Let S be the value of
n=1∑∞nd(n)+∑m=1ν2(n)(m−3)d(2mn),
where d(n) is the number of divisors of n and ν2(n) is the exponent of 2 in the prime factorization of n. If S can be expressed as (lnm)n for positive integers m and n, find 1000n+m.Proposed by Robin Park