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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
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
Sequence in context: A038666 A075625 A006094 * A099620 A045969 A100513
KEYWORD
nonn
AUTHOR
STATUS
approved