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A180118
a(n) = Sum_{k=1..n} (k+2)!/k! = Sum_{k=1..n} (k+2)*(k+1).
1
0, 6, 18, 38, 68, 110, 166, 238, 328, 438, 570, 726, 908, 1118, 1358, 1630, 1936, 2278, 2658, 3078, 3540, 4046, 4598, 5198, 5848, 6550, 7306, 8118, 8988, 9918, 10910, 11966, 13088, 14278, 15538, 16870, 18276, 19758, 21318, 22958, 24680, 26486, 28378, 30358
OFFSET
0,2
COMMENTS
In general, sequences of the form a(n) = sum((k+x+2)!/(k+x)!,k=1..n) have a closed form a(n) = n*(11+12*x+3*x^2+3*x*n+6*n+n^2)/3.
This sequence is related to A033487 by A033487(n) = n*a(n)-sum(a(i), i=0..n-1). - Bruno Berselli, Jan 24 2011
The minimal number of multiplications (using schoolbook method) needed to compute the matrix chain product of a sequence of n+1 matrices having dimensions 1 X 2, 2 X 3, ..., (n+1) X (n+2), respectively. - Alois P. Heinz, Jan 27 2017
LINKS
B. Berselli, A description of the recursive method in Comments lines: website Matem@ticamente (in Italian).
FORMULA
a(n) = +4*a(n-1)-6*a(n-2)+4*a(n-3)-1*a(n-4) for n>=4.
a(n) = n*(n^2+6*n+11)/3.
From Bruno Berselli, Jan 24 2011: (Start)
G.f.: 2*x*(3-3*x+x^2)/(1-x)^4. [corrected by Georg Fischer, May 10 2019]
Sum(a(k), k=0..n) = 2*A005718(n) for n>0. (End)
MATHEMATICA
f[n_]:=n*(n^2 + 6 n + 11)/3; f[Range[0, 60]] (* Vladimir Joseph Stephan Orlovsky, Feb 10 2011*)
CoefficientList[Series[2*x*(3 - 3*x + x^2)/(1 - x)^4, {x, 0, 50}], x] (* Vaclav Kotesovec, May 10 2019 *)
Table[Sum[(k+1)(k+2), {k, n}], {n, 0, 50}] (* or *) LinearRecurrence[{4, -6, 4, -1}, {0, 6, 18, 38}, 50] (* Harvey P. Dale, Apr 21 2020 *)
PROG
(Magma) [n*(n^2+6*n+11)/3: n in [0..45]]; // Vincenzo Librandi, Jun 15 2011
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Gary Detlefs, Aug 10 2010
STATUS
approved