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%I #7 Mar 11 2014 01:32:49
%S 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,
%T 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,
%U 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
%N a(n) = number of distinct values k of substrings in binary n where both k is prime and A030101(k) is prime.
%C A030101(m) is the decimal value of {m written in binary and then the digits reversed}.
%e 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.
%e 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.
%Y Cf. A030101, A074832.
%K base,nonn
%O 0,8
%A _Leroy Quet_, Dec 02 2009
%E Extended by _Ray Chandler_, Dec 18 2009
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