OFFSET
0,1
COMMENTS
The sequence is well-defined in that for each n the set of numbers with n nonprime substrings is not empty and finite. Proof of existence: Define m(n):=2*sum_{j=i..k} 8^j, where k:=floor((sqrt(8n+1)-1)/2), i:= n-(k(k+1)/2). For n=0,1,2,3,... the m(n) in base-8 representation are 2, 22, 20, 222, 220, 200, 2222, 2220, 2200, 2000, 22222, 22220, .... m(n) has k+1 digits and (k-i+1) 2’s. Thus, the number of nonprime substrings of m(n) is ((k+1)(k+2)/2)-k-1+i=(k(k+1)/2)+i=n. This proves the statement of existence. Proof of finiteness: Each 4-digit base-8 number has at least 1 nonprime substring. Hence, each 4(n+1)-digit number has at least n+1 nonprime substrings. Consequently, there is a boundary b < 8^(4n+3) such that all numbers > b have more than n nonprime substrings. It follows, that the set of numbers with n nonprime substrings is finite.
LINKS
Hieronymus Fischer, Table of n, a(n) for n = 0..50
FORMULA
EXAMPLE
a(0) = 491, since 491 = 753_8 (base-8) is the greatest number with zero nonprime substrings in base-8 representation.
a(1) = 3933 = 7535_8 has 1 nonprime substring in base-8 representation (=7535_8). All the other base-8 substrings are prime substrings. 3933 is the greatest such number with 1 nonprime substring.
a(2) = 24303 = 57357_8 has 15 substrings in base-8 representation, exactly 2 of them are nonprime substrings (57357_8 and 735_8), and there is no greater number with 2 nonprime substrings in base-3 representation.
a(3) = 32603 = 77533_8 has 15 substrings in base-8 representation, only 3 of them are nonprime substrings (33_8, 77_8, and 7753_8), and there is no greater number with 3 nonprime substrings in base-8 representation.
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Hieronymus Fischer, Dec 20 2012
STATUS
approved