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a(n) = (n^3 + 9*n + 14*n + 9)/3.
4

%I #22 Sep 08 2022 08:46:23

%S 3,11,21,35,55,83,121,171,235,315,413,531,671,835,1025,1243,1491,1771,

%T 2085,2435,2823,3251,3721,4235,4795,5403,6061,6771,7535,8355,9233,

%U 10171,11171,12235,13365,14563,15831,17171,18585,20075,21643,23291,25021,26835

%N a(n) = (n^3 + 9*n + 14*n + 9)/3.

%C For n >= 6, a(n) is the number of evaluating points on the hypersphere in R^n in Stoyanovas's degree 7 cubature rule.

%H Vincenzo Librandi, <a href="/A322595/b322595.txt">Table of n, a(n) for n = 0..1000</a>

%H Ronald Cools, <a href="http://nines.cs.kuleuven.be/ecf/">Encyclopaedia of Cubature Formulas</a>

%H Ronald Cools, <a href="https://doi.org/10.1016/S0377-0427(99)00229-0">Monomial cubature rules since "Stroud": a compilation - part 2</a>, Journal of Computational and Applied Mathematics - Numerical evaluation of integrals Vol. 112 (1999), 21-27.

%H Srebra B. Stoyanova, <a href="https://doi.org/10.1016/S0377-0427(97)00094-0">Cubature of the seventh degree of accuracy for the hypersphere</a>, Journal of Computational and Applied Mathematics Vol. 84 (1997), 15-21.

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).

%F a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), n >= 4.

%F a(n) = 2*binomial(n + 1, 3) + 6*binomial(n + 1, 2) + 2*binomial(n + 1, 1) + 1.

%F G.f.: (3 - x - 5*x^2 + 5*x^3)/(1 - x)^4. [Corrected by _Georg Fischer_, May 23 2019]

%F E.g.f.: (1/3)*(9 + 24*x + 12*x^2 + x^3)*exp(x).

%t Table[(n^3 + 9*n + 14*n + 9)/3, {n, 0, 50}]

%t LinearRecurrence[{4,-6,4,-1},{3,11,21,35},50] (* _Harvey P. Dale_, Aug 19 2020 *)

%o (Maxima) makelist((n^3 + 9*n + 14*n + 9)/3, n, 0, 50);

%o (Magma) [(n^3 + 9*n + 14*n + 9)/3: n in [0..45]]; // _Vincenzo Librandi_, Jun 05 2019

%Y First differences: A027693.

%Y Cf. A000292, A161680, A174794, A321124, A322594.

%K nonn,easy

%O 0,1

%A _Franck Maminirina Ramaharo_, Dec 19 2018