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From the 2006-07 Mandelbrot competition by Sam Vandervelde, which asked for the smallest composite number in this collection.

There is no 29-digit number with this property, because to have 29 factors it must be of the form p^28, but no number of that form has 29 digits.

Comments from Farideh Firoozbakht and David W. Wilson, Dec 14 2006: (Start)

"If p is a prime greater than 23 then a(p) = 0. Proof. Suppose a(p) = M > 0. If p is prime then M must be a p-digit number of the form q^(p-1) where q is prime. But if q <= 7 the number of digits of q^(p-1) is less than p and if q > 7 & p > 23 the number of digits of q^(p-1) is greater than p. Hence if p is a prime greater than 23, M doesn't exist.

"But for many numbers n greater than 29, a(n) > 0. For example 10^53999 < a(54000) <= 11^2*1009^5*(10^18+9)^2999. Proof : if n = 11^2*1009^5*(10^18+9)^2999 then n has exactly 54000 divisors d_k (k=1,2, ..., 54000) and each d_k has exactly k digits. Hence a(54000) exists and a(54000) is a 54000-digit number less than n+1.

"In fact if 0 < m <= 3000 then a(18m) exists and a(18m) <= 11 * 101^2 * 1000003^2 * (10^18+9)^(m-1). The right hand side is an 18m-digit number for 1 <= m <= 243040916832487184.

"On the other hand, under generous assumptions about the size of prime gaps, we have a(2^m) <= Product_{0 <= k < m} nextPrime(10^(2^k)), where the right side has 2^m digits, which would provide an infinitude of numbers with precisely one divisor of every possible length." (End)

Further comments from Farideh Firoozbakht, Dec 17 2006: Perhaps for each natural number m we have a(2^m) = Product_{0 <= k < m} NextPrime(10^(2^k)), namely a(2^m) = a(2^(m-1))* NextPrime(10^(2^(m-1))). This would give a(2^1) = 11, a(2^2) = 1111 = 11*101, a(2^3) = 11117777 = 11*101*10007, a(2^4) = 1111777777824439 = 11*101*10007*100000007, a(2^5) = 11117777778244457818444447290779 =11*101*10007*100000007* 10000000000000061.