 9, 58, 28, 80, 5, 30, 55, 19, 27, 19, 56, 37, 21, 27, 87, 44, 44, 48, 38, 18, 58, 42, 5, 110, 26, 112, 140, 38, 45, 32, 144, 102, 59, 5, 139, 225, 39, 44, 22, 180, 86, 114, 34, 23, 133, 41, 115

1,2

Let r(x) = A010846(x), the number of m <= x such that rad(m) | x, where rad = A007947.

Let row k of A162306 contain { m : rad(m) | k, m <= k }. Thus r(k) is the length of row k of A162306.

a(n) is the length of row A372111(n) of A162306.

Analogous to A371909, which instead regards A109890 and A109735.

Michael De Vlieger, <a href="/A372322/b372322_1.txt">Table of n, a(n) for n = 1..10000</a>

Let s(x) = A372111(x) and let r(x) = A010846(x).

a(1) = 1 since r(s(1)) = r(1) = 1.

a(2) = 2 since r(s(2)) = r(3) = 2. For prime p, r(p) = card({1, p}) = 2.

a(3) = 5 since r(s(3)) = r(6) = 5. r(6) = card({1, 2, 3, 4, 6}) = 5.

a(4) = 6 since r(s(4)) = r(10) = 6. r(10) = card({1, 2, 4, 5, 8, 10}) = 6.

a(5) = 5 since r(s(5)) = r(15) = 5. r(15) = card({1, 3, 5, 9, 15}) = 5.

a(6) = 11 since r(s(6)) = r(24) = 11. r(24) = card({1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24}) = 11, etc.

nn = 68; c[_] := False;

rad[x_] := rad[x] = Times @@ FactorInteger[x][[All, 1]];

f[x_] := Select[Range[x], Divisible[x, rad[#]] &];

Array[Set[{a[#], c[#]}, {#, True}] &, 2]; s = a[1] + a[2];

{1}~Join~Reap[Do[r = f[s]; k = SelectFirst[r, ! c[#] &];

Sow[Length[r]]; c[k] = True;

s += k, {i, 3, nn}] ][[-1, 1]]

Cf. A007947, A010846, A124652, A162306, A371909, A372111, A372323.

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Michael De Vlieger, May 05 2024

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