OFFSET
0,2
COMMENTS
The sequence is well-defined in that for each n the set of numbers with n prime substrings is not empty. Proof: Define m(0):=1, m(1):=2 and m(n+1):=7*m(n)+2 for n>0. This results in m(n)=2*sum_{j=0..n-1} 7^j = (7^n - 1)/3 or m(n)=1, 2, 22, 222, 2222, 22222,…, (in base-7) for n=0,1,2,3,…. Evidently, for n>0 m(n) has n 2’s and these are the only prime substrings in base-7 representation. This is why every substring of m(n) with more than one digit is a product of two integers > 1 (by definition) and can therefore not be prime number.
No term is divisible by 7.
LINKS
Hieronymus Fischer, Table of n, a(n) for n = 0..40
FORMULA
a(n) > 7^floor(sqrt(8*n-7)-1)/2), for n>0.
a(n) <= (7^n - 1)/3, n>0.
a(n+1) <= 7*a(n) + 2.
EXAMPLE
a(1) = 2 = 2_7, since 2 is the least number with 1 prime substring in base-7 representation.
a(2) = 16 = 22_7, since 16 is the least number with 2 prime substrings in base-7 representation (2 times 2_7=2).
a(3) = 17 = 23_7, since 17 is the least number with 3 prime substrings in base-7 representation (2_7, 3_7, and 23_7).
a(5) = 121 = 232_7, since 121 is the least number with 5 prime substrings in base-7 representation (2 times 2_7, 3_7, 23_7=17, and 32_7=23).
a(6) = 509 = 1325_7, since 509 is the least number with 6 prime substrings in base-7 representation (2_7, 3_7, 5_7, 25_7=19, 32_7=23, and 1325_7=509).
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Hieronymus Fischer, Nov 22 2012
STATUS
approved