OFFSET
1,2
COMMENTS
Also numbers such that the number of prime substrings is A000217(m-1) = m(m-1)/2, where m is the number of digits.
The sequence is finite. Proof: Let p be a number >= 10^17 and let m = 9k+j be the number of digits of p, where k = floor(m/9) >= 2 and j = m mod 9. Since each 9-digit number has at least 15 nonprime substrings, it follows that p has at least 15k = 9k + 6k > 9k + j = m nonprime substrings (since 6k >= 12> j for k >= 2). Consequently, no number >= 10^17 can be a term of the sequence.
The last term is a(858)=3733739. Proof: Each 9-digit number has at least 15 nonprime substrings, thus, the numbers 10^8 <= p < 10^14 also have at least 15 nonprime substrings and therefore cannot be terms of the sequence. Same is true for numbers 10^14 <= p < 10^17 since each 6-digit number has at least 4 nonprime substrings, and thus each number with >= 15 digits has at least 15+4 = 19 nonprime substrings. Since each 8-digit number has at least 10 nonprime substrings, it follows that the last term of the sequence must be less than 10^7. By direct search we find a(858) = 3733739.
LINKS
Hieronymus Fischer, Table of n, a(n) for n = 1..858 (full sequence)
EXAMPLE
a(1) = 1, since 1 has 1 nonprime substrings.
a(43) = 117, since 117 has 3 digits and also 3 nonprime substrings (1, 1, 117).
CROSSREFS
KEYWORD
nonn,fini,full,base
AUTHOR
Hieronymus Fischer, Aug 26 2012
EXTENSIONS
Typo in example corrected, Hieronymus Fischer, Sep 11 2012
STATUS
approved