|
|
A204555
|
|
The number of subsets of the numbers {1,2,3...,n} consisting of at most 3 elements and at most two of those are even.
|
|
0
|
|
|
1, 2, 4, 8, 15, 26, 41, 63, 89, 126, 166, 222, 279, 358, 435, 541, 641, 778, 904, 1076, 1231, 1442, 1629, 1883, 2105, 2406, 2666, 3018, 3319, 3726, 4071, 4537, 4929, 5458, 5900, 6496, 6991, 7658, 8209, 8951, 9561, 10382, 11054, 11958, 12695, 13686, 14491
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,2
|
|
COMMENTS
|
This sequence has first six terms same as Cake numbers (A000125) after that it is different. The difference can be explained by duplicated tetrahedral numbers.
|
|
LINKS
|
|
|
FORMULA
|
a(n) = {(14*n^3+15*n^2+49*n+111)-(3*n^2-15*n+15)(-1)^n}/96.
G.f. ( 1+x-x^2+x^3+4*x^4+2*x^5-x^6 ) / ( (1+x)^3*(x-1)^4 ). - R. J. Mathar, Jan 19 2012
a(0)=1, a(1)=2, a(2)=4, a(3)=8, a(4)=15, a(5)=26, a(6)=41, a(n)=a(n-1)+ 3*a(n-2)-3*a(n-3)-3*a(n-4)+3*a(n-5)+a(n-6)-a(n-7). - Harvey P. Dale, Apr 17 2012
|
|
EXAMPLE
|
a(7) = ((14*7^3+15*7^2+49*7+111)-(3*7^2-15*7+15)(-1)^7)/96 = 63.
|
|
MAPLE
|
seq(binomial(n, 3)+binomial(n, 2)+binomial(n, 1)+binomial(n, 0)- binomial(floor(n/2), 3) , n=0..29);
|
|
MATHEMATICA
|
Table[Total[Table[Binomial[n, i], {i, 0, 3}]]-Binomial[Floor[n/2], 3], {n, 0, 60}] (* or *) LinearRecurrence[{1, 3, -3, -3, 3, 1, -1}, {1, 2, 4, 8, 15, 26, 41}, 60] (* Harvey P. Dale, Apr 17 2012 *)
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|