A230260 Number of decompositions of 2n into unordered sums of two primes with an even sum of Hamming weights.

0, 1, 1, 1, 1, 0, 1, 0, 1, 2, 2, 1, 3, 1, 1, 2, 2, 0, 2, 1, 1, 2, 3, 3, 2, 2, 2, 2, 3, 3, 1, 0, 3, 2, 2, 4, 4, 2, 4, 3, 2, 2, 5, 3, 2, 4, 3, 2, 3, 3, 1, 4, 5, 3, 4, 2, 5, 5, 5, 5, 3, 2, 4, 3, 2, 2, 5, 2, 5, 5, 5, 7, 4, 2, 4, 2, 4, 5, 4, 1, 4, 5, 5, 8, 5, 4, 4, 4, 5, 4, 4, 4, 6, 5, 3, 8, 4, 2, 7, 5

Offset: 1

Author

Juri-Stepan Gerasimov, Oct 14 2013

Keywords

• nonn
• base

Comments

Number of ways to write 2*n = p + q with A000120(p) + A000120(q) = 2*k for p, q primes and some k.

A045917(n) = a(n) + (number of decompositions of 2n into unordered sums of two primes where Hamming weight of concatenation of this primes is equal to 2*m+1).

A045917(n) - a(n) = b(n): 0, 0, 0, 0, 1, 1, 1, 2, 1, 0, 1, 2, 0, 1, ... .

b(n) = 0 for n: 1, 2, 3, 4, 10, 13, 16, 19, 34, 43, 46, 49, 64, 82, 94.

Strengthening of Goldbach's conjecture: b(n) > 0 for all n > 94.

If 2*a(n) = A045917(n) then n: 1, 5, 7, 9, 14, 17, 25, 30, 33, 50, 57, 76, 77, 92, ... .

a(n) = 0 for n = 2*4^m, m>0 since 2*2*4^m in binary is 1 followed by an even number of zeros, and so 4^m-x and x (because they are binary complement of each other) together always have 2m+1 one bits, as long as x is odd. - Ralf Stephan, Oct 16 2013

Crossrefs

Cf. A000120, A002375, A045917, A171637.

Programs

PARI

a(n)=my(s);forprime(p=2,n, if((hammingweight(2*n-p)+hammingweight(p))%2==0 && isprime(2*n-p), s++)); s \\ Charles R Greathouse IV, Oct 14 2013