OFFSET
1,1
COMMENTS
Subsequence of A020338 (Doublets: base-10 representation is the juxtaposition of two identical strings).
The prime factors of 10^m + 1 are also prime factors of 10^km + 1, where k is odd, so a(n) and a(km) have those prime factors in common.
Expanding on the comment in A083359 regarding finding large terms, we can generate large doublet terms in the following way:
-- for any composite number M = 10^m + 1 try to compose a prime number p of length m from the prime factors of M. Generally this will require the factors to overlap to reduce the length to m.
-- the factors can wrap around to the beginning of p. For example, if M has a factor of 137 then p can be of the form 7...13.
-- the term is formed by concatenating p with itself to form a(n) = p||p. The resulting number will consist entirely of the concatenation of its prime factors with allowed overlap as in A083359.
EXAMPLE
With m = 4: 10^4 + 1 = 10001 = 73 * 137. We can form prime p = 1373 which concatenates with itself to give a(1) = 13731373 = 73 * 137 * 1373. We can also form the prime p = 3137 which gives a(2). The number 7313 also contains all the prime factors of 10001 but it is not prime.
With m = 33: 10^33 + 1 = 7*11*11*13*23*4093*8779*599144041*183411838171, there are 4932 m-digit numbers that contain all the factors, of which 227 of them are prime. Each of these primes generates a term in the sequence with 66 digits, the smallest of which is 112359914404134093877918341183817112359914404134093877918341183817. This is A324262(33).
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Deron Stewart, Mar 13 2019
STATUS
approved