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A262397
a(n) = floor(A261327(n)/9).
6
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, 21, 93, 25, 107, 28, 121, 32, 136, 36, 152, 40, 169, 44, 187, 49, 205, 53, 225, 58, 245, 64, 267, 69, 289, 75, 312, 81, 336, 87, 361, 93, 387, 100, 413, 106, 441
OFFSET
0,6
COMMENTS
Hexasections:
0, 1, 4, 9, 16, 25, 36, ... = A000290(n)
0, 5, 19, 40, 69, 107, 152, ... = c(n)
0, 1, 5, 11, 18, 28, 40, ... = d(n+1)
1, 9, 25, 49, 81, 121, 169, ... = A016754(n)
0, 2, 7, 13, 21, 32, 44, ... = A240438(n+1)
3, 13, 32, 59, 93, 136, 187, ... = e(n+1).
The six sequences have the signature (2, -1, 1, -2, 1), that is, the signature of a(n) without the 0's.
It appears that d(n+1) and A240438(n+1) are connected via the following scheme.
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:
..., 18, 11, 5, 1, 0, 0, 2, 7, 13, 21, 32, ...
And consider the first and second differences of this sequence:
..., -7, -6, -4, -1, 0, 2, 5, 6, 8, 11, 12, ...
..., 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, ...
In the first differences, we get A047234(n+1) and A047267(n+1). And in the second differences, we get A010882(n).
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:
..., 69, 40, 19, 5, 0, 3, 13, 32, 59, ...
that are respectively:
..., -29, -21, -14, -5, 3, 10, 19, 27, 34, ...
..., 8, 7, 9, 8, 7, 9, 8, 7, 9, ... .
Is it possible to find a direct definition for a(n)?
LINKS
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,-2,0,1).
FORMULA
a(n) = (A261327(n) - A261327(n) mod 9)/9.
From Colin Barker, Sep 25 2015: (Start)
a(n) = floor((n^2+4)/36) for n even.
a(n) = floor((n^2+4)/9) for n odd.
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)).
(End)
EXAMPLE
a(0) = floor(1/9) = 0, a(1)= floor (5/9) = 0, a(2) = floor(2/9) = 0, a(3)= floor (13/9) = 1.
MATHEMATICA
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 *)
PROG
(PARI) a(n) = numerator((n^2+4)/4)\9; \\ Michel Marcus, Sep 22 2015
(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
(PARI) a(n)=if(n%2, n^2+4, (n/2)^2+1)\9 \\ Charles R Greathouse IV, Oct 16 2015
KEYWORD
nonn,easy,less
AUTHOR
Paul Curtz, Sep 21 2015
EXTENSIONS
New name suggested by Michel Marcus, Sep 22 2015
STATUS
approved