A306700 Decimal expansion of the constant S_2 = Sum_{j>=1} prime(2*j)!/prime(2*j + 1)!.

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

Offset: 0

Author

Marco Ripà, Mar 05 2019

Keywords

• cons
• nonn
• more

Comments

The constant S_2 is connected to the gap between the j-th and (j+1)-th primes.

Together with the constant S_1 (see A306658), S_2 involves the prime gaps, since twin primes produce the heaviest terms of the summation in comparison to their next and previous addend.

On Mar 07 2019, the first 2445000000 prime numbers were used and from Rosser's theorem we obtain:

0.05166662288423 < S_2 < 0.05166662288424 + Sum_{j>=1222500000} 1/((2*j*log(2*j) + log(log(2*j)) - 1) * (2*j*log(2*j) + log(log(2*j)) - 2)) < 0.05166662288424 + 3.22757*10^(-13) < 0.05166662288457.

Examples

			S_2 = 0.0516666228842...
		

Links

• Wikipedia, Rosser's theorem

Crossrefs

Cf. A000040, A306658 (S_1), A306700, A306780.

Programs

Mathematica

b = 0; Do[f = Prime[Range[n - 999999, n]]; Do[b += N[1/Product[k, {k, f[[i]] + 1, f[[i + 1]]}], 100], {i, 1, 1000000, 2}]; Print[n, ": ", N[b, 100]], {n, 1000001, 100000001, 1000000}]; b

PARI

suminf(j=1, prime(2*j)!/prime(2*j + 1)!) \\ Michel Marcus, Apr 02 2019

Formula

Sum_{j>=1} prime(2*j)!/prime(2*j + 1)! = Sum_{j>=1} 1/(Product{k=prime(2*j) + 1, prime(2*j + 1)} k) = 1/(5*4) + 1/(11*10*9*8) + 1/(17*16*15*14) + ...