Clock Sequence

Consider the infinite repeating sequence of digits:
1234321234321234321…
Amazingly, you can break this sequence of digits into a sequence of integers such that the sum of the digits in the $n$-th value is $n$.
The sequence goes as follows:
1, 2, 3, 4, 32, 123, 43, 2123, 432, 1234, 32123, …
Let $v_n$ be the $n$-th value in this sequence. For example, $v_2=2$, $v_5=32$ and $v_{11}=32123$.
Let $S(n)$ be $v_1+v_2+\cdots+v_n$. For example, $S(11)=36120$, and $S(1000)\bmod 123454321=18232686$.
Find $S(10^{14})\bmod 123454321$.

To solve this problem, we will first create a function to generate the sequence of integers, and then calculate the sum $S(n) \mod 123454321$ up to $n = 10^{14}$.

First, let’s create a function to generate the sequence of integers as follows:

“`python
def generate_sequence(n):
sequence = []
current_number = ”
for i in range(n):
current_number += str(i % 4 + 1)
sequence.append(int(current_number))
return sequence
“`

This function takes an input `n` and generates the sequence up to that `n` by repeatedly appending the next digit from the repeating sequence.

Next, let’s calculate the sum `S(n) mod 123454321` using this generated sequence:

“`python
def calculate_sum(modulus):
sequence = generate_sequence(modulus)
total_sum = sum(sequence)
return total_sum % 123454321
“`

This function takes an input `modulus` and uses the `generate_sequence` function to generate the sequence up to `modulus`. It then calculates the sum of the sequence and returns this sum modulo `123454321`.

Finally, let’s calculate `S(10^14) mod 123454321`:

“`python
result = calculate_sum(10**14)
print(result)
“`

This will print the value of `S(10^14) mod 123454321`.

Note: Computing `S(10^14)` directly may take a long time and consume a lot of memory. One way to optimize this is to notice that the sequence repeats every 3^14 = 4782969 digits. Thus, we can calculate the sum of one repeating block (4782969 digits) and multiply it by the number of blocks in 10^14, and then add the remaining digits.

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