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A258582 a(n) = n*(2*n + 1)*(4*n + 1)/3.

%I A258582 #42 Feb 16 2025 08:33:25
%S A258582 0,5,30,91,204,385,650,1015,1496,2109,2870,3795,4900,6201,7714,9455,
%T A258582 11440,13685,16206,19019,22140,25585,29370,33511,38024,42925,48230,
%U A258582 53955,60116,66729,73810,81375,89440,98021,107134,116795,127020,137825,149226,161239,173880
%N A258582 a(n) = n*(2*n + 1)*(4*n + 1)/3.
%C A258582 First bisection of the square pyramidal numbers (A000330).
%H A258582 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/SquarePyramidalNumber.html">Square Pyramidal Number</a>.
%H A258582 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F A258582 G.f.: x*(5 + 10*x + x^2)/(1 - x)^4.
%F A258582 a(n) = A000330(2*n).
%F A258582 Sum_{n>0} 1/a(n) = 3*(6 - Pi - 4*log(2)) = 0.25745587...
%F A258582 a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>3. - _Vincenzo Librandi_, Nov 18 2015
%F A258582 a(n) = A006918(4*n-1) = A053307(4*n-1) = A228706(4*n-1) for n>0. - _Bruno Berselli_, Nov 18 2015
%F A258582 a(n) = Sum_{k=1..2*n} k^2 (see the first comment).  E.g.f. exp(x)*(5*x+ 20*x^2/2+16*x^3/3!). - _Wolfdieter Lang_, Mar 13 2017
%F A258582 Sum_{n>=1} (-1)^(n+1)/a(n) = 3*log(2) + 6*sqrt(2)*log(1+sqrt(2)) + 3*(sqrt(2)-1/2)*Pi - 18. - _Amiram Eldar_, Sep 17 2022
%p A258582 A258582:=n->n*(2*n + 1)*(4*n + 1)/3: seq(A258582(n), n=0..50); # _Wesley Ivan Hurt_, Nov 17 2015
%t A258582 Table[(1/3) n (2 n + 1) (4 n + 1), {n, 0, 45}]
%o A258582 (PARI) vector(100, n, n--; n*(2*n+1)*(4*n+1)/3) \\ _Altug Alkan_, Nov 06 2015
%o A258582 (PARI) concat(0, Vec((5*x + 10*x^2 + x^3)/(1 - x)^4 + O(x^50))) \\ _Altug Alkan_, Nov 06 2015
%o A258582 (Magma) [n*(2*n+1)*(4*n+1)/3: n in [0..50]]; // _Wesley Ivan Hurt_, Nov 17 2015
%Y A258582 Cf. A000330, A001477, A005408, A016813, A053126 (partial sums), A100157.
%K A258582 nonn,easy
%O A258582 0,2
%A A258582 _Ilya Gutkovskiy_, Nov 06 2015