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

%I #64 Sep 08 2022 08:46:02

%S 1,11,54,170,415,861,1596,2724,4365,6655,9746,13806,19019,25585,33720,

%T 43656,55641,69939,86830,106610,129591,156101,186484,221100,260325,

%U 304551,354186,409654,471395,539865,615536,698896,790449,890715,1000230,1119546,1249231

%N a(n) = n*(1 + n)*(3 - 4*n + 4*n^2)/6.

%C Antidiagonal sums of the convolution array A213838.

%C The sequence is the binomial transform of (1, 10, 33, 40, 16, 0, 0, 0, ...). - _Gary W. Adamson_, Jul 31 2015

%C From _Mircea Dan Rus_, Jul 11 2020: (Start)

%C a(n) is also the number of rectangles in a square biscuit of order n, which is obtained by stacking 2n-1 rows with their centers vertically aligned which consist successively of 1, 3, ..., 2n-3, 2n-1, 2n-3, ..., 3, 1 consecutive unit lattice squares. The order 2 and 3 square biscuits are shown below which contain 11 and 54 rectangles respectively.

%C __

%C __ __|__|__

%C __|__|__ __|__|__|__|__

%C |__|__|__| |__|__|__|__|__|

%C |__| |__|__|__|

%C |__|

%C (End)

%H Clark Kimberling, <a href="/A213840/b213840.txt">Table of n, a(n) for n = 1..200</a>

%H Teofil Bogdan and Mircea Rus, <a href="https://ssmr.ro/gazeta/gmb/2020/6/articol.pdf">Numărând dreptunghiuri pe foaia de matematică</a> (in Romanian). Gazeta Matematică, seria B, 2020 (6-7-8), pp. 281-288.

%H Teofil Bogdan and Mircea Dan Rus, <a href="https://arxiv.org/abs/2007.13472">Counting the lattice rectangles inside Aztec diamonds and square biscuits</a>, arXiv:2007.13472 [math.CO], 2020.

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

%F a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).

%F G.f.: x*(1 + 3*x)^2/(1 - x)^5.

%F From _Mircea Dan Rus_, Aug 26 2020: (Start)

%F a(n) = A000332(n+3) + 6*A000332(n+2) + 9*A000332(n+1).

%F a(n) = A002417(n) + 3*A002417(n-1). (End)

%p A213840:=n->n*(1 + n)*(3 - 4*n + 4*n^2)/6: seq(A213840(n), n=1..50); # _Wesley Ivan Hurt_, Sep 16 2017

%t Table[n (1 + n) (3 - 4 n + 4 n^2)/6, {n, 50}] (* or *) LinearRecurrence[{5, -10, 10, -5, 1}, {1, 11, 54, 170, 415}, 40] (* _Vincenzo Librandi_, Aug 01 2015 *)

%o (Magma) [n*(1+n)*(3-4*n+4*n^2)/6: n in [1..60]]; // _Vincenzo Librandi_, Aug 01 2015

%Y Cf. A000332, A002417, A213838.

%Y First differences of A271870. - _J. M. Bergot_, Aug 29 2016

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, Jul 05 2012

%E Edited (with simpler definition) by _N. J. A. Sloane_, Sep 19 2017