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%I #15 Apr 29 2022 08:43:56
%S 1,2,2,2,2,3,2,2,2,3,2,4,2,3,3,2,2,4,2,4,3,3,2,4,2,3,2,4,2,4,2,2,3,3,
%T 3,3,2,3,3,4,2,4,2,4,4,3,2,4,2,4,3,4,2,4,3,4,3,3,2,6,2,3,4,2,3,4,2,4,
%U 3,4,2,4,2,3,4,4,3,4,2,4,2,3,2,6,3,3,3,4,2,6,3,4,3,3,3,4,2,4,4,3,2,4,2,4,4
%N Number of distinct prime signatures represented among the unitary divisors of n.
%C a(n) = number of members m of A025487 such that d(m^k) divides d(n^k) for all values of k. (Here d(n) represents the number of divisors of n, or A000005(n).)
%C a(n) depends only on prime signature of n (cf. A025487).
%H Antti Karttunen, <a href="/A182860/b182860.txt">Table of n, a(n) for n = 1..10000</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/UnitaryDivisor.html">Unitary Divisor</a>
%H <a href="/index/Eu#epf">Index entries for sequences computed from exponents in factorization of n</a>
%F a(n) = A000005(A181819(n)).
%F If the canonical factorization of n into prime powers is Product p^e(p), then the formula for d(n^k) is Product_p (ek + 1). (See also A146289, A146290.)
%F a(n) = A064553(A328830(n)). - _Antti Karttunen_, Apr 29 2022
%e 60 has 8 unitary divisors (1, 3, 4, 5, 12, 15, 20 and 60). Primes 3 and 5 have the same prime signature, as do 12 (2^2*3) and 20 (2^2*5); each of the other four numbers listed is the only unitary divisor of 60 with its particular prime signature. Since a total of 6 distinct prime signatures appear among the unitary divisors of 60, a(60) = 6.
%t Table[Length@ Union@ Map[Sort[FactorInteger[#] /. {p_, e_} /; p > 0 :> If[p == 1, 0, e]] &, Select[Divisors@ n, CoprimeQ[#, n/#] &]], {n, 105}] (* _Michael De Vlieger_, Jul 19 2017 *)
%o (PARI)
%o A181819(n) = {my(f=factor(n)); prod(k=1, #f~, prime(f[k, 2])); }; \\ From A181819
%o A182860(n) = numdiv(A181819(n)); \\ _Antti Karttunen_, Jul 19 2017
%Y Cf. A000005, A034444, A064553, A085082, A146289, A146290, A182861, A182862, A212180, A328830.
%K nonn
%O 1,2
%A _Matthew Vandermast_, Jan 14 2011
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