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A295835
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Numbers k == 3 (mod 4) such that 2^((k-1)/2), 3^((k-1)/2) and 5^((k-1)/2) are congruent to 1 (mod k).
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1
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71, 191, 239, 311, 359, 431, 479, 599, 719, 839, 911, 1031, 1151, 1319, 1439, 1511, 1559, 1871, 2039, 2111, 2351, 2399, 2591, 2711, 2879, 2999, 3119, 3191, 3359, 3671, 3719, 3911, 4079, 4271, 4391, 4679, 4751, 4799, 4871, 4919, 5039, 5231, 5279, 5351, 5399, 5471
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OFFSET
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1,1
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COMMENTS
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There are very few composite numbers in this sequence: The probability of catching a pseudoprime number (A001567) with this definition is estimated at 1 in 263 billion.
Composite numbers in the sequence include the Carmichael numbers 131314855918751, 23282264781147191, 70122000249565031, 104782993259720471, 583701149409931151, 870012810301712351. - Robert Israel, Nov 28 2017
With the exception of the pseudoprimes, it seems that this is a subsequence of A139035. Primes of this form (A139035) have two special properties. 1. There exists a smallest m of the form m = (A139035 - 1)/2 such that 2^m == 1 (mod A139035). 2. m is odd. The core of this definition is based on these two properties. The term 2^((k-1)/2) == 1 (mod n) is based on the first property, while the term k == 3 (mod 4) is based on the second property. The terms 3^((k-1)/2) == 1 (mod n) and 5^((k-1)/2) == 1 (mod n) I just tried freely to Fermat.
Prime terms are congruent to 71 or 119 modulo 120. - Jianing Song, Nov 22 2018
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LINKS
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MAPLE
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filter:= proc(n) [2&^((n-1)/2), 3&^((n-1)/2), 5&^((n-1)/2)] mod n = [1, 1, 1] end proc:
select(filter, [seq(i, i=3..10000, 4)]); # Robert Israel, Nov 28 2017, corrected Feb 26 2018
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MATHEMATICA
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fQ[n_] := PowerMod[{2, 3, 5}, (n - 1)/2, n] == {1, 1, 1}; Select[3 + 4Range@ 1500, fQ] (* Michael De Vlieger, Nov 28 2017 and slightly modified by Robert G. Wilson v, Feb 26 2018 based on the renaming *)
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PROG
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(PARI) is(n) = n%4==3 && Mod(2, n)^(n\2)==1 && Mod(3, n)^(n\2)==1 && Mod(5, n)^(n\2)==1 && Mod(2, n)^logint(n+1, 2)!=1 \\ Charles R Greathouse IV, Nov 28 2017
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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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