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A354109
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Numbers that are neither an odd prime power nor twice an odd prime power.
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4
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1, 2, 4, 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, 124, 126, 128, 129, 130, 132, 133, 135
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OFFSET
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1,2
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COMMENTS
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Terms (1, 2, 4) followed by A033949, positive integers that do not have a primitive root.
Also numbers n for which A353768(n) and A353768(A267099(n)) are equal. Proof: if n is an odd prime power or twice such a number, then the odd prime factor in A267099(n) is in the opposite side of 4k+1 / 4k+3 divide of that of the odd prime factor of n, and subtracting one from it will give a number of the form 4k+0 in the other case, and 4k+2 in the other case, and either 4k != 4k+2 (mod 4) when the prime factor is unitary, or then 4k*(4k+1) != (4k+2)*(4k+3) (mod 4), when the odd prime has exponent > 1, so none of such n occur in this sequence. On the other hand, if n has more than two distinct odd prime factors, p and q, then (p-1)(q-1) == 0 (mod 4), or if n is a multiple of 4, then as phi(4) = 2 and phi(2^k) == 0 (mod 4) for k > 2, and with (p-1) giving at least one instance of factor 2, then both A267099(n) and n are guaranteed to be multiples of 4, regardless of whether p (and q) is (are) of the form 4k+1 or 4k+3.
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LINKS
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FORMULA
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MATHEMATICA
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q[n_] := ! (OddQ[n] && PrimePowerQ[n]) && ! (OddQ[n/2] && PrimePowerQ[n/2]); Select[Range[135], q] (* Amiram Eldar, May 20 2022 *)
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PROG
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(PARI)
A354108(n) = ((n && !bitand(n, n-1)) || !isprimepower(n/(2-(n%2))));
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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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