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A079841
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Largest prime that is a concatenation of partitions of n; or 0 if no such number exists.
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0
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0, 11, 3, 211, 311, 0, 22111, 311111, 0, 11131111, 1121111111, 0, 111211111111, 2111111111111, 0, 11110111, 101111111, 0, 1111111111111111111, 2111111111111111111
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OFFSET
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1,2
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COMMENTS
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a(1) = a(3k+3)= 0. Conjecture: a(n) is nonzero for all other values of n.
When a repunit (10^n - 1)/9 = A002275(n) is a repunit prime (A004022) then a(n) = A004022(n), as is the case for n = 2, 19, 23, 317, 1031, etcetera. See also: A000041 number of partitions of n (the partition numbers). For an a(n) = 0 one must have checked up to A000041 partitions for primality and chosen the maximum, or otherwise eliminated all possibilities by proof. E.g. a(9) = 0 as all integers in the partition must be single digits, whose sum must be 9, hence any such concatenation must be divisible by 9. - Jonathan Vos Post, Mar 20 2006
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LINKS
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Table of n, a(n) for n=1..20.
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EXAMPLE
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a(4) = 211 though 31 and 13 are also prime.
a(16) = 11110111 = Concatenate(1,1,1,10,1,1,1), which shows the limitations of sum-of-digits divisibility test.
a(17) = 101111111 = Concatenate(10,1,1,1,1,1,1,1).
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CROSSREFS
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Cf. A000040, A000041, A004022, A002275.
Sequence in context: A157883 A038317 A157781 * A069869 A110782 A088073
Adjacent sequences: A079838 A079839 A079840 * A079842 A079843 A079844
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KEYWORD
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base,more,nonn
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AUTHOR
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Amarnath Murthy (amarnath_murthy(AT)yahoo.com), Feb 16 2003
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EXTENSIONS
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More terms from Jonathan Vos Post, Mar 20 2006
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STATUS
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approved
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