OFFSET
1,3
COMMENTS
Row n begins with 1, ends with n, and has A010846(n) terms.
From Michael De Vlieger, Jul 08 2014: (Start)
Prime p has {1, p} and A010846(p) = 2.
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
Numbers 1 <= k <= n such that k | n^e with e >= 0. - Michael De Vlieger, May 29 2018
LINKS
T. D. Noe and Michael De Vlieger, Rows n = 1..1000 of triangle, flattened (first rows n=1..200 from T. D. Noe)
Michael De Vlieger, Plot k in row n at (x,y) = (k,-n) for n = 1..2^10.
FORMULA
Row n of this sequence is {k <= n : rad(k) | n }, where rad = A007947. - Michael De Vlieger, Jun 18 2024
EXAMPLE
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}.
MAPLE
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
MATHEMATICA
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]] ] ];
Array[f, 120] (* Michael De Vlieger, Jun 18 2024 *)
CROSSREFS
KEYWORD
AUTHOR
T. D. Noe, Jun 30 2009
STATUS
approved