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A219037
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Numbers k such that k divides 2^k + 2 and (k-1) divides 2^k + 1.
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3
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OFFSET
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1,1
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COMMENTS
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Also, numbers k such that 2^k == k-2 (mod k*(k-1)).
The sequence is infinite: if m is in this sequence, then so is 2^m + 2.
No other terms below 10^20.
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REFERENCES
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W. Sierpinski, 250 Problems in Elementary Number Theory. New York: American Elsevier, 1970. Problem #18.
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LINKS
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Table of n, a(n) for n=1..4.
Kin Y. Li et al., Solution to Problem 323, Mathematical Excalibur 14(2), 2009, p. 3.
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FORMULA
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Conjecture: a(n+1) = 2^a(n) + 2 for all n.
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CROSSREFS
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Intersection of A006517 and A055685.
Cf. A217468, A216822, A171959.
Sequence in context: A167006 A087331 A097419 * A156458 A244494 A136268
Adjacent sequences: A219034 A219035 A219036 * A219038 A219039 A219040
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KEYWORD
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nonn,hard,more
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AUTHOR
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Max Alekseyev, Nov 10 2012
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STATUS
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approved
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