A056793 Number of divisors of lcm(1..n).

1, 2, 4, 6, 12, 12, 24, 32, 48, 48, 96, 96, 192, 192, 192, 240, 480, 480, 960, 960, 960, 960, 1920, 1920, 2880, 2880, 3840, 3840, 7680, 7680, 15360, 18432, 18432, 18432, 18432, 18432, 36864, 36864, 36864, 36864, 73728, 73728, 147456, 147456, 147456, 147456

Offset: 1

Author

Labos Elemer, Aug 28 2000

Keywords

• nonn

Comments

The ratio a(n)/a(n-1) equals 1 if n is a member of A024619, equals 2 if n is prime, and is a noninteger value if n is in A025475. The noninteger ratio never seems to exceed 3/2, but appears to equal 3/2 if n is a member of A001248. The noninteger ratio conforms to the formula 1/(1 - 1/n), which has 1 for limit and only 2 as single integer solution. In terms of coordinates (x,y), the lower values are (1/(1-1/n), 2^(n-1)) for n > 2. - Eric Desbiaux, Jul 28 2013

Conjectured partial sums of A101207. - Sean A. Irvine, Jun 25 2022

Examples

			n = 20: lcm(1..20) = 2*2*2*2*3*3*5*7*11*13*17*19 = 232792560 and d(232792560) = 5*3*64 = 960.
		

Links

• Nathaniel Johnston, Table of n, a(n) for n = 1..1000
• Angad Singh, Note 106.01, The number of divisors of the LCM of the first n natural numbers, The Mathematical Gazette, Vol. 106, No. 565 (2022), pp. 116-117.

Crossrefs

Cf. A000005, A000720, A003418, A024619, A101207.

Programs

Maple

A056793 := proc(n)
    numtheory[tau](lcm($1..n)) ;
end proc; # Nathaniel Johnston, Jun 26 2011

Mathematica

Table[DivisorSigma[0, LCM @@ Range[n]], {n, 50}]
Table[Product[Floor[Log[Prime[i], n]] + 1, {i, PrimePi[n]}], {n, 100}] (* Wei Zhou, Jun 25 2011 *)

PARI

a(n)=n+=.5;prod(e=1,log(n)\log(2),(1+1/e)^primepi(n^(1/e))) \\ Charles R Greathouse IV, Jun 06 2013

Python

from math import lcm
from sympy import divisor_count
from itertools import accumulate, count, islice
def agen(): yield from map(divisor_count, accumulate(count(1), lcm))
print(list(islice(agen(), 46))) # Michael S. Branicky, Jun 25 2022

Formula

a(n) = d(lcm(1..n)) = A000005(A003418(n)).

a(n) = Product_{prime p <= n} (floor(log(n)/log(p)) + 1). - Wei Zhou, Jun 25 2011

a(n) = Product_{k>=1} (1+1/k)^pi(n^(1/k)), where pi(n) = A000720(n) (Singh, 2022). - Amiram Eldar, Aug 19 2023