cp's OEIS Frontend

The next term (a(6)) has 127 digits. - Harvey P. Dale, Dec 18 2022

Prime versus probable prime status and proofs are given in the author's table.

a(10) > 2^16. - Lucas A. Brown, Apr 18 2021

a(10) > 2*10^5. - Tyler Busby, Feb 01 2023

a(n) = 3 if n is in A056251.

From Chai Wah Wu, Nov 05 2019 : (Start)

Theorem: a(2^r*s) >= 10^(2^r-1) for all r >= 0, s > 0.

Proof: Note that if k has m digits, then k//A002275(n)//k = k*(10^(n+m)+1) + A002275(n)*10^m which is a multiple of gcd(A002275(n),10^(n+m)+1).

Next, since 10^(2^r) - 1 = (10^(2^(r-1) -1))*(10^(2^(r-1) + 1)) and 9 does not divide 10^n+1, by induction it is easy to see that 10^(2^w) + 1 is a divisor of A002275(2^r) for 1 <= w < r. Since A002275(2^r) is a divisor of A002275(2^r*s), 10^(2^w) + 1 is also a divisor of A002275(2^r*s).

For 1 <= m < 2^r, let t be the 2-adic valuation of m, i.e. 0 <= t = A007814(m) < r.

Then 10^(2^r*s+m)+1 = 10^(2^t*q)+1 = (10^(2^t))^q + 1 for some odd number q.

Since the sum of two odd powers a^q+b^q is divisible by a+b, this implies that 10^(2^r*s+m)+1 is divisible by 10^(2^t)+1.

This means that for n = 2^r*s and 1 <= m < 2^r, gcd(A002275(n),10^(n+m)+1) >= 10^(2^t)+1 > 1, i.e. k//A002275(n)//k is not prime.

Thus a(2^r*s) must have at least 2^r digits, i.e. a(2^r*s) >= 10^(2^r-1). QED

As a consequence, a(n) >= 10^(A006519(n)-1). This result is still true if some of the digits of k are leading zeros.

(End)