A043000 Number of digits in all base-b representations of n, for 2 <= b <= n.

2, 4, 7, 9, 11, 13, 16, 19, 21, 23, 25, 27, 29, 31, 35, 37, 39, 41, 43, 45, 47, 49, 51, 54, 56, 59, 61, 63, 65, 67, 70, 72, 74, 76, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99, 101, 103, 106, 108, 110, 112, 114, 116, 118, 120, 122, 124, 126

Offset

2,1

Comments

From A.H.M. Smeets, Dec 14 2019: (Start)

a(n)-a(n-1) >= 2 due to the fact that n = 10_n, so there is an increment of at least 2. If n can be written as a perfect power m^s, an additional +1 comes to it for the representation of n in each base m.

For instance, for n = 729 we have 729 = 3^6 = 9^3 = 27^2, so there is an additional increment of 3. For n = 1296 we have 1296 = 6^4 = 36^2, so there is an additional increment of 2. For n = 4096 we have 4096 = 2^12 = 4^6 = 8^4 = 16^3= 64^2, so there is an additional increment of 5. (End)

Links

A.H.M. Smeets, Table of n, a(n) for n = 2..20000

Vaclav Kotesovec, Plot of a(n)/(2*n) for n = 2..1000000

Formula

a(n) = Sum_{i=2..n} floor(log_i(i*n)); a(n) ~ 2*n. - Vladimir Shevelev, Jun 03 2011 [corrected by Vaclav Kotesovec, Apr 05 2021]

a(n) = A070939(n) + A081604(n) + A110591(n) + ... + 1. - R. J. Mathar, Jun 04 2011

From Ridouane Oudra, Nov 13 2019: (Start)

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

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

a(n) = n - 1 + A255165(n). (End)

If n is in A001597 then a(A001597(m)) - a(A001597(m)-1) = 2 + A253642(m), otherwise a(n) - a(n-1) = 2. - A.H.M. Smeets, Dec 14 2019

Example

5 = 101_2 = 12_3 = 11_4 = 10_5. Thus a(5) = 3+2+2+2 = 9.

Maple

A043000 := proc(n) add( nops(convert(n, base, b)), b=2..n) ; end proc: # R. J. Mathar, Jun 04 2011

Mathematica

Table[Total[IntegerLength[n, Range[2, n]]], {n, 2, 60}] (* Harvey P. Dale, Apr 23 2019 *)

Prog

(Magma) [&+[Floor(Log(i, i*n)):k in [2..n]]:n in [1..70]]; // Marius A. Burtea, Nov 13 2019
(Python)
def count(n, b):
    c = 0
    while n > 0:
        n, c = n//b, c+1
    return c
n = 0
while n < 50:
    n = n+1
    a, b = 0, 1
    while b < n:
        b = b+1
        a = a + count(n, b)
    print(n, a) # A.H.M. Smeets, Dec 14 2019
(PARI) a(n)=sum(b=2, n, #digits(n, b)) \\ Jeppe Stig Nielsen, Dec 14 2019
(PARI) a(n)= n-1 +sum(b=2, n, logint(n, b)) \\ Jeppe Stig Nielsen, Dec 14 2019
(PARI) a(n) = {2*n-2+sum(i=2, logint(n, 2), sqrtnint(n, i)-1)} \\ David A. Corneth, Dec 31 2019
(PARI) first(n) = my(res = vector(n)); res[1] = 2; for(i = 2, n, inc = numdiv(gcd(factor(i+1)[, 2]))+1; res[i] = res[i-1]+inc); res \\ David A. Corneth, Dec 31 2019

Crossrefs

Cf. A001597, A043306, A068953, A191322, A253642, A255165.

Sequence in context: A287723 A284589 A020904 * A160822 A111495 A187686

Adjacent sequences: A042997 A042998 A042999 * A043001 A043002 A043003

Keyword

nonn,easy,base

Author

Clark Kimberling

Status

approved