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A243822 Number of "semidivisors" of n, numbers m < n that do not divide n but divide n^e for some integer e > 1. 33

%I #51 May 19 2016 15:40:18

%S 0,0,0,0,0,1,0,0,0,2,0,2,0,2,1,0,0,4,0,2,1,3,0,3,0,3,0,2,0,10,0,0,2,4,

%T 1,5,0,4,2,3,0,11,0,3,2,4,0,5,0,6,2,3,0,8,1,3,2,4,0,14,0,4,2,0,1,14,0,

%U 4,2,12,0,6,0,5,3,4,1,15,0,4,0,5,0,16,1,5,3,3,0,20,1,4,3,5,1,8,0,7,2,6

%N Number of "semidivisors" of n, numbers m < n that do not divide n but divide n^e for some integer e > 1.

%C Semidivisors m < n are products of primes restricted to the prime divisors of n, however they have multiplicities of at least one prime divisor that exceeds the multiplicity of the corresponding prime divisor in n. As regular numbers in base n, the unit fractions of semidivisors have terminating expansions in base n (see Hardy & Wright). Semidivisors must be m < n, while the set of regular numbers in base n include the sets of semidivisors and divisors and can be larger than n.

%C a(n) = 0 for each composite n that are perfect prime powers p^e, since any semidivisor m must be p^a, with a < e, and all such possible p^a divide p^e.

%C a(n) > 0 for all composites n that are not perfect prime powers, since all squarefree semiprimes n = p * (p + 2) must have at least p^2 as an m that does not divide n. This is because p < sqrt(n), regardless of the magnitude of n. Further prime factors of n only reduce the relative size of the minimum p, ensuring that there will be a p_min^2 that does not divide n.

%H Michael De Vlieger, <a href="/A243822/b243822.txt">Table of n, a(n) for n = 1..10000</a>

%H M. De Vlieger, <a href="http://dx.doi.org/10.1145/2077808.2077809">Exploring Number Bases as Tools</a>, ACM Inroads, March 2012, Vol. 3, No. 1, pp. 4-12.

%H M. De Vlieger, <a href="http://www.vincico.com/proof/neutral.html">Neutral Numbers</a>

%F a(n) = A010846(n) - A000005(n).

%F a(n) = A045763(n) - A243823(n).

%F a(n) = (Sum_{1<=k<=n, gcd(n,k)=1} mu(k)*floor(n/k)) - tau(n). - _Michael De Vlieger_, May 10 2016, after _Benoit Cloitre_ at A010846.

%e For n = 10 with prime divisors {2, 5}, the numbers {1, 2, 4, 5, 8, 10} are regular (cf. A010846). Subtract the set of divisors {1, 2, 5, 10} to obtain the set of semidivisors of 10, {4, 8}. So a(10) = 2.

%e Note that 4 divides 10^2 and 8 divides 10^3. The set of numbers less than 10 that are neither divisors nor totatives is {4, 6, 8}; 6 is a semitotative of 10, while {4, 8} are semidivisors.

%t f[n_] := Block[{g, a}, g[x_] := First /@ FactorInteger@ x; a = g@ n; Length@ Select[Select[Range@ n, Complement[g@ #, a] == {} &], LCM[#, n] != n &]]; f /@ Range@ 120 (* _Michael De Vlieger_, Sep 15 2015 *)

%t Table[Total[MoebiusMu[#] Floor[n/#] &@ Select[Range@ n, CoprimeQ[#, n] &]] - DivisorSigma[0, n], {n, 120}] (* _Michael De Vlieger_, May 10 2016, faster *)

%Y Cf. A045763, A010846, A000005, A243823.

%K nonn

%O 1,10

%A _Michael De Vlieger_, Jun 11 2014

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