

A000125


Cake numbers: maximal number of pieces resulting from n planar cuts through a cube (or cake): C(n+1,3)+n+1.
(Formerly M1100 N0419)


64



1, 2, 4, 8, 15, 26, 42, 64, 93, 130, 176, 232, 299, 378, 470, 576, 697, 834, 988, 1160, 1351, 1562, 1794, 2048, 2325, 2626, 2952, 3304, 3683, 4090, 4526, 4992, 5489, 6018, 6580, 7176, 7807, 8474, 9178, 9920, 10701, 11522, 12384, 13288, 14235, 15226
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OFFSET

0,2


COMMENTS

Note that a(n) = a(n1) + A000124(n1). This has the following geometrical interpretation: Define a number of planes in space to be in general arrangement when
(1) no two planes are parallel,
(2) there are no two parallel intersection lines,
(3) there is no point common to four or more planes.
Suppose there are already n1 planes in general arrangement, thus defining the maximal number of regions in space obtainable by n1 planes and now one more plane is added in general arrangement. Then it will cut each of the n1 planes and acquire intersection lines which are in general arrangement. (See the comments on A000124 for general arrangement with lines.) These lines on the new plane define the maximal number of regions in 2space definable by n1 straight lines, hence this is A000124(n1). Each of this regions acts as a dividing wall, thereby creating as many new regions in addition to the a(n1) regions already there, hence a(n)=a(n1)+A000124(n1).  Peter C. Heinig (algorithms(AT)gmx.de), Oct 19 2006
More generally, we have: A000027(n) = binomial(n,0) + binomial(n,1) (the natural numbers), A000124(n) = binomial(n,0) + binomial(n,1) + binomial(n,2) (the Lazy Caterer's sequence), a(n) = binomial(n,0) + binomial(n,1) + binomial(n,2) + binomial(n,3) (Cake Numbers).  Peter C. Heinig (algorithms(AT)gmx.de), Oct 19 2006
If Y is a 2subset of an nset X then, for n>=3, a(n3) is the number of 3subsets of X which do not have exactly one element in common with Y.  Milan Janjic, Dec 28 2007
a(n) is the number of compositions (ordered partitions) of n+1 into four or fewer parts or equivalently the sum of the first four terms in the nth row of Pascal's triangle.  Geoffrey Critzer, Jan 23 2009
{a(k): 0 <= k < 4} = divisors of 8.  Reinhard Zumkeller, Jun 17 2009
a(n) is also the maximum number of different values obtained by summing n consecutive positive integers with all possible 2^n sign combinations. This maximum is first reached when summing the interval [n, 2n1].  Olivier Gérard, Mar 22 2010
a(n) contains only 5 perfect squares > 1: 4, 64, 576, 676000, and 75203584. The incidences of > 0 are given by A047694.  Frank M Jackson, Mar 15 2013
Given n tiles with two values  an A value and a B value  a player may pick either the A value or the B value. The particular tiles are [n, 0], [n1, 1], ..., [2, n2] and [1, n1]. The sequence is the number of different final A:B counts. For example, with n=4, we can have final total [5, 3] = [4, _] + [_, 1] + [_, 2] + [1, _] = [_, 0] + [3, _] + [2, _] + [_, 3], so a(4) = 2^4  1 = 15. The largest and smallest final A+B counts are given by A077043 and A002620 respectively.  Jon Perry, Oct 24 2014


