A374084 a(n) = (1 + (n+1)^2 + (n-2)*(-1)^n)/2.

3, 5, 8, 14, 17, 27, 30, 44, 47, 65, 68, 90, 93, 119, 122, 152, 155, 189, 192, 230, 233, 275, 278, 324, 327, 377, 380, 434, 437, 495, 498, 560, 563, 629, 632, 702, 705, 779, 782, 860, 863, 945, 948, 1034, 1037, 1127, 1130, 1224, 1227, 1325, 1328, 1430, 1433, 1539

Offset: 1

Author

Wesley Ivan Hurt, Jun 27 2024

Keywords

• nonn
• easy

Comments

Fill an array with the natural numbers n = 1,2,... along diagonals in alternating 'down' and 'up' directions. a(n) is column 2 of the boustrophedon-style array (see example).

In general, column k is given by (1+(t-1)^2+(n-k)*(-1)^t)/2, t = n+k. Here, k=2.

Examples

			       [ 1] [ 2] [ 3] [ 4] [ 5] [ 6] [ 7] [ 8] [ 9] [10] [11] [12]
  [ 1]   1    3    4   10   11   21   22   36   37   55   56   78   ...
  [ 2]   2    5    9   12   20   23   35   38   54   57   77   ...
  [ 3]   6    8   13   19   24   34   39   53   58   76   ...
  [ 4]   7   14   18   25   33   40   52   59   75   ...
  [ 5]  15   17   26   32   41   51   60   74   ...
  [ 6]  16   27   31   42   50   61   73   ...
  [ 7]  28   30   43   49   62   72   ...
  [ 8]  29   44   48   63   71   ...
  [ 9]  45   47   64   70   ...
  [10]  46   65   69   ...
  [11]  66   68   ...
  [12]  67   ...
        ...
		

Links

• Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Crossrefs

Cf. A128918 (Column 1) n>0, this sequence (Column 2).

Column 2 of the table in A056011.

Row 2 of the rectangular array in A056023.

Programs

Magma

[(1 + (n+1)^2 + (n-2)*(-1)^n)/2: n in [1..80]];

Mathematica

CoefficientList[Series[-(2*x^3 - 3*x^2 + 2*x + 3)/((x + 1)^2*(x - 1)^3), {x, 0, 50}], x]
k := 2; Table[(1 + (n + k - 1)^2 + (n - k) (-1)^(n + k))/2, {n, 60}]

Python

def A374084(n): return (n*(n+1)+4 if n&1 else n*(n+3))>>1 # Chai Wah Wu, Jul 07 2024

Formula

a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5).

G.f.: -x*(2*x^3-3*x^2+2*x+3)/((x+1)^2*(x-1)^3).

a(n) = A128918(n+1) - (-1)^n.

E.g.f.: (2 + x)*(exp(x)*x + 2*sinh(x))/2. - Stefano Spezia, Jun 29 2024