

A175045


a(n) = number of distinct values k of substrings in binary n where both k is prime and A030101(k) is prime.


1



0, 0, 0, 1, 0, 1, 1, 2, 0, 0, 1, 3, 1, 3, 2, 2, 0, 1, 0, 1, 1, 1, 3, 5, 1, 1, 3, 4, 2, 5, 2, 3, 0, 0, 1, 2, 0, 2, 1, 2, 1, 2, 1, 4, 3, 4, 5, 6, 1, 2, 1, 1, 3, 4, 4, 6, 2, 2, 5, 6, 2, 6, 3, 3, 0, 0, 0, 2, 1, 2, 2, 4, 0, 1, 2, 4, 1, 3, 2, 2, 1, 2, 2, 4, 1, 1, 4, 6, 3, 3, 4, 4, 5, 7, 6, 7, 1, 2, 2, 2, 1, 4, 1, 2, 3
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OFFSET

0,8


COMMENTS

A030101(m) is the decimal value of {m written in binary and then the digits reversed}.


LINKS

Table of n, a(n) for n=0..104.


EXAMPLE

23 in binary is 10111. Looking at the distinct values of substrings: 0, 0 in decimal, is not a prime. 1, 1 in decimal, is not a prime. 10 = 2 in decimal, is a prime, but its reversal is 1, which is not a prime. 11 is 3 in decimal, which is prime; and its reversal, 11, is prime in decimal too. 101 is 5 in decimal, a prime; and its reversal, 101, is prime in decimal too. 111 is 7 in decimal, a prime, and its reversal is 7, which is also a prime. 1011 is 11 in decimal, which is prime, and its reversal, 1101, is 13, which is also a prime. And 10111 itself is 23 in decimal, a prime, and its reversal is 11101, which is 29 in decimal, also a prime. There are therefore 5 values of substrings that are prime and their binary digit reversals are prime, so a(23) = 5.
10010 is 18 in decimal. Note that "010" is a substring with a decimal value of 2, and the reversal of this substring is also 2. However, this substring does not count towards the substrings being enumerated because we first take the value k of the substring, then take A030101(k) to see if both are prime. And A030101(2) = 1, not 2.


CROSSREFS

Cf. A030101, A074832.
Sequence in context: A291701 A286352 A332898 * A292377 A216238 A157608
Adjacent sequences: A175042 A175043 A175044 * A175046 A175047 A175048


KEYWORD

base,nonn


AUTHOR

Leroy Quet, Dec 02 2009


EXTENSIONS

Extended by Ray Chandler, Dec 18 2009


STATUS

approved



