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A347907
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Numbers k such that 2^(4*k-1) == 1 (mod k).
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3
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1, 7, 511, 10033, 242959, 1265839, 1838599, 4138729, 4446631, 10561159, 13179319, 19926007, 21224239, 38356159, 65746249, 72161239, 82180303, 87563239, 88323689, 98352799, 124563313, 153394537, 158525689, 219011569, 248520769, 348485359, 498260329, 636381799, 638395369, 685333399, 689463889
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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 4*k-1, where ord(2,k) is the multiplicative order of 2 modulo k.
Numbers k such that 2*k is in A130421.
Terms > 7 must be composite, since for odd primes p we have 2^(4*p-1) == 8 (mod p). If k > 1 is a term, then 4*k-1 must also be composite, since ord(2,k) | (4*k-1) and ord(2,k) <= eulerphi(k) <= k-1 < 4*k-1.
If k > 1 is a term, then (2^(4*k-1) - 1)/k is composite. Proof: since 4*k-1 is composite, write 4*k-1 = u*v, u >= v > 1. Proof: since 4*k-1 is composite, write 4*k-1 = u*v, u >= v > 1, then (2^(4*k-1) - 1)/k = (2^u - 1)*(2^(u*(v-1)) + ... + 2^u + 1)/k. Since k | 2^(4*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(4*k-1) - 1)/k > 1, so (2^(4*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+2) == 5 (mod t) (t = 1, 3, 9, 871, 2043, 2119, 8769, ...).
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LINKS
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FORMULA
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EXAMPLE
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7 is a term since 7 divides 2^27 - 1.
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MATHEMATICA
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Join[{1}, Parallelize[Select[Range[69*10^7], PowerMod[2, 4#-1, #]==1&]]] (* Harvey P. Dale, Apr 16 2023 *)
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PROG
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(PARI) isA347907(k) = if(k%2 && (!isprime(k) || k==7), Mod(2, k)^(4*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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