A134681 Number of digits of all the divisors of n.
1, 2, 2, 3, 2, 4, 2, 4, 3, 5, 3, 7, 3, 5, 5, 6, 3, 7, 3, 8, 5, 6, 3, 10, 4, 6, 5, 8, 3, 11, 3, 8, 6, 6, 5, 12, 3, 6, 6, 11, 3, 11, 3, 9, 8, 6, 3, 14, 4, 9, 6, 9, 3, 11, 6, 11, 6, 6, 3, 18, 3, 6, 8, 10, 6, 12, 3, 9, 6, 12, 3, 17, 3, 6, 9, 9, 6, 12, 3, 15, 7, 6, 3, 18, 6, 6, 6, 12, 3, 18, 6, 9, 6, 6, 6, 18
Offset: 1
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Programs
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Maple
A134681 := proc(n) add(A055642(d),d=numtheory[divisors](n)) ; end proc: seq(A134681(n),n=1..80) ; # R. J. Mathar, Feb 21 2025
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Mathematica
Array[Total[IntegerLength[Divisors[#]]]&,100] (* Harvey P. Dale, Jun 08 2013 *)
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PARI
a(n) = sumdiv(n, d, #digits(d)); \\ Michel Marcus, Sep 01 2023
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Python
from sympy import divisors def a(n): return sum(len(str(d)) for d in divisors(n, generator=True)) print([a(n) for n in range(1, 97)]) # Michael S. Branicky, Nov 03 2023
Formula
From Sida Li, Sep 01 2023: (Start)
a(n) = Sum_{d divides n} (floor(log_10(d))+1).
log_10(Product_{d divides n} d) <= a(n) <= log_10(Product_{d divides n} d) + sigma_0(n), where sigma_0(n) = A000005(n).
Equivalently, sigma_0(n)*log_10(n)/2 <= a(n) <= sigma_0(n)*log_10(n)/2 + sigma_0(n), obtained by formula in A007955.
For x >= 5, c2*log(x)^2 + c1*log(x) + c0 <= (1/x)*Sum_{n<=x} a(n) <= c2*log(x)^2 + (c1+1)*log(x) + 2*c0, where c2 = 1/(2*log(10)), c1 = (gamma-1)/log(10), c0 = 2*gamma-1, and gamma is Euler's constant. This is obtained by hyperbola trick for Sum_{n<=x} sigma_0(n), and Abel partial summation on Sum_{n<=x} sigma_0(n)*log(n). (End)
Extensions
New name from Jaroslav Krizek, Jun 15 2011
Comments