A375637 Positive numbers k such that k! does not have nontrivial infinitary divisors that are factorials.

1, 2, 6, 10, 18, 20, 24, 30, 34, 35, 36, 46, 48, 49, 54, 66, 68, 69, 72, 78, 81, 86, 87, 90, 91, 92, 96, 102, 108, 114, 116, 117, 120, 121, 126, 130, 142, 143, 150, 155, 156, 161, 166, 171, 172, 180, 184, 190, 192, 198, 204, 205, 212, 216, 222, 228, 232, 238, 240

Offset: 1

Author

Amiram Eldar, Aug 22 2024

Keywords

• nonn

Comments

The trivial infinitary divisors of a number m are 1 and m itself. Therefore if k >=2 then k! has at least 2 infinitary divisors that are factorials, 1 and k!.

Numbers k such that A375636(k) <= 2, or equivalently, 1 and numbers k such that A375636(k) = 2.

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000142, A037445, A077609, A375635, A375636.

Programs

Mathematica

expQ[e1_, e2_] := Module[{m = Length[e2], ans = 1}, Do[If[BitAnd[e1[[i]], e2[[i]]] < e2[[i]], ans = 0; Break[]], {i, 1, m}]; ans];
e[n_] := e[n] = FactorInteger[n!][[;; , 2]]; q[n_] := Sum[expQ[e[n], e[m]], {m, 2, n}] <= 1; Select[Range[240], q]

PARI

isexp(e1, e2) = {my(m = #e2, ans = 1); for(i=1,m,if(bitand(e1[i], e2[i]) < e2[i], ans = 0; break)); ans;}
e(n) = factor(n!)[,2];
is(n) = sum(m = 2, n, isexp(e(n), e(m))) <= 1;