

A046127


Maximal number of regions into which space can be divided by n spheres.


7



0, 2, 4, 8, 16, 30, 52, 84, 128, 186, 260, 352, 464, 598, 756, 940, 1152, 1394, 1668, 1976, 2320, 2702, 3124, 3588, 4096, 4650, 5252, 5904, 6608, 7366, 8180, 9052, 9984, 10978, 12036, 13160, 14352, 15614, 16948, 18356, 19840, 21402, 23044
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OFFSET

0,2


COMMENTS

If Y is a 2subset of an nset X then, for n >= 2, a(n2) is equal to the number of 2subsets and 4subsets of X having exactly one element in common with Y.  Milan Janjic, Dec 28 2007


REFERENCES

L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 73, Problem 4.
A. M. Yaglom and I. M. Yaglom: Challenging Mathematical Problems with Elementary Solutions. Vol. I. Combinatorial Analysis and Probability Theory. New York: Dover Publications, Inc., 1987, p. 13, #45 (First published: San Francisco: HoldenDay, Inc., 1964).


LINKS

T. D. Noe, Table of n, a(n) for n = 0..1000
Eric Weisstein's World of Mathematics, Space Division by Spheres.


FORMULA

n hyperspheres divide R^k into at most C(n1, k) + Sum_{i=0..k} C(n, i) regions.
a(n) = n*(n^2  3*n + 8)/3 (n >= 0).
From Philip C. Ritchey, Dec 09 2017: (Start)
The above identity proved as closed form of the following summation and its corresponding recurrence relation:
a(n) = Sum_{i=1..n} (i*(i3) + 4).
a(n) = a(n1) + n*(n3) + 4, a(0) = 0. (End)
From Colin Barker, Jan 28 2012: (Start)
a(n) = 4*a(n1)  6*a(n2) + 4*a(n3)  a(n4).
G.f.: 2*x*(1  2*x + 2*x^2)/(1  x)^4. (End)


MATHEMATICA

Join[{0}, Table[n (n^23n+8)/3, {n, 50}]] (* Harvey P. Dale, Apr 21 2011 *)


PROG

(Python)
def a(n):
....return n*(n**2  3*n + 8)//3
# Philip C. Ritchey, Dec 10 2017


CROSSREFS

Cf. A014206 (dim 2), A046127 (dim 3), A059173 (dim 4), A059174 (dim 5). See also A000124, A000125. A row of A059250.
Sequence in context: A018469 A098904 A248846 * A271480 A226454 A347775
Adjacent sequences: A046124 A046125 A046126 * A046128 A046129 A046130


KEYWORD

nonn,easy,nice


AUTHOR

Eric W. Weisstein


STATUS

approved



