Please follow the forum guidelines and post your code using code tags. Explain in better detail what you want. You will also have to get rid of the delays as they are blocking and nothing else happens until the delay has timed out.
look this over
i only had 4 LEDs to test with
each pattern function uses a timer to do the next thing
the LED pins are defined in an array and an idx used to access the pin
// LED patterns
const byte PinLeds [] = { 10, 11, 12, 13 };
const int Nleds = sizeof (PinLeds);
enum { Off = HIGH, On = LOW };
unsigned long msec;
// -----------------------------------------------------------------------------
// sequentially turn LEDs on, then all off
void
pattern10 (void)
{
static unsigned long msecLst;
static int idx;
const int IdxN = 2;
const int Idx0 = 0;
if (msec - msecLst >= 500) {
msecLst = msec;
if (IdxN < idx) {
idx = Idx0;
Serial.println ("all off");
for (int n = Idx0; n <= IdxN; n++)
digitalWrite (PinLeds [n], Off); // all off
}
else {
Serial.println (PinLeds [idx]);
digitalWrite (PinLeds [idx], On);
idx++;
}
}
}
// -----------------------------------------------------------------------------
// toggle pin
void
pattern13 (void)
{
static unsigned long msecLst;
if (msec - msecLst >= 300) {
msecLst = msec;
digitalWrite (PinLeds [3], ! digitalRead (PinLeds [3]));
}
}
// -----------------------------------------------------------------------------
void loop ()
{
msec = millis ();
pattern13 ();
pattern10 ();
}
// -----------------------------------------------------------------------------
void setup ()
{
Serial.begin (9600);
for (int n = 0; n < Nleds; n++) {
pinMode (PinLeds [n], OUTPUT);
digitalWrite (PinLeds [n], Off);
}
}