cp's OEIS Frontend

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

%I A230260 #26 Dec 09 2014 03:34:41
%S A230260 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,
%T A230260 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,
%U A230260 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
%N A230260 Number of decompositions of 2n into unordered sums of two primes with an even sum of Hamming weights.
%C A230260 Number of ways to write 2*n = p + q with A000120(p) + A000120(q) = 2*k for p, q primes and some k.
%C A230260 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).
%C A230260 A045917(n) - a(n) = b(n): 0, 0, 0, 0, 1, 1, 1, 2, 1, 0, 1, 2, 0, 1, ... .
%C A230260 b(n) = 0 for n: 1, 2, 3, 4, 10, 13, 16, 19, 34, 43, 46, 49, 64, 82, 94.
%C A230260 Strengthening of Goldbach's conjecture: b(n) > 0 for all n > 94.
%C A230260 If 2*a(n) = A045917(n) then n: 1, 5, 7, 9, 14, 17, 25, 30, 33, 50, 57, 76, 77, 92, ... .
%C A230260 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
%o A230260 (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
%Y A230260 Cf. A000120, A002375, A045917, A171637.
%K A230260 nonn,base
%O A230260 1,10
%A A230260 _Juri-Stepan Gerasimov_, Oct 14 2013