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%I #24 Sep 08 2022 08:46:20
%S 1,14,54,135,271,476,764,1149,1645,2266,3026,3939,5019,6280,7736,9401,
%T 11289,13414,15790,18431,21351,24564,28084,31925,36101,40626,45514,
%U 50779,56435,62496,68976,75889,83249,91070,99366,108151,117439,127244,137580,148461,159901
%N a(n) = Sum_{k = n..2*n+1} k^2.
%C Inverse binomial transform is 1, 13, 27, 14, 0, 0, 0, ... (0 continued).
%H Colin Barker, <a href="/A299646/b299646.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F O.g.f.: (1 + 10*x + 4*x^2 - x^3)/(1 - x)^4.
%F E.g.f.: (6 + 78*x + 81*x^2 + 14*x^3)*exp(x)/6.
%F a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
%F a(n) = (n + 2)*(14*n^2 + 11*n + 3)/6. Therefore:
%F a(6*k + r) = 504*k^3 + 18*(14*r + 13)*k^2 + (42*r^2 + 78*r + 25)*k + a(r), with 0 <= r <= 5. Example: for r=5, a(6*k + 5) = (6*k + 7)*(84*k^2 + 151*k + 68).
%p seq((n + 2)*(14*n^2 + 11*n + 3)/6, n=0..50); # _Peter Luschny_, Feb 21 2018
%t Table[(n + 2) (14 n^2 + 11 n + 3)/6, {n, 0, 50}]
%t (* Second program: *)
%t LinearRecurrence[{4, -6, 4, -1}, {1, 14, 54, 135}, 41] (* _Jean-François Alcover_, Feb 21 2018 *)
%o (Sage) [(n+2)*(14*n^2+11*n+3)/6 for n in (0..50)]
%o (Maxima) makelist((n+2)*(14*n^2+11*n+3)/6, n, 0, 50);
%o (Magma) [(n+2)*(14*n^2+11*n+3)/6: n in [0..50]];
%o (GAP) List([0..50], n -> (n+2)*(14*n^2+11*n+3)/6);
%o (PARI) a(n)=(n+2)*(14*n^2+11*n+3)/6 \\ _Charles R Greathouse IV_, Feb 21 2018
%o (PARI) Vec((1 + 10*x + 4*x^2 - x^3)/(1 - x)^4 + O(x^60)) \\ _Colin Barker_, Feb 22 2018
%Y Subsequence of A008854, A047388, A174070 (after 1).
%Y Cf. A050409: Sum_{k = n..2*n} k^2; A050410: Sum_{k = n..2*n-1} k^2.
%K nonn,easy
%O 0,2
%A _Bruno Berselli_, Feb 20 2018
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