0,1

A positive integer m is said to be n-round if it is divisible by all primes p satisfying p^(n+1) < m, or equivalently if all positive integers t < m satisfying GCD(t,m)=1 are divisible by at most n primes (counting multiplicities). Using the fact that p_(t+1)<2*p_t (p_t the (t)th prime) it is easy to prove that there are only finitely many n-round numbers for each n. 1-round numbers are usually called very round (A048597).

T. D. Noe, Table of n, a(n) for n=0..100

a(4)=2042040 as follows. Certainly it is 4-round since it is <= 19^5 and divisible by all primes < 19. Also it is > 17^5, hence the largest 4-round number must be a multiple of 510510 = 2.3.5.7.11.13.17. But no 4-round number can be > 19^5 (since it is easy to prove that if p is a prime >= 19 and q is the next prime after p then 2.3.5....p > q^5 ). Thus 2042040, being the largest multiple of 510510 which is <= 19^5, must be the largest 4-round number.

Table[k=1; While[prod=Times@@Prime[Range[k]]; prod<Prime[k+1]^(n+1), k++ ]; prod=prod/Prime[k]; prod*Floor[Prime[k]^(n+1)/prod], {n, 0, 100}] - T. D. Noe (noe(AT)sspectra.com), Sep 21 2006

Cf. A048597, A122936 (2-round numbers), A122937 (3-round numbers).

Sequence in context: A114938 A082653 A140174 this_sequence A132104 A144501 A162841

Adjacent sequences: A089013 A089014 A089015 this_sequence A089017 A089018 A089019

easy,nonn

Paul Boddington (psb(AT)maths.warwick.ac.uk), Nov 04 2003

More terms from T. D. Noe (noe(AT)sspectra.com), Sep 21 2006