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%I #20 Aug 15 2021 21:52:31
%S 0,4,32,124,320,716,1328,2340,3792,5852,8544,12260,16864,22916,30272,
%T 39188,49824,62948,78080,96348,117232,141260,168480,200292,235680,
%U 276100,321056,371484,427024,489900,558112,634724,718432,810116,909600,1018388,1135136,1263828,1402304,1551908
%N Number of triangles in an n X n unit grid that have minimal possible area (of 1/2).
%H Ray Chandler, <a href="/A088658/b088658.txt">Table of n, a(n) for n = 1..1000</a>
%H N. J. A. Sloane, <a href="/A115004/a115004.txt">Families of Essentially Identical Sequences</a>, Mar 24 2021 (Includes this sequence)
%F a(n+1) = 4*A115004(n).
%F a(n) = 4*(n-1)^2 + 4*Sum_{i=2..n-1} (n-i)*(2n-i)*phi(i). - _Chai Wah Wu_, Aug 15 2021
%e a(2)=4 because 4 (isosceles right) triangles with area 1/2 can be placed on a 2 X 2 grid.
%t z[n_] := Sum[(n - i + 1)(n - j + 1) Boole[GCD[i, j] == 1], {i, n}, {j, n}];
%t a[n_] := 4 z[n - 1];
%t Array[a, 40] (* _Jean-François Alcover_, Mar 24 2020 *)
%o (Python)
%o from sympy import totient
%o def A088658(n): return 4*(n-1)**2 + 4*sum(totient(i)*(n-i)*(2*n-i) for i in range(2,n)) # _Chai Wah Wu_, Aug 15 2021
%Y Cf. A045996.
%Y The following eight sequences are all essentially the same. The simplest is A115004(n), which we denote by z(n). Then A088658(n) = 4*z(n-1); A114043(n) = 2*z(n-1)+2*n^2-2*n+1; A114146(n) = 2*A114043(n); A115005(n) = z(n-1)+n*(n-1); A141255(n) = 2*z(n-1)+2*n*(n-1); A290131(n) = z(n-1)+(n-1)^2; A306302(n) = z(n)+n^2+2*n. - _N. J. A. Sloane_, Feb 04 2020
%K nonn
%O 1,2
%A Yuval Dekel (dekelyuval(AT)hotmail.com), Nov 21 2003
%E a(7)-a(28) from _Ray Chandler_, May 03 2011
%E Corrected and extended by _Ray Chandler_, May 18 2011
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