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A174002 a(n) = n*binomial(n+4, 4). 1
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

Table of n, a(n) for n=0..33.

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

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

Cf. A033488, A027480.

Sequence in context: A258582 A288679 A071252 * A030506 A062990 A018213

Adjacent sequences:  A173999 A174000 A174001 * A174003 A174004 A174005

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 October 21 23:02 EDT 2018. Contains 316431 sequences. (Running on oeis4.)