REFERENCES

V. I. Arnold (ed.), Arnold's Problems, Springer, 2004, comments on Problem 199011 (p. 75), pp. 503510. Numbers N_3.
R. B. Banks, Slicing Pizzas, Racing Turtles and Further Adventures in Applied Mathematics, Princeton Univ. Press, 1999. See p. 27.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 72, Problem 2.
H. E. Dudeney, Amusements in Mathematics, Nelson, London, 1917, page 177.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
T. H. Stickels, Mindstretching Puzzles. Sterling, NY, 1994 p. 85.
W. A. Whitworth, DCC Exercises in Choice and Chance, Stechert, NY, 1945, p. 30.
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
A. M. Baxter, L. K. Pudwell, Ascent sequences avoiding pairs of patterns, 2014.
D. A. Lind, On a class of nonlinear binomial sums, Fib. Quart., 3 (1965), 292298.
Svante Linusson, The number of Msequences and fvectors, Combinatorica, vol 19 no 2 (1999) 255266.
Alexsandar Petojevic, The Function vM_m(s; a; z) and Some WellKnown Sequences, Journal of Integer Sequences, Vol. 5 (2002), Article 02.1.7
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992.
Simon Plouffe, 1031 Generating Functions and Conjectures, Université du Québec à Montréal, 1992.
D. J. Price, Some unusual series occurring in ndimensional geometry, Math. Gaz., 30 (1946), 149150.
L. Pudwell, A. Baxter, Ascent sequences avoiding pairs of patterns, 2014.
Luis Manuel Rivera, Integer sequences and kcommuting permutations, arXiv preprint arXiv:1406.3081 [math.CO], 2014
H. P. Robinson, Letter to N. J. A. Sloane, Aug 16 1971, with attachments
Eric Weisstein's World of Mathematics, Cake Number
Eric Weisstein's World of Mathematics, Cube Division by Planes
Eric Weisstein's World of Mathematics, Cylinder Cutting
Eric Weisstein's World of Mathematics, Space Division by Planes
R. Zumkeller, Enumerations of Divisors [From Reinhard Zumkeller, Jun 17 2009]
Index entries for linear recurrences with constant coefficients, signature (4,6,4,1).


FORMULA

a(n) = (n+1)*(n^2n+6)/6 = (n^3 + 5*n + 6) / 6.
G.f.: (12*x+2x^2)/(1x)^4;  [Simon Plouffe in his 1992 dissertation.]
E.g.f.: (1+x+x^2/2+x^3/6)*exp(x).
a(n) = binomial(n,3)+binomial(n,2)+binomial(n,1)+binomial(n,0). [Peter C. Heinig (algorithms(AT)gmx.de), Oct 19 2006]
Paraphrasing the previous comment: the sequence is the binomial transform of [1,1,1,1,0,0,0,...].  Gary W. Adamson, Oct 23 2007
From Ilya Gutkovskiy, Jul 18 2016: (Start)
a(n) = 4*a(n1)  6*a(n2) + 4*a(n3)  a(n1).
a(n) = Sum_{k=0..n} A152947(k+1).
Inverse binomial transform of A134396.
Sum_{n>=0} a(n)/n! = 8*exp(1)/3. (End)


EXAMPLE

a(4)=15 because there are 15 compositions of 5 into four or fewer parts. a(6)=42 because the sum of the first four terms in the 6th row of Pascal's triangle is 1+6+15+20=42.  Geoffrey Critzer, Jan 23 2009
For n=5, (1, 3, 5, 7, 9, 11, 13, 17, 19, 21, 23, 25, 35) and their opposite are the 26 different sums obtained by summing 5,6,7,8,9 with any sign combination.  Olivier Gérard, Mar 22 2010


MAPLE

A000125 := n>(n+1)*(n^2n+6)/6;


MATHEMATICA

Table[(n^3+5n+6)/6, {n, 0, 50}] (* or *) LinearRecurrence[{4, 6, 4, 1}, {1, 2, 4, 8}, 50] (* Harvey P. Dale, Jan 19 2013 *)


PROG

(PARI) a(n)=(n^2+5)*n/6+1 \\ Charles R Greathouse IV, Jun 15 2011
(MAGMA) [(n^3+5*n+6)/6: n in [0..50]]; // Vincenzo Librandi, Nov 08 2014
(PARI) Vec((12*x+2*x^2)/((1x)^4) + O(x^100)) \\ Altug Alkan, Oct 16 2015


CROSSREFS

Cf. A000124, A003600.
Bisections give A100503, A100504.
Row sums of A077028.
A005408, A000124, A016813, A086514, A058331, A002522, A161701  A161705, A000127, A161706  A161708, A080856, A161710  A161713, A161715, A006261.  Reinhard Zumkeller, Jun 17 2009
Cf. A063865.  Olivier Gérard, Mar 22 2010
Cf. A051601.  Bruno Berselli, Aug 02 2013
Cf. A077043, A002620.
Sequence in context: A262146 A089140 A204555 * A129961 A133551 A114226
Adjacent sequences: A000122 A000123 A000124 * A000126 A000127 A000128


KEYWORD

nonn,easy,nice,changed


AUTHOR

N. J. A. Sloane


EXTENSIONS

More terms from James A. Sellers, Feb 22 2000


STATUS

approved



