A274320 Least inverse of A073454: Smallest m such that m divided by the primes up to m have exactly n repeated residues.

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

Offset: 1

Author

Charles R Greathouse IV, Jun 17 2016

Keywords

• nonn

Comments

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

Examples

			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.
		

Links

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

Crossrefs

Cf. A073453, A073454.

Programs

PARI

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