login
This site is supported by donations to The OEIS Foundation.

 

Logo


Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A069038 Expansion of x*(1+x)^4/(1-x)^6. 9
0, 1, 10, 51, 180, 501, 1182, 2471, 4712, 8361, 14002, 22363, 34332, 50973, 73542, 103503, 142544, 192593, 255834, 334723, 432004, 550725, 694254, 866295, 1070904, 1312505, 1595906, 1926315, 2309356, 2751085, 3258006, 3837087, 4495776 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,3

COMMENTS

Figurate numbers based on the 5-dimensional regular convex polytope, variously called the 5-dimensional hyperoctahedron, or the 5-dimensional cross-polytope, which is represented by the Schlaefli symbol {3, 3, 3, 4}. It is the dual of the 5-dimensional hypercube. Hyun Kwang Kim asserts that every nonnegative integer can be represented by the sum of no more than 14 of these 5-crosspolytope numbers. - Jonathan Vos Post, Nov 16 2004

If Y_i (i=1,2,3,4) are 2-blocks of a (n+4)-set X then a(n-4) is the number of 9-subsets of X intersecting each Y_i (i=1,2,3,4). - Milan Janjic, Oct 28 2007

Starting with 1 = binomial transform of [1, 9, 32, 56, 48, 16, 0, 0, 0,...], where (1, 9, 32, 56, 48, 16) = row 5 of the Chebyshev triangle A081277. Also = row 5 of the array in A142978. - Gary W. Adamson, Jul 19 2008

Starting with the term 1 this is the self-convolution of A001844(n). - Anton Zakharov, Sep 02 2016

REFERENCES

H. S. M. Coxeter, Regular Polytopes, New York: Dover, 1973.

LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Milan Janjic, Two Enumerative Functions

Hyun Kwang Kim, On Regular Polytope Numbers, Proc. Amer. Math. Soc., 131 (2002), 65-75.

J. V. Post, Table of polytope numbers, Sorted, Through 1,000,000.

J. V. Post, Math Pages.

Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

FORMULA

Recurrence: a(n) = 6*a(n-1)-15*a(n-2)+20*a(n-3)-15*a(n-4)+6*a(n-5)-a(n-6).

a(n) = 5-crosspolytope(n) = n*(2*n^4 + 10*n^2 + 3)/15. E.g. a(5) = 501 because 5*(2*5^4 + 10*5^2 + 3)/15 = 501. - Jonathan Vos Post, Nov 16 2004

a(n) = C(n+4,5) + 4 C(n+3,5) + 6 C(n+2,5) + 4 C(n+1,5) + C(n,5)

sum(1/(((1/15)*n*(2*n^4+10*n^2+3)*n!)),n=1..infinity)=hypergeom([1, 1, 1+I*sqrt(10-2*sqrt(19))*(1/2), 1-I*sqrt(10-2*sqrt(19))*(1/2), 1+I*sqrt(10+2*sqrt(19))*(1/2), 1-I*sqrt(10+2*sqrt(19))*(1/2)], [2, 2, 2+I*sqrt(10-2*sqrt(19))*(1/2), 2-I*sqrt(10-2*sqrt(19))*(1/2), 2+I*sqrt(10+2*sqrt(19))*(1/2), 2-I*sqrt(10+2*sqrt(19))*(1/2)], 1)=1.05351734968093116819345664995829700099916... [Stephen Crowley, Jul 14 2009]

MAPLE

al:=proc(s, n) binomial(n+s-1, s); end; be:=proc(d, n) local r; add( (-1)^r*binomial(d-1, r)*2^(d-1-r)*al(d-r, n), r=0..d-1); end; [seq(be(5, n), n=0..100)];

MATHEMATICA

CoefficientList[Series[x (1 + x)^4/(1 - x)^6, {x, 0, 32}], x] (* Michael De Vlieger, Sep 02 2016 *)

PROG

(MAGMA) [n*(2*n^4 + 10*n^2 + 3)/15: n in [0..40]]; // Vincenzo Librandi, May 22 2011

(PARI) concat(0, Vec(x*(1+x)^4/(1-x)^6 + O(x^99))) \\ Altug Alkan, Sep 02 2016

CROSSREFS

Cf. A000332, A014820, A005900.

Cf. A142978, A081277.

Cf. A001844.

Sequence in context: A143855 A124162 A077044 * A213563 A030183 A224327

Adjacent sequences:  A069035 A069036 A069037 * A069039 A069040 A069041

KEYWORD

nonn,easy

AUTHOR

Vladeta Jovovic, Apr 03 2002

STATUS

approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent | More pages
The OEIS Community | Maintained by The OEIS Foundation Inc.

License Agreements, Terms of Use, Privacy Policy .

Last modified October 21 11:34 EDT 2017. Contains 293693 sequences.