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A347906
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Numbers k such that 2^(2*k-1) == 1 (mod k).
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6
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1, 80519, 107663, 1284113, 1510313, 3933023, 4557713, 24849833, 71871113, 80646143, 98058097, 104832833, 106694033, 146987033, 168204191, 188997463, 205428713, 332693873, 333681761, 336327863, 380284847, 533039513, 552913169, 711999113, 725943719, 805031663, 1000519033, 1069441313, 1476327353, 1610020913
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OFFSET
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1,2
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COMMENTS
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Odd numbers k such that ord(2,k) divides 2*k-1, where ord(2,k) is the multiplicative order of 2 modulo k.
Numbers k such that 2*k is in A006935. For k > 1, k is a term if and only if 2*k is an even pseudoprime to base 2.
Terms > 1 must be composite, since for odd primes p we have 2^(2*p-1) == 2 (mod p). If k > 1 is a term, then 2*k-1 must also be composite, since ord(2,k) | (2*k-1) and ord(2,k) <= eulerphi(k) <= k-1 < 2*k-1.
If k > 1 is a term, then (2^(2*k-1) - 1)/k is composite. Proof: since 2*k-1 is composite, write 2*k-1 = u*v, u >= v > 1, then (2^(2*k-1) - 1)/k = (2^u - 1)*(2^(u*(v-1)) + ... + 2^u + 1)/k. Since k | 2^(2*k-1) - 1, there exist positive integers a,b such that a*b = k and that a | 2^u - 1 and b | 2^(u*(v-1)) + ... + 2^u + 1. Note that (2^u - 1)/a, (2^(u*(v-1)) + ... + 2^u + 1)/b >= (2^u - 1)/k >= (2^sqrt(2*k-1) - 1)/k > 1, so (2^(2*k-1) - 1)/k is the product of two integers > 1, so it is composite.
2^t - 1 is a term if and only if 2^(t+1) == 3 (mod t) (t = 1, 111481, 465793, ... in A296370).
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LINKS
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FORMULA
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EXAMPLE
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80519 is a term since 80519 divides 2^161037 - 1 (the multiplicative order of 2 modulo 80519 is 261, which is a divisor of 161037). Note that 2 * 80519 = 161038 = A006935(2) is the smallest even pseudoprime to base 2.
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PROG
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(PARI) isA347906(k) = if(k%2 && !isprime(k), Mod(2, k)^(2*k-1)==1, 0)
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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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