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%I #8 Mar 03 2015 00:51:33
%S 0,1,4,0,4,7,0,4,5,0,1,6,0,9,3,0,6,5,0,8,5,0,3,2,0,6,3,0,3,9,0,5,5,0,
%T 3,6,0,5,2,0,8,6,0,8,6,0,2,12,0,8,1,0,2,7,0,2,1,0,4,5,0,7,5,0,8,6,0,7,
%U 6,0,3,9,0,4,11,0,4,5,0,5,2
%N Let S_n be the set of the integers having alternating bit sum equal to n. There are a(n) primes among the smallest 2n+1 numbers of S_n.
%H W. Bomfim, <a href="http://oeis.org/w/images/2/26/A184907.txt">A method to find the first 2n+1 integers having alternating bit sum equal to n</a>
%e The smallest 2n+1 = 5 numbers of the set S_2 of the integers having alternating bit sum 2, are 5, 17, 20, 23, and 29, so a(2)=4.
%o (PARI)II()={i = (4^n - 1)/3 - 2^(2*n-2) + 2^(2*n); if(isprime(i),an++)};
%o III()={w = 2^(2*n-2); for(j=1, n-1, i += w; w /= 4; i -= w; if(isprime(i), an++;))};
%o IV()={i+=3; if(isprime(i), an++); w=2; for(j=1, n-1, i -= w; w *= 4; i+=w; if(isprime(i),an++))};
%o print1("0, 1, 4, ");for(n=3,80, an=0; II(); III(); IV(); print1(an,", ")) \\ _Washington Bomfim_, Jan 25 2011
%Y Cf. A065359, A065084, A002450.
%K nonn
%O 0,3
%A _Washington Bomfim_, Jan 25 2011
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