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A162306
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Irregular triangle in which row n contains the numbers <= n whose prime factors are a subset of prime factors of n.
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37
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1, 1, 2, 1, 3, 1, 2, 4, 1, 5, 1, 2, 3, 4, 6, 1, 7, 1, 2, 4, 8, 1, 3, 9, 1, 2, 4, 5, 8, 10, 1, 11, 1, 2, 3, 4, 6, 8, 9, 12, 1, 13, 1, 2, 4, 7, 8, 14, 1, 3, 5, 9, 15, 1, 2, 4, 8, 16, 1, 17, 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 1, 19, 1, 2, 4, 5, 8, 10, 16, 20, 1, 3, 7, 9, 21, 1, 2, 4, 8, 11, 16, 22, 1, 23
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OFFSET
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1,3
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COMMENTS
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Row n begins with 1, ends with n, and has A010846(n) terms.
Prime p has {1, p} and A010846(p) = 2.
Prime power p^e has {1, p, ..., p^e} and A010846(p^e) = A000005(p^e) = e + 1.
Composite c that are not prime powers have A010846(c) = A000005(c) + A243822(c), where A243822(c) is nonzero positive, since the minimum prime divisor p of c produces at least one semidivisor (e.g., p^2 < c). Thus these have the set of divisors of c and at least one semidivisor p^2. For squareful c that are not prime powers, p^2 may divide c, but p^3 does not. The minimum squareful c = 12, 2^3 does not divide 12 yet is less than 12 and is a product of the minimum prime divisor of 12. All other even squareful c admit a power of 2 that does not divide c, since there must be another prime divisor q > 2. (end)
Numbers 1 <= k <= n such that (floor(n^k/k) - floor((n^k - 1)/k)) = 1. - Michael De Vlieger, May 26 2016
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LINKS
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FORMULA
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EXAMPLE
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n = 6: {1, 2, 3, 4, 6}.
n = 7: {1, 7}.
n = 8: {1, 2, 4, 8}.
n = 9: {1, 3, 9}.
n = 10: {1, 2, 4, 5, 8, 10}.
n = 11: {1, 11}.
n = 12: {1, 2, 3, 4, 6, 8, 9, 12}.
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MAPLE
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A:= proc(n) local F, S, s, j, p;
F:= numtheory:-factorset(n);
S:= {1};
for p in F do
S:= {seq(seq(s*p^j, j=0..floor(log[p](n/s))), s=S)}
od;
S
end proc; map(op, [seq(A(n), n=1..100)]); # Robert Israel, Jul 15 2014
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MATHEMATICA
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pf[n_] := If[n==1, {}, Transpose[FactorInteger[n]][[1]]]; SubsetQ[lst1_, lst2_] := Intersection[lst1, lst2]==lst1; Flatten[Table[pfn=pf[n]; Select[Range[n], SubsetQ[pf[ # ], pfn] &], {n, 27}]]
(* Second program: *)
f[x_, y_ : 0] :=
Block[{m, n, nn, j, k, p, t, v, z},
n = Abs[x]; nn = If[y == 0, n, y];
If[n == 1, {1},
z = Length@
MapIndexed[Set[{p[#2], m[#2]}, {#1, 0}] & @@
{#1, First[#2]} &, FactorInteger[n][[All, 1]] ];
k = Times @@ Array[p[#]^m[#] &, z]; Set[{v, t}, {1, False}];
Union@ Reap[Do[Set[t, k > nn];
If[t, k /= p[v]^m[v]; m[v] = 0; v++; If[v > z, Break[]],
v = 1; Sow[k] ]; m[v]++; k *= p[v], {i, Infinity}] ][[-1, 1]] ] ];
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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