%I #48 Dec 05 2023 14:38:33
%S 0,2,3,6,12,15,20,30,35,42,56,63,72,90,99,110,132,143,156,182,195,210,
%T 240,255,272,306,323,342,380,399,420,462,483,506,552,575,600,650,675,
%U 702,756,783,812,870,899
%N A142705 (numerators of 1/4-1/(4n^2)) sorted to natural order.
%C A198442(n) without indices 4*n+2.
%C a(n)/A130823(n+1) = 0, 2,3,2, 4,5,4, 6,7,6, 8,9,8, ... (equal to A133310+1, after 0; see also A008611).
%C -1, 0, 2, 3, is divisible by 1 (for a(-1)=-1),
%C 3, 6, 12, 15, 3,
%C 15, 20, 30, 35 5,
%C 35, 42, 56, 63 7,
%C 63, 72, 90, 99 9,
%C 99, 110, 132, 143, 11, etc.
%C First column: A000466(n),
%C second column: A002943(n),
%C third column: A002939(n+1),
%C fourth column: A000466(n+1).
%C a(n) is also the numerator of 1/4-1/(4*n+2)^2: 0/1, 2/9, 3/16, 6/25, 12/49, 15/64, 20/81, 30/121, 35/144, 42/169, 56/225,...
%C The n-th denominator is equal to 4*a(n) + A146325(n+2).
%C Note that the differences of a(n-1): 1, 2, 1, 3, 6, 3, 5, 10, 5, 7, 14, 7, 9, 18, 9, 11, 22,... (from A043547 by pairs and 2*n+1) has the same recurrence.
%C (Of course every sequence which obeys a linear recurrence with constant coefficients has first differences that obey the same linear recurrence. - _R. J. Mathar_, Jun 14 2013)
%H Paolo Xausa, <a href="/A226023/b226023.txt">Table of n, a(n) for n = 0..10000</a>
%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,2,-2,0,-1,1).
%F a(n) = floor( (2*n + 1)/3 ) * floor( (2*n + 5)/3 ) = A004396(n) * A004396(n+2).
%F Recurrences: a(n) = 3*a(n-3) -3*a(n-6) +a(n-9) = a(n-1) +2*a(n-3) -2*a(n-4) -a(n-6) +a(n-7).
%F a(n+15) - a(n) = 10*A042968(n+8).
%F a(n+1) - a(n-2) = 2*A042968(n) with a(-2)=0, a(-1)=-1.
%F G.f.: x*(2+x+3*x^2+2*x^3+x^4-x^5)/((1-x)^3 * (1+x+x^2)^2). [_Ralf Stephan_, May 24 2013]
%p A226023 := proc(n)
%p option remember;
%p if n <=6 then
%p op(n+1,[0,2,3,6,12,15,20]) ;
%p else
%p procname(n-1)+2*procname(n-3)-2*procname(n-4)-procname(n-6)+procname(n-7) ;
%p end if;
%p end proc: # _R. J. Mathar_, Jun 28 2013
%t A226023[n_]:=Floor[(2n+1)/3]Floor[(2n+5)/3];
%t Array[A226023,100,0] (* _Paolo Xausa_, Dec 05 2023 *)
%Y Trisections: A002939, A000466, A002943.
%K nonn,easy
%O 0,2
%A _Paul Curtz_, May 23 2013