OFFSET
0,2
COMMENTS
The hexasections of A262397(n) are
0, 1, 4, 9, 16, 25, 36, ... = A000290(n)
0, 5, 19, 40, 69, 107, 152, ... = a(n)
0, 1, 5, 11, 18, 28, 40, ... = A240438(n+1)
1, 9, 25, 49, 81, 121, 169, ... = A016754(n)
0, 2, 7, 13, 21, 32, 44, ... = A262523(n)
3, 13, 32, 59, 93, 136, 187, ... = e(n+1).
The five-step recurrence in FORMULA is valuable for the six sequences.
Consider a(n) extended from right to left with their first two differences:
..., 59, 32, 13, 3, 0, 5, 19, 40, 69, ...
..., -27, -19, -10, -3, 5, 14, 21, 29, 38, ...
..., 8, 9, 7, 8, 9, 7, 8, 9, 7, ... .
From 0,the first row is
1) from right to left: e(n)
2) from left to right: a(n).
a(n) and e(n) are companions.
The third row is of period 3.
The last digit of a(n) is of period 15; the same is true of e(n).
LINKS
Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,-1,1,-2,1).
FORMULA
a(n) = 2*a(n-1) - a(n-2) + a(n-3) - 2*a(n-4) - a(n-5), n> 4.
a(-n) = e(n).
a(-n) + a(n) = 8*n^2.
a(n+2) - 2*a(n+1) + a(n) = period 3:repeat 9, 7, 8.
a(n+3) - a(n-3) = 8*(1 + 6*n).
a(n+7) - a(n-7) = 40*(2 + 3*n).
a(2n+1) = -a(2n) + 6*n + 3.
a(2n+2) = -a(2n+1) + 4*(n+1).
a(3n) = 4*n*(9*n+1) = 8*A022267(n), a(3n+1) = 36*n^2 +28*n +5, a(3n+2) = 36*n^2 +52*n +19.
G.f.: -x*(x+1)*(3*x^2+4*x+5) / ((x-1)^3*(x^2+x+1)). - Colin Barker, Oct 08 2015
MATHEMATICA
a[0] = 0; a[1] = 5; a[2] = 19; a[n_] := a[n] = a[n - 3] + 24 (n - 3) + 40; Table[a@ n, {n, 0, 46}] (* Michael De Vlieger, Oct 09 2015 *)
PROG
(PARI) vector(100, n, n--; 4*n^2 + (4*(n+1)-3)\3) \\ Altug Alkan, Oct 07 2015
(PARI) concat(0, Vec(-x*(x+1)*(3*x^2+4*x+5)/((x-1)^3*(x^2+x+1)) + O(x^100))) \\ Colin Barker, Oct 08 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Paul Curtz, Oct 07 2015
STATUS
approved