OFFSET
0,2
COMMENTS
Number of ways of placing of an even number of indistinguishable objects in 6 distinguishable boxes with condition that in each box can be at most n objects.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,-14,14,0,-14,14,-6,1).
FORMULA
a(n) = ((n + 1)^6 + (1+(-1)^n)/2 )/2.
G.f.: (x^2 + 10*x + 1)*(x^4 + 16*x^3 + 26*x^2 + 16*x + 1) / ( (1+x)*(1-x)^7 ). - R. J. Mathar, Jun 06 2011
a(2n) = A175113(n). - R. J. Mathar, Jun 07 2011
EXAMPLE
a(1)=32 compositions of even natural numbers in 6 parts <= 1 are
:(0,0,0,0,0,0)--> 6!/(6!0!) = 1
:(0,0,0,0,1,1)--> 6!/(4!2!) = 15
:(0,0,1,1,1,1)--> 6!/(2!4!) = 15
:(1,1,1,1,1,1)--> 6!/(0!6!) = 1
a(2)=365 compositions of even natural numbers in 6 parts <= 2 are
:(0,0,0,0,0,0)--> 6!/(6!0!0!) = 1
:(0,0,0,0,1,1)--> 6!/(4!2!0!) = 15
:(0,0,0,0,0,2)--> 6!/(5!0!1!) = 6
:(0,0,1,1,1,1)--> 6!/(2!4!0!) = 15
:(0,0,0,1,1,2)--> 6!/(3!2!1!) = 60
:(0,0,0,0,2,2)--> 6!/(4!0!2!) = 15
:(0,1,1,1,1,2)--> 6!/(1!4!1!) = 30
:(0,0,0,2,2,2)--> 6!/(3!0!3!) = 20
:(0,0,1,1,2,2)--> 6!/(2!2!2!) = 90
:(1,1,1,1,1,1)--> 6!/(0!6!0!) = 1
:(0,1,1,2,2,2)--> 6!/(1!2!3!) = 60
:(0,0,2,2,2,2)--> 6!/(2!0!4!) = 15
:(1,1,1,1,2,2)--> 6!/(0!4!2!) = 15
:(0,2,2,2,2,2)--> 6!/(1!0!5!) = 6
:(1,1,2,2,2,2)--> 6!/(0!2!4!) = 15
:(2,2,2,2,2,2)--> 6!/(0!0!6!) = 1
MATHEMATICA
Table[1/2*((n + 1)^6 + (1 + (-1)^n)*1/2), {n, 0, 25}]
PROG
(Magma) [((n + 1)^6 + (1+(-1)^n)/2 )/2: n in [0..40]]; // Vincenzo Librandi, Jun 16 2011
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Adi Dani, Jun 03 2011
STATUS
approved