A043000 -id:A043000 - cp's OEIS frontend

A043306 Sum of all digits in all base-b representations for n, for 2 <= b <= n.

1, 3, 4, 8, 10, 16, 17, 21, 25, 35, 34, 46, 52, 60, 58, 74, 73, 91, 92, 104, 114, 136, 128, 144, 156, 168, 171, 199, 193, 223, 221, 241, 257, 281, 261, 297, 315, 339, 333, 373, 367, 409, 416, 430, 452, 498, 472, 508, 515, 547, 556, 608, 598, 638, 634, 670, 698, 756, 717, 777

Offset: 2

Clark Kimberling

			5 = 101_2 = 12_3 = 11_4 = 10_5. Thus a(5) = 2 + 3 + 2 + 1 = 8.

Alois P. Heinz, Table of n, a(n) for n = 2..1000
Robin Fissum, Digit sums and the number of prime factors of the factorial n!=1.2...n, Bachelor's project in BMAT, Norwegian University of Science and Technology, 2020.
Jan-Christoph Puchta and Jürgen Spilker, Altes und Neues zur Quersumme, Math. Semesterber., Vol. 49, No. 2 (2002), pp. 209-226; alternative link.
Vladimir Shevelev, Compact integers and factorials, Acta Arith., Vol. 126, No. 3 (2007), pp. 195-236 (see p. 205).

Cf. A014837, A043000, A068953, A191251, A191322, A191350, A309531.

Row sums of A240236.

Table[Sum[Total[First[RealDigits[n, i]]], {i, 2, n}], {n, 2, 80}] (* Carl Najafi, Aug 16 2011 *)

a(n) = sum(i=2, n, vecsum(digits(n, i))); \\ Michel Marcus, Jan 03 2017

a(n) = sum(b=2, n, sumdigits(n, b)); \\ Michel Marcus, Aug 18 2017

from sympy.ntheory.digits import digits
def a(n): return sum(sum(digits(n, b)[1:]) for b in range(2, n+1))
print([a(n) for n in range(2, 62)]) # Michael S. Branicky, Apr 04 2022

From Vladimir Shevelev, Jun 03 2011: (Start)
a(n) = (n-1)*n - Sum_{i=2..n} (i-1)*Sum_{r>=1} floor(n/i^r).
a(n) <= (n-1)^2*log(n+1)/log(n).
Problem: find a better upper estimate. (End)
From Amiram Eldar, Apr 16 2021: (Start)
a(n) = A014837(n) + 1.
a(n) ~ (1-Pi^2/12)*n^2 + O(n^(3/2)) (Fissum, 2020). (End)

A255165 a(n) = Sum_{k=2..n} floor(log(n)/log(k)), n >= 1.

Original entry on oeis.org

0, 1, 2, 4, 5, 6, 7, 9, 11, 12, 13, 14, 15, 16, 17, 20, 21, 22, 23, 24, 25, 26, 27, 28, 30, 31, 33, 34, 35, 36, 37, 39, 40, 41, 42, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72

Offset: 1

Richard R. Forberg, Feb 15 2015

The sum jumps up by 2 or more where n is a power of one or more k < n, otherwise it gains 1 with each increase in n.
First differences = A089723.
This calculation is analogous to that used for the sum of the number of divisors for all integers <= n in A006218.
a(n)+n gives the number of digits in the representations of n from base 2 to base n+1. - Christina Steffan, Dec 06 2015
Without floor, Sum_{k=2..n} log(n)/log(k) ~ n * (1 + 1/log(n) + 2/log(n)^2 + 6/log(n)^3 + 24/log(n)^4 + 120/log(n)^5 + ...). - Vaclav Kotesovec, Apr 06 2021

			The first jump is at n = 4 where, in the summation, log(4)/log(2), as it reaches a new floor.
Note: Possible complications exist calculating the floor function on ratios of logs that produce exact integers (e.g., in Mathematica). Adding an infinitesimal amount to n solves it.

