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. Proof: Define m(n):=2*sum_{j=i..k} 7^j, where k:=floor((sqrt(8*n+1)-1)/2), i:= n-A000217(k). For n=0,1,2,3,... the m(n) in base-7 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, which proves the statement.
If p is a number with k prime substrings and d digits (in base-7 representation), p != 1 (mod 7), m>=d, than b := p*7^(m-d) has m*(m+1)/2 - k nonprime substrings, and a(A000217(n)-k) <= b.
LINKS
Hieronymus Fischer, Table of n, a(n) for n = 0..210
FORMULA
a(n) >= 7^floor((sqrt(8*n-7)-1)/2) for n>0, equality holds if n=1 or n+1 is a triangular number (cf. A000217).
a(A000217(n)-1) = 7^(n-1), n>1.
a(A000217(n)) = floor(400 * 7^(n-4)), n>0.
a(A000217(n)) = 111…111_7 (with n digits), n>0.
a(A000217(n)-k) >= 7^(n-1) + k-1, 1<=k<=n, n>1.
a(A000217(n)-k) = 7^(n-1) + p, where p is the minimal number >= 0 such that 7^(n-1) + p, has k prime substrings in base-7 representation, 1<=k<=n, n>1.
EXAMPLE
a(0) = 2, since 2 = 2_7 is the least number with zero nonprime substrings in base-7 representation.
a(1) = 1, since 1 = 1_7 is the least number with 1 nonprime substring in base-7 representation.
a(2) = 7, since 7 = 10_7 is the least number with 2 nonprime substrings in base-7 representation (these are 0 and 1).
a(3) = 8, since 8 = 11_7 is the least number with 3 nonprime substrings in base-7 representation (1, 1 and 11).
a(4) = 51, since 51 = 102_7 is the least number with 4 nonprime substrings in base-7 representation, these are 0, 1, 02, and 102 (remember, that substrings with leading zeros are considered to be nonprime).
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Hieronymus Fischer, Dec 12 2012
STATUS
approved