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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):=4*m(n)+2 for n>0. This results in m(n)=2*sum_{j=0..n-1} 4^j = 2*(4^n - 1)/3 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-4 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 4. a(1) = 2 is the only even term.

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):=5*m(n)+2 for n>0. This results in m(n)=2*sum_{j=0..n-1} 5^j = (5^n - 1)/2 or m(n)=1, 2, 22, 222, 2222, 22222,…, (in base-5) 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-5 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 a prime number.

No term is divisible by 5.

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):=6*m(n)+2 for n>0. This results in m(n)=2*sum_{j=0..n-1} 6^j = 2*(6^n - 1)/5 or m(n)=1, 2, 22, 222, 2222, 22222, …, (in base-6) 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-6 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 6.

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.

Only numbers > 100 are considered, since all 2-digit primes are trivial members. See A069488 for the sequence with prime terms > 100.

The sequence is infinite (for example, consider the continued concatenation of ‘11’ or of ‘13’: 111, 1111, 11111, ..., 131, 1313, 13131, ... are members).

Infinitely many terms are palindromic.

Only numbers > 10000 are considered, since all 4-digit primes are trivial members.

By definition, each term of the sequence with more than 5 digits is built up by an overlapped union of previous terms, i.e., a(254)=182339 has the two embedded previous terms a(26)=18233 and a(208)=82339.

The sequence is finite, the last term is 934919 (n=263). Proof of finiteness: Let p be a number with more than 6 digits. By the argument above, each 6-digit substring must be a previous term. The only 6-digit terms are 182339, 349199, 432713, 487793, 511933, 654799, 782339, 787793, 917333, 934919 (n=254..263, see b-file). As can be directly verified, none of them can be extended to a 7-digit number with the desired property.

The sequence is finite. Proof: Each 5-digit number has at least 2 nonprime substrings. Thus, each number with more than 5 digits has >= 2 nonprime substrings, too. Consequently, there is a boundary b<10^4, such that all numbers > b have at least 2 nonprime substrings.

The first term is a(1)=1=A213302(1). The last term is a(51)=3797=A213300(1).

The sequence is finite. Proof: Each 6-digit number has at least 4 nonprime substrings. Thus, each number with more than 6 digits has >= 4 nonprime substrings, too. Consequently, there is a boundary b<10^5, such that all numbers > b have more than 2 nonprime substrings.

The first term is a(1)=11=A213302(2). The last term is a(130)=37337=A213300(2).

The sequence is finite. Proof: Each 6-digit number has at least 4 nonprime substrings. Thus, each number with more than 6 digits has >= 4 nonprime substrings, too. Consequently, there is a boundary b<10^5, such that all numbers > b have more than 3 nonprime substrings.

The first term is a(1)=10=A213302(3). The last term is a(310)=73373=A213300(3).

The sequence is finite. Proof: Each 6-digit number has at least 4 nonprime substrings, and each 4-digit number has at least 1 nonprime substring. Thus, each 10-digit number has at least 5 nonprime substrings. Consequently, there is a boundary b, such that all numbers >= b have more than 4 nonprime substrings.

The first term is a(1)=103=A213302(4). The last term is a(653)=373379=A213300(4).