Michael De Vlieger, <a href="/A372323/b372323_1.txt">Table of n, a(n) for n = 3..10000</a>
Michael De Vlieger, <a href="/A372323/b372323_1.txt">Table of n, a(n) for n = 3..10000</a>
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2, 4, 4, 4, 5, 7, 5, 8, 8, 2, 10, 8, 12, 11, 13, 6, 13, 6, 6, 9, 8, 11, 4, 8, 16, 5, 6, 7, 13, 12, 7, 10, 19, 15, 16, 17, 9, 6, 15, 10, 3, 11, 8, 18, 28, 14, 14, 10, 30, 28, 15, 4, 20, 33, 13, 12, 6, 22, 18, 21, 12, 11, 29, 12, 11, 8, 24, 18, 8, 14, 17, 32, 33
3,1
Let b(x) = A124652(x) and let s(x) = A372111(x), where A372111 contains partial sums of A124652.
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.
Let T(k,j) represent the j-th term in row k of irregular triangle A162306.
a(n) = j is the position of b(n) in row s(n-1) of A162306.
b(n) = T(s(n-1), a(n)).
Analogous to A371910, which instead regards A109890 and A109735.
Michael De Vlieger, <a href="/A372323/b372323_1.txt">Table of n, a(n) for n = 3..10000</a>
Let b(x) = A124652(x) and let s(x) = A372111(x), where A372111 contains partial sums of A124652.
a(3) = 2 since b(3) = 3 is the 2nd term in row s(3) = 3 of A162306, {1, [3]}.
a(4) = 4 since b(4) = 4 is the 4th term in row s(4) = 6 of A162306, {1, 2, 3, [4], 6}.
a(5) = 4 since b(5) = 5 is T(s(n-1), 4) = T(10, 4), {1, 2, 4, [5], 8, 10}.
a(6) = 4 since b(6) = 9 is T(s(n-1), 4) = T(15, 4), {1, 3, 5, [9], 15}.
a(7) = 5 since b(7) = 6 is T(s(n-1), 5) = T(24, 5), {1, 2, 3, 4, [6], 8, 9, 12, 16, 18, 24}, etc.
Table relating this sequence to b = A124652, s = A372111, r = A372322, and A162306.
n b(n) s(n-1) a(n) r(n) row s(n-1) of A162306
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3 3 3 2 2 {1, [3]}
4 4 6 4 5 {1, 2, 3, [4], 6}
5 5 10 4 6 {1, 2, 4, [5], 8, 10}
6 9 15 4 5 {1, 3, 5, [9], 15}
7 6 24 5 11 {1, 2, 3, 4, [6], ..., 24}
8 8 30 7 18 {1, 2, 3, 4, 5, 6, [8], ..., 30}
9 16 38 5 8 {1, 2, 4, 8, [16], 19, 32, 38}
10 12 54 8 16 {1, 2, 3, 4, 6, 8, 9, [12], ..., 54}
11 11 66 8 22 {1, 2, 3, 4, 6, 8, 9, [11], ..., 66}
12 7 77 2 5 {1, [7], 11, 49, 77}
13 14 84 10 28 {1, 2, 3, 4, ..., 12, [14], ..., 84}
14 28 98 8 13 {1, 2, 4, 7, ..., 16, [28], ..., 98}
nn = 75; 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];
Reap[Do[r = f[s]; k = SelectFirst[r, ! c[#] &];
Sow[FirstPosition[r, k][[1]]]; c[k] = True;
s += k, {i, 3, nn}] ][[-1, 1]]
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Michael De Vlieger, May 05 2024
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