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A274320
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Least inverse of A073454: Smallest m such that m divided by the primes up to m have exactly n repeated residues.
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1
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6, 15, 35, 95, 187, 259, 671, 903, 905, 1273, 1967, 2938, 3161, 4382, 6004, 6005, 9718, 11049, 12371, 14194, 16181, 17285, 20842, 27242, 27257, 31937, 35758, 35767, 50407, 54071, 56345, 59917, 59923, 75898, 86833, 86839, 106999, 116651, 116653, 134027, 134034, 134041, 156138, 171613, 173499, 188170, 194554, 194555, 228122, 253291, 253327, 260374, 302371, 302395, 302396, 346837, 368983, 376262, 376267, 376268, 376270
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OFFSET
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1,1
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COMMENTS
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Trivially a(n) >= prime(n+1). I would like to see a better lower bound.
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LINKS
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Charles R Greathouse IV, Table of n, a(n) for n = 1..134
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EXAMPLE
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The primes up to 15 are (2, 3, 5, 7, 11, 13) and 15 mod each of these primes leaves residues of (1, 0, 0, 1, 4, 2). There are two duplicates (1 appears twice and so does 0) and no smaller number has this property, so a(2) = 15.
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PROG
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(PARI) a(n)=my(P=List(), m=1); while(#P-#Set(apply(p->m%p, P)) != n, if(isprime(m++), listput(P, m))); m
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CROSSREFS
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Cf. A073453, A073454.
Sequence in context: A038666 A075625 A006094 * A099620 A045969 A100513
Adjacent sequences: A274317 A274318 A274319 * A274321 A274322 A274323
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KEYWORD
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nonn
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AUTHOR
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Charles R Greathouse IV, Jun 17 2016
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STATUS
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approved
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