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Search: id:A048109
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| 8, 24, 27, 40, 54, 56, 88, 104, 120, 125, 135, 136, 152, 168, 184, 189, 232, 248, 250, 264, 270, 280, 296, 297, 312, 328, 343, 344, 351, 375, 376, 378, 408, 424, 440, 456, 459, 472, 488, 513, 520, 536, 552, 568, 584, 594, 616, 621, 632, 664, 680, 686, 696
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OFFSET
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1,1
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COMMENT
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For these terms the number of divisors should be a special power of two because ud(n)=2^r and nud(n)=ud(n). In particular the exponent of 2 is 1+A001221(n), the number of distinct prime factors + 1. Thus this is a subsequence of A036537 where A000005(A036537(n)) = 2^s; here s=1+A001221(n)
Let us introduce a function D(n)=sigma_0(n)/(2^(alfa(1)+...+alfa(r)), sigma_0(n) number of divisors of n (A000005), prime factorization of n=p(1)^alfa(1) * ... * p(r)^alfa(r), alfa(1)+...+alfa(r) is sequence (A086436). This function splits the set of positive integers into subsets, according to the value of D(n). Squarefree numbers (A005117) has D(n)=1, other numbers are "deviated" from the squarefree ideal and have 0 < D(n) < 1. So for D(n)=1/2 we have A048109, D(n)=3/4 we have A067295. [From Ctibor O. Zizka (c.zizka(AT)email.cz), Sep 21 2008]
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FORMULA
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d[ x ] = 2^(r[ x ]+1) or A000005[ x ]=2^(A(001221[ x ])+1)=2*A034444[ x ].
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EXAMPLE
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n=88=2*2*2*11 has 8 divisors of which 4 is unitary divisors- because of 2 distinct prime factors - and has 4 non unitary divisors: U={1,88,11,8} and NU = {2,44,4,22}.
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CROSSREFS
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A000005, A001221, A034444, A036537, A048106, A048107.
Sequence in context: A052349 A029607 A060476 this_sequence A068781 A038524 A162829
Adjacent sequences: A048106 A048107 A048108 this_sequence A048110 A048111 A048112
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KEYWORD
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nonn
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AUTHOR
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Labos E. (labos(AT)ana.sote.hu)
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