A100682 Floor of 4th root of pentatope numbers.

0, 1, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 6, 7, 7, 8, 8, 9, 9, 10, 10, 11, 11, 11, 12, 12, 13, 13, 14, 14, 15, 15, 16, 16, 16, 17, 17, 18, 18, 19, 19, 20, 20, 21, 21, 21, 22, 22, 23, 23, 24, 24, 25, 25, 25, 26, 26, 27, 27, 28, 28, 29, 29, 30, 30, 30, 31, 31

Offset: 0

Author

Jonathan Vos Post, Dec 06 2004

Keywords

• easy
• nonn

Comments

Conjecture: a(n) = floor((n - 3/2)/24^(1/4)) for n not in {0, 1, 6, 17, 2403, 5318}. - Charles R Greathouse IV, May 01 2012

Examples

			a(3) = 1 because floor((3*4*5*6/24)^(1/4)) = floor(15^(1/4)) = floor(1.96798967) = 1.
		

References

• J. H. Conway and R. K. Guy, The Book of Numbers, pp. 55-57, Copernicus Press, NY, 1996.

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..10000
• Hyun Kwang Kim, On Regular Polytope Numbers, Proc. Amer. Math. Soc., 131 (2003), 65-75.
• Jonathan Vos Post, Table of Polytope Numbers, Sorted, Through 1,000,000.
• Eric Weisstein's World of Mathematics, Pentatope Number

Crossrefs

Cf. A000332, A100009, A007501, A099179.

Programs

Magma

[Floor(Binomial(n+3, 4)^(1/4)): n in [3..70]]; // Vincenzo Librandi, Dec 14 2015

Maple

a:= n-> floor(binomial(n+3, 4)^(1/4)):
seq(a(n), n=0..70);  # Alois P. Heinz, Dec 14 2015

PARI

a(n)=binomial(n+3,4)^(1/4)\1 \\ Charles R Greathouse IV, May 01 2012

PARI

a(n)=sqrtnint(binomial(n+3,4),4) \\ Charles R Greathouse IV, Dec 14 2015

Python

from math import comb
from sympy import integer_nthroot
def A100682(n): return integer_nthroot(comb(n+3,4),4)[0] # Chai Wah Wu, Oct 02 2024

Formula

a(n) = floor((A000332(n+3))^(1/4)) = floor(Ptop(n)^(1/4)) = floor(C(n+3, 4)^1/4) = floor((n * (n+1) * (n+2) * (n+3)/4!)^(1/4)).

a(n) = 0.4518... * n + O(1). - Charles R Greathouse IV, Dec 14 2015