A175045 a(n) = number of distinct values k of substrings in binary n where both k is prime and A030101(k) is prime.

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

Offset: 0

Author

Leroy Quet, Dec 02 2009

Keywords

• base
• nonn

Comments

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

Examples

			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.

Extensions

Extended by Ray Chandler, Dec 18 2009