OFFSET
1,1
COMMENTS
Let p be a prime. Then the n-th number with p divisors is equal to prime(n)^(p-1). - Omar E. Pol, May 06 2008
LINKS
R. J. Mathar, Table of n, a(n) for n = 1..1000
OEIS Wiki, Index entries for number of divisors
FORMULA
a(n) = A000040(n)^10, i.e. tenth power of n-th prime. - Henry Bottomley, Aug 20 2001
From Amiram Eldar, Jan 24 2021: (Start)
Product_{n>=1} (1 - 1/a(n)) = 1/zeta(10) = 93555/Pi^10 = 1/A013668. (End)
MATHEMATICA
(Prime@Range@30)^10 (* Vladimir Joseph Stephan Orlovsky, Apr 11 2011 *)
PROG
(Sage)
[p**10 for p in prime_range(100)]
# Zerinvary Lajos, May 16 2007
(Magma) [p^10: p in PrimesUpTo(300)]; // Vincenzo Librandi, Mar 27 2014
(PARI) is(n)=isprimepower(n)==10 \\ Charles R Greathouse IV, Jun 19 2016
(Python)
from sympy import prime
def A030629(n): return prime(n)**10 # Chai Wah Wu, Sep 13 2024
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved