A306658 Decimal expansion of the constant S_1 = Sum_{j>=1} prime(2*j - 1)!/prime(2*j)!.

3, 6, 7, 5, 5, 5, 4, 4, 2, 0, 1, 9, 2, 6

Offset: 0

Author

Marco Ripà, Mar 03 2019

Keywords

• cons
• nonn
• more

Comments

Derived from an open problem published in 2012, the constant S_1 is connected to the gap between the j-th and (j+1)-th primes.

Together with the constant S_2 (see A306700), S_1 involves the prime gaps, since twin primes produce the heaviest terms of the summation in comparison to their next and previous addend.

On Mar 06 2019, the first 4200000000 prime numbers were used and using Rosser's theorem we get:

0.367555442019264 < S_1 < 0.367555442019265 + Sum_{j>=2100000000} 1/((2*j*log(2*j) + log(log(2*j)) - 1) * (2*j*log(2*j) + log(log(2*j)) - 2)) < 0.367555442019265 + 1/1.94576*10^(-13) < 0.367555442019460.

Examples

			S_1 = 0.36755544201926...
		

References

• M. Ripà, Congetture su interrogativi inediti: tra speculazioni, voli pindarici e riflessioni spicciole, Jun 2012. ISBN 9788863699463

Links

• Wikipedia, Rosser's theorem

Crossrefs

Cf. A000040, A306700.

Programs

Mathematica

a = 0; Do[f = Prime[Range[n - 999999, n]]; Do[a += N[1/Product[k, {k, f[[i]] + 1, f[[i + 1]]}], 100], {i, 1, 1000000, 2}]; Print[n, ": ", N[a, 100]], {n, 1000000, 100000000, 1000000}]; a

PARI

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

Formula

Sum_{j>=1} prime(2*j - 1)!/prime(2*j)! = 1/3 + Sum_{j>=2} 1/(Product{k=prime(2*j - 1) + 1, prime(2*j)} k) = 1/3 + 1/(7*6) + 1/(13*12) + ...