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a(n) = n*(n+1)*(n+2)*(n^2+2*n+17)/120.
4

%I #35 Feb 08 2022 22:13:46

%S 1,5,16,41,91,182,336,582,957,1507,2288,3367,4823,6748,9248,12444,

%T 16473,21489,27664,35189,44275,55154,68080,83330,101205,122031,146160,

%U 173971,205871,242296,283712,330616,383537,443037,509712,584193,667147,759278,861328,974078

%N a(n) = n*(n+1)*(n+2)*(n^2+2*n+17)/120.

%C Antidiagonal sums of the array of pyramidal numbers shown in Table 2 of Sardelis and Valahas paper (see A261720).

%C This is the case j = 3 of (n^2 + (j-1)*n + (j+1)^2 + 1)*binomial(n+j-1, j)/((j+1)*(j+2)), where j is the space dimension: a(n) = (n^2+2*n+17)*binomial(n+2,3)/20.

%C The sequence is the binomial transform of (1, 4, 7, 7, 4, 1, 0, 0, 0, ...). - _Gary W. Adamson_, Aug 26 2015

%H D. A. Sardelis and T. M. Valahas, <a href="http://arxiv.org/abs/0805.4070">On Multidimensional Pythagorean Numbers</a>, arXiv:0805.4070 [math.GM], 2008.

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (6,-15,20,-15,6,-1).

%F G.f.: x*(1 - x + x^2)/(1 - x)^6.

%t Table[n (n + 1) (n + 2) (n^2 + 2n + 17)/120, {n, 40}]

%t LinearRecurrence[{6,-15,20,-15,6,-1},{1,5,16,41,91,182},40] (* _Harvey P. Dale_, Mar 18 2018 *)

%o (Magma) [n*(n+1)*(n+2)*(n^2+2*n+17)/120: n in [1..40]]; // _Vincenzo Librandi_, Apr 18 2015

%Y Cf. A257200, A257201.

%Y For another version of the array, see A080851.

%K nonn,easy

%O 1,2

%A _Luciano Ancora_, Apr 18 2015