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A323221 a(n) = n*(n + 5)*(n + 7)/6 + 1.

%I A323221 #22 Jan 29 2019 04:33:54
%S A323221 1,9,22,41,67,101,144,197,261,337,426,529,647,781,932,1101,1289,1497,
%T A323221 1726,1977,2251,2549,2872,3221,3597,4001,4434,4897,5391,5917,6476,
%U A323221 7069,7697,8361,9062,9801,10579,11397,12256,13157,14101,15089,16122,17201,18327,19501
%N A323221 a(n) = n*(n + 5)*(n + 7)/6 + 1.
%C A323221 a(n) is related to the total angular defect of certain polytopes. See Hilton and Pedersen, Cor. 1; compare A275874.
%H A323221 P. Hilton and J. Pedersen, <a href="https://www.e-periodica.ch/digbib/view?pid=ens-001:1981:27::136#485">Descartes, Euler, Poincaré, Pólya and Polyhedra</a>, L'Enseign. Math., 27 (1981), 327-343.
%F A323221 Let I(n) denote the set of all tuples of length n with elements from {0, 1, 2, 3} with sum <= 3 and C(m) denote the m-th Catalan number. Then for n > 0
%F A323221 a(n) = Sum_{(j1,...,jn) in I(n)} C(j1)*C(j2)*...*C(jn).
%F A323221 a(n) = [x^n] (3*x^3 - 8*x^2 + 5*x + 1)/(x - 1)^4.
%F A323221 a(n) = n! [x^n] exp(x)*(x^3 + 15*x^2 + 48*x + 6)/6.
%F A323221 a(n) = a(n - 1)*(n*(n + 5)*(n + 7) + 6)/(n*(n + 2)*(n + 7) - 18) for n > 0.
%F A323221 a(n) = A323224(n, 4).
%F A323221 a(n) = A275874(n+4) + 1.
%e A323221 For n = 2 the sum formula gives:
%e A323221 I(2) = {{0,0}, {0,1}, {1,0}, {0,2}, {1,1}, {2,0}, {0,3}, {1,2}, {2,1}, {3,0}};
%e A323221 a(2) = 1 + 1 + 1 + 2 + 1 + 2 + 5 + 2 + 2 + 5 = 22.
%p A323221 a := n -> n*(35 + 12*n + n^2)/6 + 1:
%p A323221 seq(a(n), n = 0..45);
%t A323221 a[n_] := n (35 + 12 n + n^2)/6 + 1;
%t A323221 Table[a[n], {n, 0, 45}]
%Y A323221 Çf. A323224 (column 4), A323233 (row 4), A034856 (first difference), A275874.
%K A323221 nonn,easy
%O A323221 0,2
%A A323221 _Peter Luschny_, Jan 25 2019