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A339870
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Composite numbers k of the form 4u+1 for which the odd part of phi(k) divides k-1.
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9
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85, 561, 1105, 1261, 1285, 2465, 4369, 6601, 8245, 8481, 9061, 9605, 10585, 16405, 16705, 17733, 18721, 19669, 21845, 23001, 28645, 30889, 38165, 42121, 43165, 46657, 54741, 56797, 57205, 62745, 65365, 74593, 78013, 83665, 88561, 91001, 106141, 117181, 124645, 126701, 134521, 136981, 141661, 162401, 171205, 176437
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OFFSET
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1,1
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COMMENTS
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Equally, squarefree composite numbers k of the form 4u+1 for which A336466(k) divides k-1. This follows because on squarefree n, A336466(n) = A053575(n).
No common terms with A016105, because 4xy + 2(x+y) + 1 does not divide 4xy + 3(x+y) + 2 for any distinct x, y >= 0 (where 4x+3 and 4y+3 are the two prime factors of Blum integers).
This can also seen by another way: If this sequence contained any Blum integers, then, because A016105 is a subsequence of A339817, we would have found a composite number n satisfying Lehmer's totient problem y * phi(n) = n-1, for some integer y > 1. But Lehmer proved that such solutions should have at least 7 distinct prime factors, while Blum integers have only two.
(End)
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LINKS
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EXAMPLE
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85 = 4*21 + 1 = 5*17, thus phi(85) = 4*16 = 64, the odd part of which is A000265(64) = 1, which certainly divides 85-1, therefore 85 is included as a term.
561 = 4*140 + 1 = 3*11*17, thus phi(561) = 2*10*16 = 320, the odd part of which is A000265(320) = 5, which divides 560, therefore 561 is included.
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MATHEMATICA
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odd[n_] := n/2^IntegerExponent[n, 2]; Select[4*Range[45000] + 1, CompositeQ[#] && Divisible[# - 1, odd[EulerPhi[#]]] &] (* Amiram Eldar, Feb 17 2021 *)
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PROG
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(PARI)
isA339870(n) = ((n>1)&&!isprime(n)&&(1==(n%4))&&!((n-1)%A000265(eulerphi(n))));
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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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