|
|
A174002
|
|
a(n) = n*binomial(n+4, 4).
|
|
3
|
|
|
0, 5, 30, 105, 280, 630, 1260, 2310, 3960, 6435, 10010, 15015, 21840, 30940, 42840, 58140, 77520, 101745, 131670, 168245, 212520, 265650, 328900, 403650, 491400, 593775, 712530, 849555, 1006880, 1186680, 1391280, 1623160, 1884960, 2179485
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,2
|
|
COMMENTS
|
This sequence can be computed from Pascal's triangle. Find the fifth number in a row and multiply it by the second number of the next row. - Alonso del Arte, Jan 21 2018
|
|
LINKS
|
|
|
FORMULA
|
a(n) = (n^5 + 10*n^4 + 35*n^3 + 50*n^2 + 24*n) / 24.
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), with a(0)=0, a(1)=5, a(2)=30, a(3)=105, a(4)=280, a(5)=630. - Harvey P. Dale, Dec 03 2011
|
|
MATHEMATICA
|
Table[n Binomial[n + 4, 4], {n, 0, 40}] (* or *) LinearRecurrence[{6, -15, 20, -15, 6, -1}, {0, 5, 30, 105, 280, 630}, 40] (* Harvey P. Dale, Dec 03 2011 *)
|
|
PROG
|
(Magma) [ (n^5+10*n^4+35*n^3+50*n^2+24*n)/24: n in [0..40] ]; // Vincenzo Librandi, Dec 28 2010
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|