On the sum of digits of $2^x$

Question

I was looking into the sum of the digits of $2^x$. When I plotted y = sum of digits of $2^x$, the relationship was very linear (as one should expect) with a correlation coefficient of $0.999587$, when I calculated to $x = 10000$. However, my question concerns the slope of the line of best fit: $1.3556600856084817$. The slope seemed to fluctuate around $1.35$. Is there any reason for this?

Here are the slopes for f(n), representing the slope of y = the sum of digits for n^x: When plotted, it appears to be logarithmic.

• f(2) = 1.3556600856084817
• f(3) = 2.1458633077830696
• f(4) = 2.709191384919649
• f(5) = 3.147413019705652
• f(6) = 3.5012234023042232
• f(7) = 3.8034815099113444
• f(8) = 4.064013020137782
• (9) = 4.293926142384194
• f(11) = 4.68565515673792
• f(12) = 4.856282415979148
• f(13) = 5.0120684909979305
• f(14) = 5.156650757838338
• f(15) = 5.29277691693154
• f(35) = 6.946651885840713
• f(99) = 8.980022856020353
• f(223) = 10.56770121473439

Does anyone have any explanation for the meaning behind these slopes?

Here is some of my sloppy python code if you want to play around with it:

import matplotlib.pyplot as plt
import numpy as np
import pandas as pd

sum_npx = []

#sum of digits of n^x
def sum_of_digits(integer_list):
    total_digit_sum = 0
    for number in integer_list:
        num_str = str(number)
        for char in num_str:
            if char.isdigit():
                total_digit_sum += int(char)
    return total_digit_sum

for _ in range(1, 10000):
    sum_npx.append(sum_of_digits([int(i) for i in str(2**_)]))

x = [a for a in range(1, 10000)]
plt.plot(np.unique(x), np.poly1d(np.polyfit(x, sum_npx, 1))(np.unique(x)))
plt.plot(x, sum_npx)
plt.show()

df = {
    "Array 1": x,
    "Array 2": sum_npx
}

coefficients = np.polyfit(x, sum_npx, 1)

# Extract the slope (coefficient for x, which is the first element)
slope = coefficients[0]

# Print the slope
print("Slope of the line of best fit:", slope)

data = pd.DataFrame(df)
print(data.corr())

Thanks in advance.

Answer 1

$2^x$ has about $$ \log_{10}(2^x) = x\log_{10}(2) \approx 0.30103x $$ digits in base $10$. The average digit is about $4.5$, so the sum of the digits is about $$ 4.5 \log_{10}(2)x \approx 1.35 x. $$

This argument generalizes to show that the sum of the base $b$ digits of $n^x$ is about $$ \frac{b-1}{2}\log_{b}(n) x $$

Answer 2

It may be worth noting how well the wedge-like shape of this plot is explained by the assumption that the digits of $2^x$ are approximately uniformly distributed (and independent). The mean and variance of the sum of these digits in this case are easily found to be as follows: $$\mu(x)=4.5\times (\text{number of digits in $2^x$})$$ $$\sigma^2(x)=8.25\times(\text{number of digits in $2^x$})$$ Computations show (in the plot on the left) that the two lines $\mu(x)\pm 3\,\sigma(x)$ almost perfectly outline the plot, with $\mu(x)$ being the center line:

The plot on the right shows the approximate normality of the sums-of-digits by using their standardized values $$Z(x)={\text{sum-of-digits$(2^x)$}-\mu(x)\over\sigma(x)},$$ the blue curve being the standard normal density function.

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SageMath code for the plot on the left:

def sum_of_digits(x, n, b):
    return sum((n^x).digits(base=b))
def mu(x, n, b):
    return ((b - 1) / 2) * (1 + floor(log(n^x, b)))
def sd(x, n, b):
    return sqrt( ((b^2 - 1) / 12) * (1 + floor(log(n^x, b))))
b = 10
n = 2
xmax = 10^4
L = [(x, sum_of_digits(x,n,b)) for x in range(xmax)]
Lmu = [(x, mu(x,n,b)) for x in range(xmax)]
Llo = [(x, mu(x,n,b) - 3.0*sd(x,n,b)) for x in range(xmax)]
Lhi = [(x, mu(x,n,b) + 3.0*sd(x,n,b)) for x in range(xmax)]
gr = line(L, color='orange')
gr += line(Lmu, linestyle=':', color='black')
gr += line(Llo, linestyle=':', color='blue')
gr += line(Lhi, linestyle=':', color='blue')
gr.show(axes_labels=[r'Integer $x$', 'Sum of digits'],
    axes_labels_size=1, frame=True, axes=False, 
    title=r'Sum of base-${%s}$ digits in ${%s}^x$ versus $x$'%(str(b),str(n)), figsize=4)