OFFSET
1,10
COMMENTS
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
PROG
(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
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Juri-Stepan Gerasimov, Oct 14 2013
STATUS
approved