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A033949
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Positive integers that do not have a primitive root.
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37
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8, 12, 15, 16, 20, 21, 24, 28, 30, 32, 33, 35, 36, 39, 40, 42, 44, 45, 48, 51, 52, 55, 56, 57, 60, 63, 64, 65, 66, 68, 69, 70, 72, 75, 76, 77, 78, 80, 84, 85, 87, 88, 90, 91, 92, 93, 95, 96, 99, 100, 102, 104, 105, 108, 110, 111, 112, 114, 115, 116, 117, 119, 120, 123
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OFFSET
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1,1
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COMMENTS
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Numbers k such that the cyclotomic polynomial Phi(k,x) is reducible over Zp for all primes p. Harrison shows that this is equivalent to k > 2 and the discriminant of Phi(k,x), A004124(k), being a square. - T. D. Noe, Nov 06 2007
Numbers k with the property that there exists a positive integer m with 1 < m < k-1 and m^2 == 1 (mod k). - Reinhard Muehlfeld, May 27 2014
Numbers k of the form a + b + 2*sqrt(a*b + 1) for positive integers a,b such that a*b + 1 is a square. Proof: If 1 < m < k - 1 and m^2 == 1 (mod k), take a = (m^2 - 1)/k and b = ((k - m)^2 - 1)/k. Conversely, if k = a + b + 2*sqrt(a*b + 1), take m = a + sqrt(a*b + 1). - Tor Gunston, Apr 24 2021
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REFERENCES
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I. Niven and H. S. Zuckerman, An Introduction to the Theory of Numbers, 4th edition, page 62, Theorem 2.25.
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LINKS
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FORMULA
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Positive integers except 1, 2, 4 and numbers of the form p^i and 2p^i, where p is an odd prime and i >= 1.
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MAPLE
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m := proc(n) local k, r; r := 1; if n = 2 then return false fi;
for k from 1 to n do if igcd(n, k) = 1 then r := modp(r*k, n) fi od; r end:
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MATHEMATICA
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Select[Range[2, 130], !IntegerQ[PrimitiveRoot[#]]&] (* Harvey P. Dale, Oct 25 2011 *)
a[n_] := Module[{j, l = {}}, While[Length[l]<n, For[j = 1+If[l=={}, 0, l // Last], True, j++, If[EulerPhi[j] > CarmichaelLambda[j], AppendTo[l, j]; Break[]]]]; l[[n]]]; Array[a, 100] (* Jean-François Alcover, May 29 2018, after Alois P. Heinz's Maple code for A277915 *)
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PROG
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(Sage)
[n for n in range(1, 100) if not Integers(n).multiplicative_group_is_cyclic()]
(Haskell)
a033949 n = a033949_list !! (n-1)
a033949_list = filter
(\x -> any ((== 1) . (`mod` x) . (^ 2)) [2 .. x-2]) [1..]
(PARI) is(n)=n>7 && (!isprimepower(if(n%2, n, n/2)) || n>>valuation(n, 2)==1) \\ Charles R Greathouse IV, Oct 08 2016
(Python)
from itertools import count, islice
from sympy.ntheory import sqrt_mod_iter
def A033949_gen(): # generator of terms
return filter(lambda n:max(filter(lambda k:k<n-1, sqrt_mod_iter(1, n))) > 1, count(3))
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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