LED emits low light when using randomSeed.

Using Arduino Programming Questions
1

I'm a beginner in Arduino and I came across this while I was familiarizing with the codes and functions. As the title says, when I use randomSeed, LEDs emits low light. Per LED, I series 1kOhm limitting resistor. I tried to change it with a lower value (In some forums they said 100 Ohms is the recommended minimum value, tried it but still gets the same result). I tried also to change the LEDs to know if they are faulty, but it seems that they are not the problem. Then I tried another sketch I found on the internet utilizing the random() function. this is the code:

int ledRed = 2;
int ledGreen = 3;
int ledYellow = 4;
int ledBlack = 5;
 
void setup() {
  Serial.begin(9600);
  pinMode(ledRed, OUTPUT);
  pinMode(ledGreen, OUTPUT);
  pinMode(ledYellow, OUTPUT);  
  pinMode(ledBlack, OUTPUT);
}
 
void loop() {
  Serial.println(random(5));
  digitalWrite(ledRed, random(5));
  digitalWrite(ledGreen, random(5));
  digitalWrite(ledYellow, random(5));
  digitalWrite(ledBlack, random(5));
  
  delay(1000);
}

The LEDs then light up normaly, emitting its normal light intensity. What could be the problem? Thanks in advance for the help.

2 
random(5)

Returns a number from zero to four.

How about

random(256)

Which will return a number between zero and 255 inclusive.

Plus, you want

analogWrite(pin, random(256));

Where did you randomSeed?

What's a black led do?

3

I see no call to randomSeed. You are clearly not talking about the code you posted.

4

Here's the code. The ledBlack is just a variable. Random name only sorry.

long randNumber;

void setup()
{
  pinMode(randNumber,OUTPUT);
  Serial.begin(9600);
  randomSeed(analogRead(0));
}

void loop()
{
  int randNumber = random(7,12);
  digitalWrite(randNumber,HIGH);
  delay(500);
  digitalWrite(randNumber,LOW);
}
5

Pins are INPUT by default. An LED on an INPUT pin will not be very bright.

6

So I should first set my 8-11 pins as OUTPUT on setup() for them to light up bright?

7

PaulS:
Pins are INPUT by default. An LED on an INPUT pin will not be very bright.

That did it. Thanks for the help good sir.

8

you may want to try your original sketch, but like this:

int ledRed = 2;
int ledGreen = 3;
int ledYellow = 4;
int ledBlack = 5;
 
void setup() {
  Serial.begin(9600);
  pinMode(ledRed, OUTPUT);
  pinMode(ledGreen, OUTPUT);
  pinMode(ledYellow, OUTPUT);  
  pinMode(ledBlack, OUTPUT);
}
 
void loop() {
  Serial.println(random(0,2));
  digitalWrite(ledRed, random(0,2));
  digitalWrite(ledGreen, random(0,2));
  digitalWrite(ledYellow, random(0,2));
  digitalWrite(ledBlack, random(0,2));
  delay(1000);
}
9