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Search: id:A069932
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| A069932 |
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Number of k, 1<=k<=n, such that phi(k) divides n. |
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1
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| 1, 2, 2, 4, 2, 5, 2, 7, 2, 5, 2, 11, 2, 5, 2, 11, 2, 9, 2, 10, 2, 5, 2, 19, 2, 5, 2, 9, 2, 11, 2, 16, 2, 5, 2, 20, 2, 5, 2, 18, 2, 9, 2, 10, 2, 5, 2, 32, 2, 7, 2, 9, 2, 13, 2, 15, 2, 5, 2, 26, 2, 5, 2, 22, 2, 11, 2, 9, 2, 7, 2, 38, 2, 5, 2, 9, 2, 9, 2, 30, 2, 5, 2, 23, 2, 5, 2, 17, 2, 17, 2, 10, 2, 5
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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Sequence does not give the number of all integers of the form phi(k) dividing n (for some n and some m>n phi(m) divides n).
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FORMULA
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Asymptotically (still conjectured) : sum(k=1, n, a(k))=C*n*ln(n)+o(nln(n)) with C=1, 5...
G.f.sum(k>=1, 1/(1-x^phi(k)))
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PROGRAM
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(PARI) for(n=1, 150, print1(sum(i=1, n, if(n%eulerphi(i), 0, 1)), ", "))
(PARI) a(n)=if(n<1, 0, polcoeff(sum(k=1, n, 1/(1-x^eulerphi(k)), x*O(x^n)), n))
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CROSSREFS
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Adjacent sequences: A069929 A069930 A069931 this_sequence A069933 A069934 A069935
Sequence in context: A005127 A140773 A133911 this_sequence A056148 A057567 A005128
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KEYWORD
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easy,nonn
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AUTHOR
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Benoit Cloitre (benoit7848c(AT)orange.fr), May 05 2002
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