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a(n) = n*(n + 1)*(5*n^2 + 5*n - 4)/12.
4

%I #20 Sep 08 2022 08:46:14

%S 0,1,13,56,160,365,721,1288,2136,3345,5005,7216,10088,13741,18305,

%T 23920,30736,38913,48621,60040,73360,88781,106513,126776,149800,

%U 175825,205101,237888,274456,315085,360065,409696,464288,524161,589645,661080,738816,823213,914641

%N a(n) = n*(n + 1)*(5*n^2 + 5*n - 4)/12.

%C Partial sums of centered 10-gonal (or decagonal) pyramidal numbers.

%C Subsequence of A204221. In fact, a(n) is of the form (k^2-1)/15 for k = 5*n*(n+1)/2-1. - _Bruno Berselli_, Nov 27 2015

%H OEIS Wiki, <a href="https://oeis.org/wiki/Figurate_numbers">Figurate numbers</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PyramidalNumber.html">Pyramidal Number</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).

%F G.f.: x*(1 + 8*x + x^2)/(1 - x)^5.

%F a(n) = Sum_{k = 0..n} A004466(k).

%F a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). - _Vincenzo Librandi_, Nov 27 2015

%t Table[n (n + 1) (5 n^2 + 5 n - 4)/12, {n, 0, 50}]

%t LinearRecurrence[{5,-10,10,-5,1},{0,1,13,56,160},40] (* _Harvey P. Dale_, Aug 14 2017 *)

%o (Magma) [n*(n+1)*(5*n^2+5*n-4)/12: n in [0..50]]; // _Vincenzo Librandi_, Nov 27 2015

%o (PARI) a(n)=n*(n+1)*(5*n^2+5*n-4)/12 \\ _Charles R Greathouse IV_, Jul 26 2016

%Y Cf. A004466 (first differences), A201106 (partial sums), A204221.

%Y Cf. similar sequences listed in A264854.

%K nonn,easy

%O 0,3

%A _Ilya Gutkovskiy_, Nov 26 2015