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A105939
a(n) = binomial(n+3,3)*binomial(n+6,3).
3
20, 140, 560, 1680, 4200, 9240, 18480, 34320, 60060, 100100, 160160, 247520, 371280, 542640, 775200, 1085280, 1492260, 2018940, 2691920, 3542000, 4604600, 5920200, 7534800, 9500400, 11875500, 14725620, 18123840, 22151360, 26898080, 32463200, 38955840, 46495680
OFFSET
0,1
COMMENTS
a(n) is the number of ordered pairs (A,B) of size 3 subsets of {1,2,...,n+6} such that A and B are disjoint. - Geoffrey Critzer, Sep 03 2013
LINKS
Ömür Deveci and Anthony G. Shannon, Some aspects of Neyman triangles and Delannoy arrays, Mathematica Montisnigri (2021) Vol. L, 36-43.
FORMULA
G.f.: 20/(1-x)^7. - Colin Barker, Jun 06 2012
E.g.f.: (d/dx)^6 (x^3/3!)^2 * exp(x). - Geoffrey Critzer, Sep 03 2013
a(n) = A000292(n+1)*A000292(n+4) = 20*A000579(n+6). - R. J. Mathar, Nov 30 2015
From Amiram Eldar, Jan 06 2021: (Start)
Sum_{n>=0} 1/a(n) = 3/50.
Sum_{n>=0} (-1)^n/a(n) = 48*log(2)/5 - 661/100. (End)
E.g.f.: (1/36)*(720 + 4320*x + 5400*x^2 + 2400*x^3 + 450*x^4 + 36*x^5 + x^6)*exp(x). - G. C. Greubel, Mar 11 2025
EXAMPLE
If n=0 then C(0+3,0)*C(0+6,3) = C(3,0)*C(6,3) = 1*20 = 20.
If n=8 then C(8+3,8)*C(8+6,3) = C(11,8)*C(14,3) = 165*364 = 60060.
MATHEMATICA
nn=25; f[x_]:=Exp[x](x^3/3!)^2; Range[0, nn]! CoefficientList[Series[ a=f''''''[x], {x, 0, nn}], x] (* Geoffrey Critzer, Sep 03 2013 *)
Table[Binomial[n+3, 3]Binomial[n+6, 3], {n, 0, 30}] (* or *) LinearRecurrence[ {7, -21, 35, -35, 21, -7, 1}, {20, 140, 560, 1680, 4200, 9240, 18480}, 30] (* Harvey P. Dale, Mar 09 2022 *)
20*Binomial[Range[0, 40] +6, 6] (* G. C. Greubel, Mar 11 2025 *)
PROG
(Magma)
A105939:= func< n | 20*Binomial(n+6, 6) >;
[A105939(n): n in [0..40]]; // G. C. Greubel, Mar 11 2025
(SageMath)
def A105939(n): return 20*binomial(n+6, 6)
print([A105939(n) for n in range(41)]) # G. C. Greubel, Mar 11 2025
CROSSREFS
Sequence in context: A236988 A358865 A134382 * A054389 A374161 A253003
KEYWORD
easy,nonn
AUTHOR
Zerinvary Lajos, Apr 27 2005
EXTENSIONS
More terms from Geoffrey Critzer, Sep 03 2013
STATUS
approved