cp's OEIS Frontend

A222084 Number of the least divisors of n whose LCM is equal to n.

%I A222084 #33 Mar 13 2018 05:16:43
%S A222084 1,2,2,3,2,3,2,4,3,3,2,4,2,3,3,5,2,5,2,4,3,3,2,6,3,3,4,4,2,4,2,6,3,3,
%T A222084 3,6,2,3,3,5,2,5,2,4,4,3,2,8,3,5,3,4,2,7,3,5,3,3,2,5,2,3,4,7,3,5,2,4,
%U A222084 3,4,2,7,2,3,5,4,3,5,2,7,5,3,2,6,3,3,3
%N A222084 Number of the least divisors of n whose LCM is equal to n.
%C A222084 If we write n as the product of its prime factors, n = p1^a1*p2^a2*p3^a3*...*pr^ar, then tau#(n) gives the number of divisors from 1 to max(p1^a1, p2^a2, p3^a3, ..., pr^ar).
%C A222084 In general tau#(n) <= tau(n).
%C A222084 Also, tau#(n) = tau(n) is A000961, tau#(n) < tau(n) is A024619.
%C A222084 For any prime number p tau(p) = tau#(p) = 2.
%C A222084 tau#(n) = 3 only for semiprimes (A001358).
%H A222084 Paolo P. Lava, <a href="/A222084/b222084.txt">Table of n, a(n) for n = 1..1000</a>
%e A222084 For n=40, the divisors are (1, 2, 4, 5, 8, 10, 20, 40), so tau(40)=8.
%e A222084 lcm(1, 2, 4, 5, 8) = 40, but lcm(1, 2, 4, 5) = 20 < 40, so tau#(40)=5.
%p A222084 with(numtheory);
%p A222084 A222084:=proc(q)
%p A222084 local a,b,c,j,n; print(1);
%p A222084 for n from 2 to q do a:=ifactors(n)[2]; b:=nops(a); c:=0;
%p A222084   for j from 1 to b do if a[j][1]^a[j][2]>c then c:=a[j][1]^a[j][2]; fi; od;
%p A222084   a:=op(sort([op(divisors(n))])); b:=nops(divisors(n));
%p A222084   for j from 1 to b do if a[j]=c then break; fi; od; print(j); od; end:
%p A222084 A222084(100000);
%t A222084 Table[Count[ Divisors[n] , q_Integer /; q <= Max[Power @@@ FactorInteger[n]]], {n, 87}] (* _Wouter Meeussen_, Feb 09 2013 *)
%o A222084 (PARI) a(n) = {my(d = divisors(n), k = 1); while (lcm(vector(k, j, d[j])) != n, k++); k;} \\ _Michel Marcus_, Mar 13 2018
%Y A222084 Cf. A000005, A000961, A001358, A003418, A005179, A024619, A034444, A077610, A222085.
%K A222084 nonn
%O A222084 1,2
%A A222084 _Paolo P. Lava_, Feb 07 2013