Comments (unedited) from  Dimitris Valianatos, April 25 2016:

 First: This formula (1) must be placed in A271974 Let product: 1) Π(1-1/v){v ε A271974, exclude first -1->3}= =(1-1/3)*(1-1/(-5))*(1-1/7)*(1-1/9)*(1-1/(-9))*(1-1/(-11))*(1-1/15)*(1-1/(-15))*(1-1/19)*(1-1/21)*(1-1/(-21))*(1-1/(-23)*...= (2/3) * (4/3) * (6/5) * (6/7) * (8/9) * (10/9) * (12/11) * (14/15) * (16/15) * (18/19) * (20/21) * (22/21) * (24/23)* ... = 1 2) Π(1-1/v){v ε A271974, exclude first -1->3}=[Π(1-1/v){v ε A271974, exclude first -1->3}]^(-1)=1 (is the reciprocal) In correspondence with Euler product Π{p,prime}(1-1/(p^s))^(-1)=Σ{n ε N}1/(n^s),s>1 call this product (1) Π(1-1/v){v ε A271974, exclude first -1->3} dimval product or MAGIC FORMULA 3) the equality for s=1 [Π(1-1/v){v ε A271974, exclude first -1->3}]^(-1)=Sum_{n >= 1, n not divisible by 2 or 3} 1/a(n) = 1, a(n)-> A272295 (I call this equality the MAGIC KEY of primes.) 4) Subtracting 1 from the sum Sum_{n >= 1, n not divisible by 2 or 3} 1/a(n) = 1, a(n)-> A272295, then the sum of terms with a positive sign was equal to the sum of terms with a negative sign. (another formula, as a direct consequence) PROOF (1): Π(1-X(Pk)/Pk) = 4/pi for k=2 to infinite, Pk is kth prime. Π(1+X(Pk)/Pk) = 2/pi for k=2 to infinite, Pk is kth prime. Where: { +1 if P=1 mod 4 X(P)={ { -1 if P=3 mod 4 For k>3 the 2 products are equals. Π(1-X(Pk)/Pk) = 3/pi Π(1+X(Pk)/Pk) = 3/pi Because for k=2 we have (1+1/3)=4/3 and (1-1/3)=2/3 So 4/pi*3/4=3/pi and 2/pi*3/2=3/pi The quotient (Π(1-X(Pk)/Pk)) / (Π (1+X(Pk)/Pk)) = ((1-1/5)/(1+1/5))*((1+1/7)/(1-1/7))*...= =(4/5)/(6/5)*(8/7)/(6/7)*... = 4/6*8/6*12/10*... = 2/3*4/3*6/5*...=(3/pi)/(3/pi)=1 for k>3 whenever proved. See the page: http://mathworld.wolfram.com/PrimeProducts.html cf. (26)...(34)