A344428 Decimal expansion of exp(-2/5).

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

Offset: 0

Author

Jianing Song, May 19 2021

Keywords

• nonn
• cons

Comments

Let f(s) = zeta(zeta(s+1)) - 1, where zeta(s) is the Riemann zeta function. Then f(s) is a strictly increasing function from (0, +oo) to (0, +oo), lim_{s->0+} f(s) = 0, lim_{s->+oo} f(s) = +oo.

Conjecture:

(i) f(s) has a unique fixed point s = A069995 - 1 in (0, +oo);

(ii) Lim_{s->+oo} f(s)/2^s = 1, lim_{s->0+} f(s)/2^(-1/s) = exp(-2/5) = A344428.

If these are true, let s_0 be any real number > alpha, s_n = zeta(s_{n-1}) for n >= 1, where alpha = A069995 is the fixed point of zeta(s) in (1, +oo), then {s_{2n}} diverges quickly to +oo, {s_{2n+1}} converges quickly to 1.

This is because the derivative of zeta(zeta(s)) - s at s = alpha is (zeta'(alpha))^2 - 1 = A344427^2 - 1 > 0, so (i) implies that zeta(zeta(s)) > s for s > alpha and zeta(zeta(s)) < s for 1 < s < alpha, hence ... > s_{2n} > s_{2n-2} > ... > s_2 > s_0 > alpha > s_1 > s_3 > ... > s_{2n+1} > ..., and it follows from (i) that lim_{n->+oo} s_{2n} = +oo, lim_{n->+oo} s_{2n+1} = 1. By definition s_n - 1 = f(s_{n-2} - 1), n >= 2. For large n, s_{2n} - 1 is approximately equal to 2^(s_{2(n-1)} - 1), and 1/(s_{2n+1} - 1) is approximately equal to exp(2/5) * 2^(1/(s_{2(n-1)+1} - 1)).

Examples

			exp(-2/5) = 0.67032004603563930074... In comparison, (zeta(zeta(0.001+1)) - 1) / 2^(-1/0.001) = 0.67022226725425164463...
		

Links

• Index entries for transcendental numbers

Crossrefs

Cf. A069995, A344427.

Programs

Mathematica

RealDigits[Exp[-2/5], 10, 100][[1]] (* Amiram Eldar, May 19 2021 *)

PARI

default(realprecision, 100); exp(-2/5)