Vaclav Kotesovec, Table of n, a(n) for n = 1..10000
Jan Mycielski, Sur les représentations des nombres naturels par des puissances à base et exposant naturels, Colloquium Mathematicum 2 (1951), 254-260. See T(n) pp. 258-259.

Cf. A006218, A089361, A089723, A043000.

[0] cat [&+[Floor(Log(n)/Log(k)):k in [2..n]]:n in [2..70]]; // Marius A. Burtea, Nov 13 2019

Table[Sum[Floor[Log[n]/Log[k]], {k, 2, n}], {n, 1, 100}]

a(n)=sum(k=2,n,log(n)\log(k)) \\ Anders Hellström, Dec 06 2015

a(n) = Sum_{k=2..n} floor(log(n)/log(k)), n >= 1.
It appears that a(n) = A089361(n) + n - 1. - Michel Marcus, Feb 17 2015
From Ridouane Oudra, Nov 13 2019: (Start)
a(n) = Sum_{i=2..n} floor(n^(1/i)).
a(n) = Sum_{i=1..floor(log_2(n))} floor(n^(1/i) - 1).
a(n) = A043000(n) - n + 1. (End)
a(n) ~ n. - Vaclav Kotesovec, Apr 06 2021

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

Original entry on oeis.org

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

Avid Rajai, Mar 28 2021

nonn

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

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

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

a(n)=sum(k=1, n, sqrtnint(n,k)) \\ Andrew Howroyd, Mar 28 2021

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

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

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.

A327737 a(n) is the sum of the lengths of the base-b expansions of n for all b with 1 <= b <= n.

Original entry on oeis.org

1, 4, 7, 11, 14, 17, 20, 24, 28, 31, 34, 37, 40, 43, 46, 51, 54, 57, 60, 63, 66, 69, 72, 75, 79, 82, 86, 89, 92, 95, 98, 102, 105, 108, 111, 115, 118, 121, 124, 127, 130, 133, 136, 139, 142, 145, 148, 151, 155, 158, 161, 164, 167, 170, 173, 176, 179, 182, 185

Offset: 1

Steve Engledow, Sep 23 2019

			a(5) = 14 because 5 has the following representations in bases 1 to 5: 11111, 101, 12, 11, 10 giving a total length of 5+3+2+2+2 = 14.
a(12) = 37 because 12 in bases 1 through 12 is 1...1 (12 1's), 1100, 110, and for bases 4 through 12 we get a 2-digit number, for a total length of 12+4+3+9*2 = 37. - _N. J. A. Sloane_, Sep 23 2019

Cf. A043000.

package main
import (
    "fmt"
    "strconv"
)
func main() {
    // Due to limitations in strconv, this will only work for the first 36 terms
    for i := 1; i <= 36; i++ {
        count := i
        for base := 2; base <= i; base++ {
            count += len(strconv.FormatInt(int64(i), base))
        }
        fmt.Printf("%d, ", count)
    }
}

a(n) = my(i=n); for(b=2, n, i+=#digits(n, b)); i \\ Felix Fröhlich, Sep 23 2019

def count(n,b):
    c = 0
    while n > 0:
        n, c = n//b, c+1
    return c
n = 0
while n < 60:
    n = n+1
    a, b = n, 1
    while b < n:
        b = b+1
        a = a + count(n,b)
    print(n,a) # A.H.M. Smeets, Sep 23 2019

a(n) = A043000(n) + n. - A.H.M. Smeets, Sep 23 2019

More terms from Felix Fröhlich, Sep 23 2019

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A043306 Sum of all digits in all base-b representations for n, for 2 <= b <= n.

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A255165 a(n) = Sum_{k=2..n} floor(log(n)/log(k)), n >= 1.

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

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A327737 a(n) is the sum of the lengths of the base-b expansions of n for all b with 1 <= b <= n.

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