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A268923
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All odd primes a(n) such that for all odd primes q smaller than a(n) the order of 2 modulo a(n)*q is a proper divisor of phi(a(n)*q)/2. The totient function phi is given in A000010.
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4
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17, 31, 41, 43, 73, 89, 97, 109, 113, 127, 137, 151, 157, 193, 223, 229, 233, 241, 251, 257, 277, 281, 283, 307, 313, 331, 337, 353, 397, 401, 409, 431, 433, 439, 449, 457, 499, 521, 569, 571, 577, 593, 601, 617, 631, 641, 643, 673, 683, 691, 727, 733, 739, 761, 769, 809, 811, 857, 881, 911, 919
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OFFSET
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1,1
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COMMENTS
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It seems that if for an odd prime p > 3 the order(2, p*3) < phi(p*3)/2 = p-1 then p is in this sequence.
Note that 2^(phi(p*q)/2) == 1 (mod p*q) for distinct odd primes p and q, due to Nagell's corollary on Theorem 64, p. 106. The products of distinct primes considered in the present sequence have order of 2 modulo p*q smaller than phi(p*q)/2.
Up to and including prime(100) = 541 the only odd primes p such that for all odd primes q smaller than p the order of 2 modulo p*q equals phi(p*q)/2 are 5, 7, and 11.
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LINKS
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EXAMPLE
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n=1: Order(2, 17*3) = 8, and 8 is a proper divisor of phi(17*3)/2 = 16;
order(2, 17*5) = 8, and 8 is a proper divisor of phi(17*5)/2 = 32;
order(2, 17*7) = 24, and 24 is a proper divisor of phi(17*7)/2 = 48;
order(2, 17*11) = 40, and 40 is a proper divisor of phi(17*11)/2 = 80;
order(2, 17*13) = 24, and 24 is a proper divisor of phi(17*13)/2 = 96.
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MATHEMATICA
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Select[Prime@ Range[3, 157], Function[p, AllTrue[Prime@ Range[2, PrimePi@ p - 1], Function[q, With[{e = EulerPhi[p q]/2}, And[Divisible[e, #], # != e]] &@ MultiplicativeOrder[2, p q]]]]] (* Michael De Vlieger, Apr 01 2016, Version 10 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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