A184907 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.

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, 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, 6, 0, 3, 9, 0, 4, 11, 0, 4, 5, 0, 5, 2

Offset: 0

Author

Washington Bomfim, Jan 25 2011

Keywords

• nonn

Comments

The smallest 2n+1 numbers in the set S_n are given by (4^(n+1)-1)/3 - (-2)^k for k in 0..2*n. - Andrew Howroyd, Dec 15 2024

Examples

			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.
		

Links

• Andrew Howroyd, Table of n, a(n) for n = 0..500
• Washington Bomfim, A method to find the first 2n+1 integers having alternating bit sum equal to n.

Crossrefs

Cf. A065359, A065084, A002450, A184908.

Programs

PARI

II()={i = (4^n - 1)/3 - 2^(2*n-2) + 2^(2*n); if(isprime(i),an++)};
III()={w = 2^(2*n-2); for(j=1, n-1, i += w; w /= 4; i -= w; if(isprime(i), an++;))};
IV()={i+=3; if(isprime(i), an++); w=2; for(j=1, n-1, i -= w; w *= 4; i+=w; if(isprime(i),an++))};
print1("0, 1, 4, ");for(n=3,80, an=0; II(); III(); IV(); print1(an,", ")) \\ Washington Bomfim, Jan 25 2011

PARI

a(n)={my(m=(4^(n+1)-1)/3); sum(k=0, 2*n, isprime(bitxor(m,1<Andrew Howroyd, Dec 15 2024

Formula

a(n) = 0 for n == 0 (mod 3). - Andrew Howroyd, Dec 15 2024