Search: id:A101113 Results 1-1 of 1 results found. %I A101113 %S A101113 1,2,2,4,2,4,2,6,4,4,2,8,2,4,4,10,2,8,2,8,4,4,2,12,4,4,6,8,2,8,2,14,4, 4, %T A101113 4,16,2,4,4,12,2,8,2,8,8,4,2,20,4,8,4,8,2,12,4,12,4,4,2,16,2,4,8,22,4, 8, %U A101113 2,8,4,8,2,24,2,4,8,8,4,8,2,20,10,4,2,16,4,4,4,12,2,16,4,8,4,4,4,28,2, 8 %N A101113 Let t(G) = number of unitary factors of the Abelian group G. Then a(n) = sum t(G) over all Abelian groups G of order exactly n. %C A101113 From Schmidt paper: Let A denote the set of all Abelian groups. Under the operation of direct product, A is a semigroup with identity element E, the group with one element. G_1 and G_2 are relatively prime if the only common direct factor of G_1 and G_2 is E. We say that G_1 and G_2 are unitary factors of G if G=G_1 X G_2 and G_1, G_2 are relatively prime. Let t(G) denote the number of unitary factors of G. This sequence is a(n) = sum_{G in A, |G| = n} t(n). %C A101113 A101113(n) = A034444(n) * A000688(n). %D A101113 Schmidt, Peter Georg, Zur Anzahl unitaerer Faktoren abelscher Gruppen. [On the number of unitary factors in Abelian groups] Acta Arith., 64 (1993), 237-248. %D A101113 Wu, J., On the average number of unitary factors of finite Abelian groups, Acta Arith. 84 (1998), 17-29. %F A101113 a(n) = 2^(number of distinct prime factors of n) * product of prime powers in factorization of n. %e A101113 The only finite Abelian group of order 6 is C6=C2xC3. The unitary divisors are C1, C2, C3 and C6. So a(6) = 4. %t A101113 Apply[Times, 2*Map[PartitionsP, Map[Last, FactorInteger[n]]]] %Y A101113 Cf. A101114. %Y A101113 Cf. A034444, A000688. %Y A101113 Sequence in context: A147848 A129089 A124315 this_sequence A055155 A085191 A165872 %Y A101113 Adjacent sequences: A101110 A101111 A101112 this_sequence A101114 A101115 A101116 %K A101113 mult,easy,nonn %O A101113 1,2 %A A101113 Russ Cox (rsc(AT)swtch.com), Dec 01 2004 Search completed in 0.001 seconds