Question

Benford’s Law is an observation about the distribution of the frequencies of the first digits of the numbers in many different data sets. It is frequently found that the first digits are not uniformly distributed, but follow the logarithmic distribution P(d) = log10 ((d + 1)/d)

That is, numbers starting with 1 are more common than those starting with 2, and so on, with those starting with 9 the least common. The probabilities follow:

1 0.301
2 0.176
3 0.125
4 0.097
5 0.079
6 0.067
7 0.058
8 0.051
9 0.046

Benford’s Law is most accurate for data sets which span several orders of magnitude and can be proved to be exact for some infinite sequences of numbers.

1 Demonstrate in Python that the first digits of the first 500 Fibonacci numbers follow Benford’s Law quite closely.

Answer 1

Please find the code below. Thumbs up if you like the solution.

# Program to display the Fibonacci sequence up to n-th term where n is provided by the user
nterms = 500
# first two terms
n1 = 0
n2 = 1
count = 0
lst=[]

# check if the number of terms is valid
if nterms <= 0:
print("Please enter a positive integer")
elif nterms == 1:
print("Fibonacci sequence upto",nterms,":")
print(n1)
else:
print("Fibonacci sequence upto",nterms,":")
while count < nterms:
print(n1,end=' , \n')
#Creating a list with the starting of each fibonacci number.

numStr = str(n1)
numStr = int(numStr[0])
lst.append(numStr)
nth = n1 + n2
# update values
n1 = n2
n2 = nth
count += 1

# Initializing the counters.

counter1 = 0
counter2 = 0
counter3 = 0
counter4 = 0
counter5 = 0
counter6 = 0
counter7 = 0
counter8 = 0
counter9 = 0

#Counting the unique numbers in the list

for items in lst:
if items == 1:
counter1 +=1
if items == 2:
counter2 +=1
if items == 3:
counter3 +=1
if items == 4:
counter4 +=1
if items == 5:
counter5 +=1
if items == 6:
counter6 +=1
if items == 7:
counter7 +=1
if items == 8:
counter8 +=1
if items == 9:
counter9 +=1

#Dividing the probabilities by 1000 as dividing by 500 would give double the value as that of values in Benford's Law

print('Probability of 1 = ', counter1/1000)
print('Probability of 2 = ', counter2/1000)
print('Probability of 3 = ', counter3/1000)
print('Probability of 4 = ', counter4/1000)
print('Probability of 5 = ', counter5/1000)
print('Probability of 6 = ', counter6/1000)
print('Probability of 7 = ', counter7/1000)
print('Probability of 8 = ', counter8/1000)
print('Probability of 9 = ', counter9/1000)

Let me know if this works for you.