Alternating Odd Numbers

Question

Is there a method or formula to get check if an odd number is an alternate odd number.

Explainer

1  
3 alt  
5  
7 alt  
9  
11 alt  
13  
15 alt...

I have tried Googling for an answer but only able to find modulo for odd/even or arrays with the values held to check against.

int half = (gridsize) / 2;
int row = 0;

for (int x = 0; x < gridsize; x++)
{
    int tilesThisRow = gridsize - (half - row);
    int drop = gridsize - tilesThisRow;

    int frontDrop;
    int backDrop;

    if(drop % 2 != 0)
    {
        //3 front
        //5 back
        //7 front
        //9 back
        //11 fron
        //13 back
        //15 front
        //17 back

        // it's this i need to alternate depending on the size
        frontDrop = (drop / 2) + 1;
        backDrop = (drop / 2);
    }
    
    for (int y = 0; y < gridsize; y++)
    {
        if (y >= frontDrop && y < gridsize - backDrop)
        {
            AddHexTile(x, y, gridOffsets);
        }
    }

    if (x < half)
    {
        row++;
    }
    else
    {
        row--;
    }
}

It's for a method i'm using to generate a hexagon hex tile grid which breaks depending on the size being used.

Pictures for clarification.

Size 7

Size 9

Answer 1

You could e.g. check for

value % 4 == 3

while using value % 4 == 1 for the others.

see fiddle

0
1 normal odd
2
3 alt odd
4
5 normal odd
6
7 alt odd
8
9 normal odd
10
11 alt odd
12
13 normal odd
14
15 alt odd
16
17 normal odd
18
19 alt odd
20
21 normal odd
22
23 alt odd
...