OFFSET
1,3
COMMENTS
Conjecture: this sequence is finite.
Comments from Owen Whitby, Jul 10 2008 (Start): If 2,3,...,q are all of the primes <= q (q=19 is sufficient for n<=22) and if a=2^a2*3^a3*...*q^aq, where ai>=0 for all i, then f(a)=(a2+1)(a3+1)...(aq+1) is the number of divisors of a and similarly f(k)=(k2+1)(k3+1)...(kq+1).
Hence f(k*a)=r(a;k)*f(a) where r(a;k)=[(a2+k2+1)(a3+k3+1)...(aq+kq+1)]/[(a2+1)(a3+1)...(aq+1)]. For a(n) find a2,a3,...,aq to minimize a (the product of the prime powers) while satisfying each of the inequalities r(a;k)>r(a;k-1) for k=2,3,...,n.
After simplification, each inequality involves only a small number of ai and examining the inequalities sequentially is fairly tractable up to at least n=20.
Number of digits in a(1) to a(20) is 1,1,1,2,2,3,3,12,12,12,12,14,14,39,51,51,51,66,66,120. a(21), a(22), a(23) exist and are <= 5.3 10^128. (End)
LINKS
Owen Whitby, Table of n, a(n) for n = 1..20
EXAMPLE
720 is valid for a(6) because the number of divisors for the first 6 multiples of 720 are 30, 36, 40, 42, 45, 48
CROSSREFS
KEYWORD
nonn
AUTHOR
J. Lowell, May 30 2008
EXTENSIONS
a(8) to a(20) from Owen Whitby, Jul 10 2008
STATUS
approved