login
Decimal expansion of lim_{x -> infinity} ((Sum_{k>=1} (k^(1/k^(1 + 1/x)) - 1)) - x^2).
2

%I #26 Jun 18 2023 10:04:47

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

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

%U 3,5,4,8,4,9,2,9,2,2,1,6,2,6,9,5,3,2,6

%N Decimal expansion of lim_{x -> infinity} ((Sum_{k>=1} (k^(1/k^(1 + 1/x)) - 1)) - x^2).

%C That this constant is less than one allows Sum_{k>=1} (k^(1/k^(1 + 1/x)) - 1) = floor(x^2), when x is the square root of any natural number greater than 1.

%C The limit converges slowly.

%H Daniel Hoyt, <a href="/A351885/a351885.pdf">Computing the limiting difference between the sum and integral of x^(1/x)</a>, 2022, pp. 1-4.

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Stieltjes_constants">Stieltjes Constants</a>.

%F Equals 1/2 - A001620 + Sum_{k>=2} (1/(k-1)^(k+1)) + Sum_{k>=3} Sum_{n>=k} (((-1)^k)*Stieltjes(n)*(n-k+1)^(k-2))/((n-k+2)!*(k-2)!).

%F From _Vaclav Kotesovec_, Jun 17 2023: (Start)

%F Equals lim_{n->oo} (Sum_{m=1..n} m^(1/m)) - n - log(n)^2/2.

%F Equals sg1 + Sum_{k>=2} (-1)^k / k! * k-th derivative of zeta(k), where sg1 is the first Stieltjes constant (see A082633). (End)

%e 0.98854960114226875064475410833997126442199868380...

%t digits = 120; d = 1; j = 2; s = StieltjesGamma[1]; While[Abs[d] > 10^(-digits - 5), d = (-1)^j / j! * Derivative[j][Zeta][j]; s += d; j++]; RealDigits[s, 10, 120][[1]] (* _Vaclav Kotesovec_, Jun 17 2023 *)

%o (Python)

%o # Gives 14 correct digits

%o from mpmath import stieltjes,fac

%o def limgen(n):

%o terms = []

%o for y in range(3, n):

%o for x in range(y, n):

%o terms.append((((-1)**y)*stieltjes(x)*(x-(y-1))**(y-2))/(fac(x-(y-2))*fac(y-2)))

%o n,o_sum = 2,0

%o while True:

%o n_term = 1/((n-1)**(n+1))

%o n_sum = o_sum + n_term

%o if o_sum == n_sum:

%o break

%o o_sum = n_sum

%o n += 1

%o return sum(terms) + 0.5 - stieltjes(0) + n_sum

%o print(str(limgen(60))[:-1])

%Y Cf. A098572, A350862, A329117, A351885, A001620, A363716.

%K nonn,cons

%O 0,1

%A _Daniel Hoyt_, Jun 16 2023