A342871 a(n) = Sum_{k=1..n} floor(n^(1/k)), n >= 1.

1, 3, 5, 8, 10, 12, 14, 17, 20, 22, 24, 26, 28, 30, 32, 36, 38, 40, 42, 44, 46, 48, 50, 52, 55, 57, 60, 62, 64, 66, 68, 71, 73, 75, 77, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 100, 102, 104, 107, 109, 111, 113, 115, 117, 119, 121, 123, 125, 127, 129, 131, 133

Offset

1,2

Links

Avid Rajai and David A. Corneth, Table of n, a(n) for n = 1..10000

Formula

Lim_{n->infinity} a(n)/n = 2.

a(n) = 2*n + sqrt(n) + O(n^(1/3)).

Lim_{n->infinity} (a(n)/n - 2)*sqrt(n) = 1.

a(n) = A043000(n) + 1 for n >= 2.

a(n) = A255165(n) + n for n >= 2.

a(n) = A089361(n) + 2*n - 1 for n >= 2.

a(n) = n + Sum_{i=1..floor(log_2(n))} floor(n^(1/i) - 1).

If n is in A001597 then a(A001597(m)) - a(A001597(m)-1) = 2 + A253642(m), otherwise a(n) - a(n-1) = 2.

2 <= a(n)/n <= 9/4 iff n >= 4.

1 <= (a(n)/n - 2)*sqrt(n) <= 27/16 iff n >= 27.

2*n + sqrt(n) < a(n) <= 2*n + (27/16)*sqrt(n) iff n >= 27.

Mathematica

Table[Sum[Floor[n^(1/k)], {k, n}], {n, 100}] (* Giorgos Kalogeropoulos, Mar 31 2021 *)

Prog

(PARI) a(n)=sum(k=1, n, sqrtnint(n, k)) \\ Andrew Howroyd, Mar 28 2021
(PARI) a(n) = if(n < 2, return(n)); my(c = logint(n, 2)); 2*n + sum(i = 2, c, sqrtnint(n, i)) - c \\ David A. Corneth, Mar 28 2021
(Python)
from sympy import integer_nthroot
def A342871(n):
    c = 0
    for k in range(1, n+1):
        m = integer_nthroot(n, k)[0]
        if m == 1:
            return c+n-k+1
        else:
            c += m
    return c # Chai Wah Wu, Apr 06 2021

Crossrefs

Cf. A043000, A255165, A089361, A001597, A253642.

Sequence in context: A360394 A193714 A184584 * A191160 A189929 A047391

Adjacent sequences: A342868 A342869 A342870 * A342872 A342873 A342874

Keyword

nonn

Author

Avid Rajai, Mar 28 2021

Status

approved