A362151 Decimal expansion of exp(zeta(2)/exp(gamma)) where gamma is the Euler-Mascheroni constant A001620.

2, 5, 1, 8, 2, 4, 9, 0, 3, 2, 5, 7, 2, 9, 6, 7, 5, 3, 9, 2, 0, 4, 0, 7, 1, 0, 5, 9, 5, 4, 2, 6, 8, 7, 0, 0, 2, 9, 3, 4, 5, 3, 5, 8, 6, 7, 8, 8, 6, 7, 7, 9, 4, 3, 8, 1, 4, 6, 2, 0, 6, 2, 1, 8, 5, 3, 6, 8, 3, 9, 3, 9, 9, 3, 4, 8, 8, 4, 6, 9, 4, 2, 4, 9, 3, 5, 1, 6, 9, 3, 4, 0, 5, 8, 5, 4, 2, 0, 8, 9, 8, 5, 0, 6, 8, 0, 4, 4, 0, 4, 2, 1, 8, 7, 3

Offset: 1

Author

Artur Jasinski, Apr 09 2023

Keywords

• cons
• nonn

Comments

Theorem: Let p(n) be the smallest prime p such that Product_{prime p<=p(n)} (1 + 1/p) >= n. Then lim_{n->oo} p(n+1)/p(n) = exp(zeta(2)/exp(gamma)).

Proof:

Follow E. C. Titchmarsh and D. R. Heath-Brown p. 67 eq. (3.15.3).

Product_{p<=x} (1+1/p) ~ log(x)*exp(gamma)/zeta(2).

For any particular integer n, it follows from the equations

n = log(x)*exp(gamma)/zeta(2) -> x_n = exp(n*exp(-gamma)*zeta(2)) and

n+1 = log(x)*exp(gamma)/zeta(2) -> x_(n+1) = exp((n+1)*exp(-gamma)*zeta(2))

that lim_{n->oo} exp((n+1)*exp(-gamma)*zeta(2))/exp(n*exp(-gamma)*zeta(2)) = exp(zeta(2)/exp(gamma)).

Convergence table:

n p(n) truncated product up to p(n) ratio p(n)/p(n-1)

22 667038311 22.0000000031301736805108740934458 2.51828570030407176

23 1679809291 23.0000000125715665307020553151962 2.51831006300326279

24 4230219377 24.0000000051805484055085694624554 2.51827359192764460

25 10652786759 25.0000000022564574124503565355420 2.51825870235476442

26 26826453991 26.0000000003663337659198715971438 2.51825692167692061

27 67555877849 27.0000000003436918565017475632101 2.51825596747390854

28 170122677001 28.0000000000496255633187331645369 2.51825129681914497

29 428411419313 29.0000000000157769668449867937821 2.51824992919951377

oo oo oo 2.51824903257296753

Examples

			2.518249032572967539204071059542687...
		

References

• E. C. Titchmarsh and D. R. Heath-Brown, The theory of the Riemann zeta-function, 2nd ed., 1986.

Crossrefs

Cf. A001620, A013661, A072997, A360895.

Programs

Mathematica

RealDigits[N[Exp[Zeta[2]/Exp[EulerGamma]], 115]][[1]]

Formula

Equals exp(A013661/exp(A001620)).

Limit_{n->oo} A072997(n+1)/A072997(n).