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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.
For n > 0: a(n) = A003506(n+4, 5).
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
G.f.: 5*x/(1-x)^6. - Colin Barker, Mar 18 2012
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
Sequence in context: A258582 A288679 A071252 * A030506 A062990 A018213
KEYWORD
nonn,easy
AUTHOR
Reinhard Zumkeller, Mar 05 2010, Mar 17 2010
EXTENSIONS
Title switched with first Formula section entry, at the suggestion of Alonso del Arte, by Jon E. Schoenfield, Jan 28 2018
STATUS
approved

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Last modified April 24 16:34 EDT 2024. Contains 371961 sequences. (Running on oeis4.)