

A274320


Least inverse of A073454: Smallest m such that m divided by the primes up to m have exactly n repeated residues.


1



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

1,1


COMMENTS

Trivially a(n) >= prime(n+1). I would like to see a better lower bound.


LINKS

Charles R Greathouse IV, Table of n, a(n) for n = 1..134


EXAMPLE

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.


PROG

(PARI) a(n)=my(P=List(), m=1); while(#P#Set(apply(p>m%p, P)) != n, if(isprime(m++), listput(P, m))); m


CROSSREFS

Cf. A073453, A073454.
Sequence in context: A038666 A075625 A006094 * A099620 A045969 A100513
Adjacent sequences: A274317 A274318 A274319 * A274321 A274322 A274323


KEYWORD

nonn


AUTHOR

Charles R Greathouse IV, Jun 17 2016


STATUS

approved



