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A051532
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The Abelian orders (or Abelian numbers): n such that every group of order n is Abelian.
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16
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1, 2, 3, 4, 5, 7, 9, 11, 13, 15, 17, 19, 23, 25, 29, 31, 33, 35, 37, 41, 43, 45, 47, 49, 51, 53, 59, 61, 65, 67, 69, 71, 73, 77, 79, 83, 85, 87, 89, 91, 95, 97, 99, 101, 103, 107, 109, 113, 115, 119, 121, 123, 127, 131, 133, 137, 139, 141, 143, 145, 149, 151, 153, 157, 159, 161
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OFFSET
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1,2
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COMMENTS
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Except for a(2)=2 and a(4)=4, all of the terms in the sequence are odd. This is because of the existence of a non-Abelian dihedral group of order 2n for each n>2.
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REFERENCES
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J. Pakianathan and K. Shankar, Nilpotent numbers, Amer. Math. Monthly, 107 (Aug. 2000), 631-634.
W. R. Scott, Group Theory, Dover, 1987, page 217.
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LINKS
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T. D. Noe, Table of n, a(n) for n = 1..1000
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FORMULA
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n must be cube-free and its prime divisors must satisfy certain congruences.
Let the prime factorization of n be p1^e1...pr^er. Then n is in this sequence if ei<3 for all i and pi^k does not equal 1 (mod pj) for all i and j and 1 <= k <= ei. - T. D. Noe, Mar 25 2007
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EXAMPLE
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a(4)=4 because every group of order 4 is Abelian.
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MATHEMATICA
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isA051532[n_] := Catch[f = FactorInteger[n]; v = f[[All, 1]]; lv = Length[v]; For[i = 1, i <= lv, i++, If[f[[i, 2]] > 2, Throw[False], v[[i]] = f[[i, 1]]^f[[i, 2]]]]; For[i = 1, i <= lv, i++, For[j = i + 1, j <= lv, j++, If[Mod[v[[i]], f[[j, 1]]] == 1 || Mod[v[[j]], f[[i, 1]]] == 1, Throw[False]]]]; Throw[True]]; Select[Range[200], isA051532] (* From Jean-François Alcover, May 03 2012, after PARI *)
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PROG
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(PARI) is(n)=my(f=factor(n), v=vector(#f[, 1])); for(i=1, #v, if(f[i, 2]>2, return(0), v[i]=f[i, 1]^f[i, 2])); for(i=1, #v, for(j=i+1, #v, if(v[i]%f[j, 1]==1|v[j]%f[i, 1]==1, return(0)))); 1 \\ Charles R Greathouse IV, Feb 13 2011
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CROSSREFS
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Cf. A003277, A064899, A060652.
Sequence in context: A092755 A032515 A024926 * A135785 A008732 A130520
Adjacent sequences: A051529 A051530 A051531 * A051533 A051534 A051535
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KEYWORD
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nonn,nice,easy,changed
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AUTHOR
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Des MacHale
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STATUS
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approved
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