cp's OEIS Frontend

If n is represented as a zerofree base-4 number (see A084544) according to n=d(m)d(m-1)...d(3)d(2)d(1)d(0) then a(n) = Sum_{j=0..m} c(d(j))*10^j, where c(k)=2,3,5,7 for k=1..4. - Hieronymus Fischer, May 30 2012

According to A153025, it seems that 5, 235 and 72335 are the only terms whose square is also a term, i.e., which are also in the sequence A275971 of square roots of the terms which are squares, listed in A191486. - M. F. Hasler, Sep 16 2016

Primes in which repeatedly deleting the least significant digit gives a prime at every step until a single-digit prime remains. The sequence ends at a(83) = 73939133 = A023107(10).

The subsequence which consists of the following "chain" of consecutive right truncatable primes: 73939133, 7393913, 739391, 73939, 7393, 739, 73, 7 yields the largest sum, compared with other chains formed from subsets of this sequence: 73939133 + 7393913 + 739391 + 73939 + 7393 + 739 + 73 + 7 = 82154588. - Alexander R. Povolotsky, Jan 22 2008

Can also be seen as a table whose n-th row lists the n-digit terms; row lengths (0 for n >= 9) are given by A050986. The sequence can be constructed starting with the single-digit primes and appending, for each p in the list, the primes within 10*p and 10(p+1), formed by appending a digit to p. - M. F. Hasler, Nov 07 2018

The terms are primes for n = 1, 2, 3, 4, 5, 6, 7, 8, 10, 21, 23, 27, 31, 43, 45, 60, 67, 82, 91, .... The further terms until index 102 are composite. For the subsequence with prime terms see A211682. - [updated by Hieronymus Fischer, Oct 02 2018]

From Hieronymus Fischer, Oct 02 2018: (Start)

For indices n > 8, prime terms satisfy n mod 24 = 1, 3, 5, 7, 10, 12, 19, 21, 23. However, this condition is not sufficient. Indeed, for n <= 200 most of those terms are proven composite unless the terms with n = 103, 106, 123, 156, 165, 175, 178, 191, 193 and 195 which are potentially prime.

The terms are composite for n > 10 and n mod 24 = 0, 2, 4, 6, 8, 9, 11, 13, 14, 15, 16, 17, 18, 20, 22 (see formula section for the details).

(End)

Cf. A213299 for the partial sums.

A211681 restricted to primes. Contains A085823. In contrast to A085823 this sequence seems to be not finite (conjecture).

Only numbers > 1000 are considered, since all 3-digit primes are trivial members. See A069489 for the sequence with prime terms > 1000.

The sequence is infinite (for example, consider the continued concatenation of '19' or of '337': 1919, 19191, 191919, ..., 3373, 33733, 337337, ... are members).

Infinitely many terms are palindromic.

A 10-automatic sequence realized by a linear recurrence relation. - Charles R Greathouse IV, Jan 04 2013

Only numbers > 1000 are considered, since all 3-digit primes are trivial members.

By definition, each term of the sequence with more than 4 digits is built up by an overlapped union of previous terms, i.e., a(59)=33739 has the two embedded previous terms a(14)=3373 and a(21)=3739.

The sequence is finite, the last term is 349199 (n=63). Proof of finiteness: Let p be a number with more than 6 digits. By the argument above, each 6-digit substring of p must be a previous term. The only 6-digit term is 349199. Thus, there is no number p with the desired property.

The sequence is well-defined in that for each n the set of numbers with n nonprime substrings is nonempty and finite. Proof of existence: Define m(n):=2*sum_{j=i..k} 10^j, where k:=floor((sqrt(8n+1)-1)/2), i:= n - k(k+1)/2. For n=0,1,2,3,... the m(n) 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. This proves existence. Proof of finiteness: Each 4-digit number has at least 1 nonprime substring. Hence each 4*(n+1)-digit number has at least n+1 nonprime substrings. Consequently, there is a boundary b < 10^(4n+3) such that all numbers > b have more than n nonprime substrings. It follows that the set of numbers with n nonprime substrings is finite.

The following statements hold true:

For all n>=0 there are minimal numbers with n nonprime substrings (cf. A213302 - A213304).

For all n>=0 there are maximal numbers with n nonprime substrings (= A213300 = this sequence).

For all n>=0 there are minimal numbers with n prime substrings (cf. A035244).

The greatest number with n prime substrings does not exist. Proof: If p is a number with n prime substrings, than 10*p is a greater number with n prime substrings.

Comment from N. J. A. Sloane, Sep 01 2012: it is a surprise that any number greater than 373 has a nonprime substring!

The sequence is well-defined in that for each n the set of numbers with n prime substrings in binary representation is not empty. Proof: A000975(n+1) has exactly n prime substrings in binary representation (see A000975).

All terms with n > 1 are odd.

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):=9*m(n)+2 for n>0. This results in m(n)=2*sum_{j=0..n-1} 9^j = (9^n - 1)/4 or m(n)=1, 2, 22, 222, 2222, 22222, …, (in base-9) 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-9 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 9.