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%I #38 Oct 16 2015 13:50:54
%S 0,0,0,1,0,3,1,5,1,9,2,13,4,19,5,25,7,32,9,40,11,49,13,59,16,69,18,81,
%T 21,93,25,107,28,121,32,136,36,152,40,169,44,187,49,205,53,225,58,245,
%U 64,267,69,289,75,312,81,336,87,361,93,387,100,413,106,441
%N a(n) = floor(A261327(n)/9).
%C Hexasections:
%C 0, 1, 4, 9, 16, 25, 36, ... = A000290(n)
%C 0, 5, 19, 40, 69, 107, 152, ... = c(n)
%C 0, 1, 5, 11, 18, 28, 40, ... = d(n+1)
%C 1, 9, 25, 49, 81, 121, 169, ... = A016754(n)
%C 0, 2, 7, 13, 21, 32, 44, ... = A240438(n+1)
%C 3, 13, 32, 59, 93, 136, 187, ... = e(n+1).
%C The six sequences have the signature (2, -1, 1, -2, 1), that is, the signature of a(n) without the 0's.
%C It appears that d(n+1) and A240438(n+1) are connected via the following scheme.
%C Let x(n) be the sequence that concatenates terms of d(n+1) in reverse order with terms of A240438(n+1), both without their index_0 term:
%C ..., 18, 11, 5, 1, 0, 0, 2, 7, 13, 21, 32, ...
%C And consider the first and second differences of this sequence:
%C ..., -7, -6, -4, -1, 0, 2, 5, 6, 8, 11, 12, ...
%C ..., 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, ...
%C In the first differences, we get A047234(n+1) and A047267(n+1). And in the second differences, we get A010882(n).
%C In the same way, c(n) and e(n+1) are connected via the first and second differences of, with both their index_0 term:
%C ..., 69, 40, 19, 5, 0, 3, 13, 32, 59, ...
%C that are respectively:
%C ..., -29, -21, -14, -5, 3, 10, 19, 27, 34, ...
%C ..., 8, 7, 9, 8, 7, 9, 8, 7, 9, ... .
%C Is it possible to find a direct definition for a(n)?
%H Colin Barker, <a href="/A262397/b262397.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_22">Index entries for linear recurrences with constant coefficients</a>, signature (0,2,0,-1,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,-2,0,1).
%F a(n) = (A261327(n) - A261327(n) mod 9)/9.
%F From _Colin Barker_, Sep 25 2015: (Start)
%F a(n) = floor((n^2+4)/36) for n even.
%F a(n) = floor((n^2+4)/9) for n odd.
%F G.f.: -x^3*(x^4 +x^3 +x^2 +x +1)*(x^12 -x^11 +x^10 -x^8 +2*x^6 -x^4 +x^2 -x +1) / ((x -1)^3*(x +1)^3*(x^2 -x +1)*(x^2 +x +1)*(x^6 -x^3 +1)*(x^6 +x^3 +1)).
%F (End)
%e a(0) = floor(1/9) = 0, a(1)= floor (5/9) = 0, a(2) = floor(2/9) = 0, a(3)= floor (13/9) = 1.
%t LinearRecurrence[{0, 3, 0, -3, 0, 1}, {1, 5, 2, 13, 5, 29}, 70]/9 // Floor (* _Jean-François Alcover_, Sep 26 2015, after _Vincenzo Librandi_ in A261327 *)
%o (PARI) a(n) = numerator((n^2+4)/4)\9; \\ _Michel Marcus_, Sep 22 2015
%o (PARI) concat([0,0,0], Vec(-x^3*(x^4 +x^3 +x^2 +x +1)*(x^12 -x^11 +x^10 -x^8 +2*x^6 -x^4 +x^2 -x +1) / ((x -1)^3*(x +1)^3*(x^2 -x +1)*(x^2 +x +1)*(x^6 -x^3 +1)*(x^6 +x^3 +1)) + O(x^100))) \\ _Colin Barker_, Sep 25 2015
%o (PARI) a(n)=if(n%2,n^2+4,(n/2)^2+1)\9 \\ _Charles R Greathouse IV_, Oct 16 2015
%Y Cf. A000290, A010882, A016754, A047234, A047267, A240438, A261327.
%K nonn,easy,less
%O 0,6
%A _Paul Curtz_, Sep 21 2015
%E New name suggested by _Michel Marcus_, Sep 22 2015
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