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a(n) = n*(n+1)*(n+2)*(n+3)*(2*n^2+6*n+7)/360.
4

%I #72 Jul 18 2026 03:44:21

%S 0,1,9,43,147,406,966,2058,4026,7359,12727,21021,33397,51324,76636,

%T 111588,158916,221901,304437,411103,547239,719026,933570,1198990,

%U 1524510,1920555,2398851,2972529,3656233,4466232,5420536,6539016,7843528,9358041,11108769,13124307

%N a(n) = n*(n+1)*(n+2)*(n+3)*(2*n^2+6*n+7)/360.

%C After 0, second bisection of A129548.

%C This sequence is also the total number of squares of all sizes in i X i subsquares in an n X n grid, whereas A000330 simply gives the number of all sizes of squares in an n X n grid. See illustrations.

%H Colin Barker, <a href="/A259181/b259181.txt">Table of n, a(n) for n = 0..1000</a>

%H Luce ETIENNE, <a href="/A259181/a259181.pdf">Illustration of initial terms</a>.

%H Feihu Liu, Guoce Xin, and Chen Zhang, <a href="https://arxiv.org/abs/2412.18744">Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS</a>, arXiv:2412.18744 [math.CO], 2024. See p. 14.

%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (7,-21,35,-35,21,-7,1).

%F a(n) = (1/6)*Sum_{i=0..n} (i+1)*(i+2)*(2*i+3)*(n-i)^2.

%F a(n) = Sum_{i=0..n} A000290(n-i)*A000330(i+1).

%F G.f.: x*(1 + x)^2 / (1 - x)^7. - _Colin Barker_, Nov 08 2015

%F a(n) = (A000539(n+1) - A000217(n+1))/30. - _Yasser Arath Chavez Reyes_, Feb 24 2024

%F Sum_{n>=1} 1/a(n) = 48*sqrt(5)*Pi*tanh(sqrt(5)*Pi/2)/7 - 2300/49. - _Amiram Eldar_, Jul 18 2026

%e a(0) = 0; a(1) = 1*1; a(2) = 4*1+1*5 = 9; a(3) = 9*1+4*5+1*14 = 43.

%t A259181[n_] := n*(n + 1)*(n + 2)*(n + 3)*(2*n*(n + 3) + 7)/360;

%t Array[A259181, 40, 0] (* _Paolo Xausa_, Jul 13 2026 *)

%o (PARI) vector(100, n, n--; n*(n+1)*(n+2)*(n+3)*(2*n^2+6*n+7)/360) \\ _Altug Alkan_, Nov 08 2015

%o (PARI) concat(0, Vec(-x*(x+1)^2 / (x-1)^7 + O(x^100))) \\ _Colin Barker_, Nov 08 2015

%Y Cf. A000217, A000290, A000330, A000539, A129548.

%Y Cf. A060060 ((1/6)*Sum_{i=0..n} (i+1)*(i+2)*(2*i+3)*i^2).

%K nonn,easy,changed

%O 0,3

%A _Luce ETIENNE_, Nov 08 2015