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A343935
Number of ways to choose a multiset of n divisors of n.
5
1, 3, 4, 15, 6, 84, 8, 165, 55, 286, 12, 6188, 14, 680, 816, 4845, 18, 33649, 20, 53130, 2024, 2300, 24, 2629575, 351, 3654, 4060, 237336, 30, 10295472, 32, 435897, 7140, 7770, 8436, 177232627, 38, 10660, 11480, 62891499, 42, 85900584, 44, 1906884, 2118760
OFFSET
1,2
FORMULA
a(n) = ((sigma(n), n)) = binomial(sigma(n) + n - 1, n) where sigma = A000005 and binomial = A007318.
EXAMPLE
The a(1) = 1 through a(5) = 6 multisets:
{1} {1,1} {1,1,1} {1,1,1,1} {1,1,1,1,1}
{1,2} {1,1,3} {1,1,1,2} {1,1,1,1,5}
{2,2} {1,3,3} {1,1,1,4} {1,1,1,5,5}
{3,3,3} {1,1,2,2} {1,1,5,5,5}
{1,1,2,4} {1,5,5,5,5}
{1,1,4,4} {5,5,5,5,5}
{1,2,2,2}
{1,2,2,4}
{1,2,4,4}
{1,4,4,4}
{2,2,2,2}
{2,2,2,4}
{2,2,4,4}
{2,4,4,4}
{4,4,4,4}
MATHEMATICA
multchoo[n_, k_]:=Binomial[n+k-1, k];
Table[multchoo[DivisorSigma[0, n], n], {n, 25}]
PROG
(Python)
from math import comb
from sympy import divisor_count
def A343935(n): return comb(divisor_count(n)+n-1, n) # Chai Wah Wu, Jul 05 2024
CROSSREFS
Diagonal n = k of A343658.
Choosing n divisors of n - 1 gives A343936.
The version for chains of divisors is A343939.
A000005 counts divisors.
A000312 = n^n.
A007318 counts k-sets of elements of {1..n}.
A009998 = n^k (as an array, offset 1).
A059481 counts k-multisets of elements of {1..n}.
A146291 counts divisors of n with k prime factors (with multiplicity).
A253249 counts nonempty chains of divisors of n.
Strict chains of divisors:
- A067824 counts strict chains of divisors starting with n.
- A074206 counts strict chains of divisors from n to 1.
- A251683 counts strict length k + 1 chains of divisors from n to 1.
- A334996 counts strict length-k chains of divisors from n to 1.
- A337255 counts strict length-k chains of divisors starting with n.
- A337256 counts strict chains of divisors of n.
- A343662 counts strict length-k chains of divisors.
Sequence in context: A071167 A194924 A205489 * A280685 A076365 A070971
KEYWORD
nonn
AUTHOR
Gus Wiseman, May 05 2021
STATUS
approved