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

0,5

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

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

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

Example

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

Maple

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

Prog

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

Crossrefs

Cf. A000332, A100009, A007501, A099179.

Sequence in context: A362131 A095773 A062108 * A075355 A287864 A347578

Adjacent sequences: A100679 A100680 A100681 * A100683 A100684 A100685

Keyword

easy,nonn

Author

Jonathan Vos Post, Dec 06 2004

Status

approved