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A335336
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Carmichael numbers k such that k+1 is divisible by gpf(k)+1, where gpf = A006530.
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1
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687979968481, 1928376089641, 2638625591701, 3148470889201, 3152088903601, 14682521533681, 19656816822721, 37333372057201, 47003559452641, 80643055074121, 129235662445121, 140940741166849, 196945133626801, 336301807660741, 345186571310209, 439931062854361
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OFFSET
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1,1
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COMMENTS
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Are there any Carmichael numbers k with exactly four prime factors such that k+1 is divisible by gpf(k)+1?
Richard J. McIntosh and Mitra Dipra found the following base 2 Fermat pseudoprimes with exactly four prime factors satisfying s-1 | k-1 and s+1 | k+1, where s is the largest prime factor of k: 988679226253951, 3143193486942417481, 44307784380481317090001.
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LINKS
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EXAMPLE
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For k = 687979968481 = 13 * 29 * 71 * 181 * 211 * 673, which is a Carmichael number, we have gpf(k) = 673. Thereafter we have gpf(k)+1 = 2 * 337 and k+1 = 2 * 337 * 347 * 911 * 3229, satisfying gpf(k)+1 | k+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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STATUS
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approved
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