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A363704 Decimal expansion of lim_{x -> infinity} ((Sum_{k>=1} (k^(1/k^(1 + 1/x)) - 1)) - x^2).

%I A363704 #26 Jun 18 2023 10:04:47
%S A363704 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 A363704 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 A363704 3,5,4,8,4,9,2,9,2,2,1,6,2,6,9,5,3,2,6
%N A363704 Decimal expansion of lim_{x -> infinity} ((Sum_{k>=1} (k^(1/k^(1 + 1/x)) - 1)) - x^2).
%C A363704 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 A363704 The limit converges slowly.
%H A363704 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 A363704 Wikipedia, <a href="https://en.wikipedia.org/wiki/Stieltjes_constants">Stieltjes Constants</a>.
%F A363704 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 A363704 From _Vaclav Kotesovec_, Jun 17 2023: (Start)
%F A363704 Equals lim_{n->oo} (Sum_{m=1..n} m^(1/m)) - n - log(n)^2/2.
%F A363704 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 A363704 0.98854960114226875064475410833997126442199868380...
%t A363704 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 A363704 (Python)
%o A363704 # Gives 14 correct digits
%o A363704 from mpmath import stieltjes,fac
%o A363704 def limgen(n):
%o A363704     terms = []
%o A363704     for y in range(3, n):
%o A363704         for x in range(y, n):
%o A363704             terms.append((((-1)**y)*stieltjes(x)*(x-(y-1))**(y-2))/(fac(x-(y-2))*fac(y-2)))
%o A363704     n,o_sum = 2,0
%o A363704     while True:
%o A363704         n_term = 1/((n-1)**(n+1))
%o A363704         n_sum = o_sum + n_term
%o A363704         if o_sum == n_sum:
%o A363704             break
%o A363704         o_sum = n_sum
%o A363704         n += 1
%o A363704     return sum(terms) + 0.5 - stieltjes(0) + n_sum
%o A363704 print(str(limgen(60))[:-1])
%Y A363704 Cf. A098572, A350862, A329117, A351885, A001620, A363716.
%K A363704 nonn,cons
%O A363704 0,1
%A A363704 _Daniel Hoyt_, Jun 16 2023