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A162668
a(n) = n*(n+1)*(n+2)*(n+3)/3.
4
0, 8, 40, 120, 280, 560, 1008, 1680, 2640, 3960, 5720, 8008, 10920, 14560, 19040, 24480, 31008, 38760, 47880, 58520, 70840, 85008, 101200, 119600, 140400, 163800, 190008, 219240, 251720, 287680, 327360, 371008, 418880, 471240, 528360, 590520
OFFSET
0,2
COMMENTS
a(n+3) is the number of equivalence classes of n-tuples from the set {1,0,-1} where the number of nonzero elements is 4 and two n-tuples are equivalent if they are negatives of each other. - Michael Somos, Oct 19 2022
LINKS
Diego Marques, The order of appearance of the product of consecutive Lucas numbers, Fibonacci Quarterly, 51 (2013), 38-43. - From N. J. A. Sloane, Mar 06 2013
FORMULA
From R. J. Mathar, Jul 13 2009: (Start)
a(n) = 8*A000332(n+3).
G.f.: 8*x/(1-x)^5. (End)
For n > 0, a(n) = 1/(Integral_{x=0..Pi/2} sin(x)^7 * cos(x)^(2*n-1) dx). - Francesco Daddi, Aug 02 2011
E.g.f.: x*(24 + 36*x + 12*x^2 + x^3)*exp(x)/3. - G. C. Greubel, Aug 27 2019
From Amiram Eldar, Nov 03 2022: (Start)
Sum_{n>=1} 1/a(n) = 1/6.
Sum_{n>=1} (-1)^(n+1)/a(n) = 4*log(2) - 8/3.
Product_{n>=1} 1-1/a(n) = 4*cos(sqrt(13)*Pi/2)*cosh(sqrt(3)*Pi/2)/(3*Pi^2). (End)
EXAMPLE
G.f. = 8*x + 40*x^2 + 120*x^3 + 280*x^4 + 560*x^5 + ... - Michael Somos, Oct 19 2022
MAPLE
seq(8*binomial(n+3, 4), n=0..40); # G. C. Greubel, Aug 27 2019
MATHEMATICA
Table[n*(n+1)*(n+2)*(n+3)/3, {n, 0, 40}] (* Vladimir Joseph Stephan Orlovsky, Jul 21 2009, modified by G. C. Greubel, Aug 27 2019 *)
8*Binomial[Range[40]+2, 4] (* G. C. Greubel, Aug 27 2019 *)
PROG
(Magma) [n*(n+1)*(n+2)*(n+3)/3: n in [0..40] ];
(PARI) binomial(n+3, 4)/8 \\ Charles R Greathouse IV, Jan 11 2012
(Sage) [8*binomial(n+3, 4) for n in (0..40)] # G. C. Greubel, Aug 27 2019
(GAP) List([0..40], n-> 8*Binomial(n+3, 4)); # G. C. Greubel, Aug 27 2019
CROSSREFS
Sequence in context: A143943 A135796 A105374 * A227733 A191903 A028596
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Jul 10 2009
EXTENSIONS
Definition factorized, offset corrected by R. J. Mathar, Jul 13 2009
STATUS
approved