OFFSET
1,1
COMMENTS
If two such numbers A050278(n_1)/(2^k_1*5^m_1) and A050278(n_2)/(2^k_2*5^m_2) are equal, then A050278(n_1) appears earlier than A050278(n_2) iff A050278(n_1)<A050278(n_2). For example, a(8)/(2^0*5^8)=a(9)/(2^1*5^8)= 4671. There are 234710 such pairs.
Note that, a(1) = 3076521984 means that min(A050278(n)/(2^k*5^m)) = 3076521984/(2^21*5^0) = 1467.
CROSSREFS
KEYWORD
nonn,base,fini
AUTHOR
Vladimir Shevelev and Peter J. C. Moses, May 12 2015
STATUS
approved