A162668 a(n) = n*(n+1)*(n+2)*(n+3)/3.
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
Examples
G.f. = 8*x + 40*x^2 + 120*x^3 + 280*x^4 + 560*x^5 + ... - _Michael Somos_, Oct 19 2022
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..10000
- 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
- Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).
Programs
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GAP
List([0..40], n-> 8*Binomial(n+3,4)); # G. C. Greubel, Aug 27 2019
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Magma
[n*(n+1)*(n+2)*(n+3)/3: n in [0..40] ];
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Maple
seq(8*binomial(n+3, 4), n=0..40); # G. C. Greubel, Aug 27 2019
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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 *) -
PARI
binomial(n+3,4)/8 \\ Charles R Greathouse IV, Jan 11 2012
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Sage
[8*binomial(n+3, 4) for n in (0..40)] # G. C. Greubel, Aug 27 2019
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)
Extensions
Definition factorized, offset corrected by R. J. Mathar, Jul 13 2009
Comments