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A337235
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Even composite integers m such that A006190(m)^2 == 1 (mod m).
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5
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4, 8, 16, 68, 1208, 1424, 3056, 3824, 3928, 20912, 52174, 63716, 88708, 123148, 161872, 582224, 887566, 17083292, 18900412, 34648888, 39991684, 44884912, 51390736, 103170448, 107825236, 132238514, 279900272, 686071244, 769252508, 3251623346, 3358311986, 3535011826
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OFFSET
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1,1
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COMMENTS
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If p is a prime, then A006190(p)^2 == 1 (mod p).
This sequence contains the even composite integers for which the congruence holds.
The generalized Lucas sequence of integer parameters (a,b) defined by U(n+2) = a*U(n+1)-b*U(n) and U(0)=0, U(1)=1, satisfies the identity U^2(p) == 1 (mod p) whenever p is prime and b=-1,1.
For a=3, b=-1, U(n) recovers A006190(n) ("Bronze" Fibonacci numbers).
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REFERENCES
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D. Andrica, O. Bagdasar, Recurrent Sequences: Key Results, Applications and Problems. Springer (to appear, 2020).
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LINKS
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MATHEMATICA
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Select[Range[3, 25000, 2], CompositeQ[#] && Divisible[Fibonacci[#, 5]*Fibonacci[#, 5] - 1, #] &]
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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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