A351885 -id:A351885 - cp's OEIS frontend

A098572 a(n) = floor(Sum_{m=1..n} m^(1/m)).

1, 2, 3, 5, 6, 7, 9, 10, 11, 13, 14, 15, 16, 18, 19, 20, 21, 22, 23, 25, 26, 27, 28, 29, 30, 31, 33, 34, 35, 36, 37, 38, 39, 40, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79

Offset: 1

Mark Hudson (mrmarkhudson(AT)hotmail.com), Sep 16 2004

			floor(1^(1/1)+2^(1/2)+3^(1/3))=3 and floor(1^(1/1)+2^(1/2)+3^(1/3)+4^(1/4))=5.

Vaclav Kotesovec, Table of n, a(n) for n = 1..10000 (terms 1..5000 from G. C. Greubel)

Cf. A351885, A363704.

[Floor((&+[k^(1/k): k in [1..n]])): n in [1..30]]; // G. C. Greubel, Feb 03 2018

A098572 := proc(p)
    option remember;
    add(root[i](i),i=1..p) ;
    floor(%) ;
end proc:

Table[Floor[Sum[k^(1/k), {k, 1, n}]], {n, 1, 50}] (* G. C. Greubel, Feb 03 2018 *)

for(n=1,30, print1(floor(sum(k=1,n, k^(1/k))), ", ")) \\ G. C. Greubel, Feb 03 2018

a(n) ~ n + log(n)^2/2 + c, where c = A363704 = sg1 + Sum_{k>=2} (-1)^k / k! * k-th derivative of zeta(k) = 0.9885496011422687506447541083399712644219986838..., where sg1 is the first Stieltjes constant (see A082633). - Vaclav Kotesovec, Jun 17 2023

A350862 Decimal expansion of Sum_{k>=1} (k^(1/k^(1 + 1/1111)) - 1).

Original entry on oeis.org

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

Offset: 7

Daniel Hoyt, Jan 19 2022

If x is a whole number greater than 1, the Sum_{k>=1} (k^(1/k^(1 + 1/sqrt(x))) - 1) = x + C, where C is a constant less than 1.
The above relation was tested for all 1 < x < 10^7.
If x = 1, the sum is A329117.
This sum demonstrates this relationship: setting sqrt(x) = 1111 generates the sum 1111^2 + C or 1234321 + C. Another example would be Sum_{k>=1} (k^(1/k^(1 + 1/sqrt(1729))) - 1) = 1729.84841430674....
Evaluating the sum at larger x values converges slower and slower. Monotonically changing extrapolation methods such as Richardson's Extrapolation must be used to compute these values.
Since the output (x + C) will be the square of the input (sqrt(x)) plus a constant less than 1, this implies that Sum_{k>=1} (k^(1/k^(1 + 1/sqrt(x))) - 1) diverges as x tends to infinity, or simplified to Sum_{k>=1} (k^(1/k) - 1).

			1234321.98281389093336864244004887748682691258771548...

Daniel Hoyt, Computing the limiting difference between the sum and integral of x^(1/x), 2022, pp. 3-4.

Cf. A329117, A308511.

digits = 120; d = 1; j = 1; s = 0; While[Abs[d] > 10^(-digits - 5), d = (-1)^j/j!*Derivative[j][Zeta][(1 + 1/1111)*j]; s += d; j++]; RealDigits[s, 10, 120][[1]] (* Vaclav Kotesovec, Jun 18 2023 *)

PARI
```
sumpos(k=1, k^(1/(k^(1 + 1/1111))) - 1)
```

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

Original entry on oeis.org

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

Offset: 0

Daniel Hoyt, Jun 16 2023

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.

The limit converges slowly.

			0.98854960114226875064475410833997126442199868380...

Daniel Hoyt, Computing the limiting difference between the sum and integral of x^(1/x), 2022, pp. 1-4.
Wikipedia, Stieltjes Constants.

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

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 *)

# Gives 14 correct digits
from mpmath import stieltjes,fac
def limgen(n):
    terms = []
    for y in range(3, n):
        for x in range(y, n):
            terms.append((((-1)**y)*stieltjes(x)*(x-(y-1))**(y-2))/(fac(x-(y-2))*fac(y-2)))
    n,o_sum = 2,0
    while True:
        n_term = 1/((n-1)**(n+1))
        n_sum = o_sum + n_term
        if o_sum == n_sum:
            break
        o_sum = n_sum
        n += 1
    return sum(terms) + 0.5 - stieltjes(0) + n_sum
print(str(limgen(60))[:-1])

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)!).
From Vaclav Kotesovec, Jun 17 2023: (Start)
Equals lim_{n->oo} (Sum_{m=1..n} m^(1/m)) - n - log(n)^2/2.
Equals sg1 + Sum_{k>=2} (-1)^k / k! * k-th derivative of zeta(k), where sg1 is the first Stieltjes constant (see A082633). (End)

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A098572 a(n) = floor(Sum_{m=1..n} m^(1/m)).

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A350862 Decimal expansion of Sum_{k>=1} (k^(1/k^(1 + 1/1111)) - 1).

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

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