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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 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 January 21 22:47 EST 2020. Contains 331129 sequences. (Running on oeis4.)