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The sequence is well defined since for each n >= 0 there is a number with n nonprime substrings.

Different from A213304, first different term is a(16).

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} 3^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-3 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-3 representation), p != 1 (mod 3), m>=d, than b := p*3^(m-d) has m*(m+1)/2 - k nonprime substrings, and a(A000217(n)-k) <= b.

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):=3*m(n)+2 for n>0. This results in m(n)=2*sum_{j=0..n-1} 3^j = 3^n - 1 or m(n)=1, 2, 22, 222, 2222, 22222, …,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-3 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 3.

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):=8*m(n)+2 for n>0. This results in m(n)=2*sum_{j=0..n-1} 8^j = 2*(8^n - 1)/7 or m(n)=1, 2, 22, 222, 2222, 22222, …, (in base-8) 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-8 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 8.

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.

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} 4^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-4 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-4 representation), m>=d, than b := p*4^(m-d) has m*(m+1)/2 - k nonprime substrings, and a(A000217(n)-k) <= b.