We are to compute the i-th term of the sequence A126071.
The sequence is defined as:
a(n) = number of bases b (with 2 <= b <= n+1) in which the representation of n is a palindrome.

However, note that the problem asks for the i-th number of the sequence. So we are to compute a(i) for the given index i.

But note: the problem says "Do your best to solve coding problems to get the ith number of the sequence."

We are to write a function A126071(i) that returns the i-th term.

Approach:
For a given integer n (which will be our index i), we want to count the number of bases b in the range [2, n+1]
for which the base-b representation of n is a palindrome.

Steps for a fixed n (which is the index, so we are computing a(n)):
For each base b in [2, n+1]:
Convert n to base b.
Check if the base-b representation is a palindrome.

However, note: the representation in base b for a number n is a list of digits. We must note that:

• The representation should not have leading zeros (unless the number is zero, but n>=1).
• A single digit is always a palindrome.

But note: the problem says "i-th number", meaning we are to compute for n = i.

However, the sequence definition:
a(n) = number of bases b (2<=b<=n+1) in which n is a palindrome.

Example:
n=1:
bases b from 2 to 2 (since n+1=2).
In base 2: 1 is written as "1", which is a palindrome -> count=1.
So a(1)=1.

n=2:
bases: 2 and 3.
base2: 2 in base2 is "10" -> not a palindrome.
base3: 2 in base3 is "2" -> palindrome.
So a(2)=1.

n=3:
bases: 2, 3, 4.
base2: 3 is "11" -> palindrome.
base3: 3 is "10" -> not.
base4: 3 is "3" -> palindrome.
So a(3)=2.

Implementation:

We need a helper function to convert n to base b and then check for palindrome.

However, note: when n is less than b, then the representation is a single digit -> always a palindrome.

Also, note: we can avoid building the entire string by using math.

Steps for converting n to base b:
digits = []
while n:
digits.append(n % b)
n //= b
then reverse digits to get the representation.

But note: we don't need the reversed list for palindrome check? We can check the list of digits without reversing?
Actually, the representation is from most significant to least. The first remainder we get is the least significant.
So the list we have is from least significant to most. We can check if the list is a palindrome without reversing?
Actually, we can reverse the list and then compare? Or we can check the list as it is?
Example: n=3, b=2:
3 % 2 = 1, then 3//2 = 1
1 % 2 = 1, then 1//2=0 -> digits = [1,1] (from bottom to top: so the representation is 11, which is the same as the reversed of [1,1]? Actually, we built it in reverse order: the first digit we get is the last digit of the representation).

Actually, we have the digits in reverse order. So the representation is the reversed list?
But for palindrome check: we want to compare the first half with the second half in the correct order.

Alternatively, we can build the representation in the correct order by inserting at the front? But that is inefficient.

Instead, we can check the list of digits (which is in reverse order) without reversing?
Actually, the representation in base b is the list of digits in the order we collected is the reverse.
So the actual representation is the reversed list of what we built?
But note: the palindrome condition is that the representation (without leading zeros) reads the same forwards and backwards.

Let D = the list of digits we built (least significant first). Then the actual representation is the reverse of D.

However, we can avoid building the entire list and then reversing by checking the list D and its reverse?
Alternatively, we can build the list and then check:

But note: there is a special case when n=0?
The problem says n is the index, and the sequence is defined for n>=1?
Also, when n=0, then the representation in base b is "0", which is a palindrome? But our loop would not run because n=0.
We are only considering n>=1.

Implementation:

We are to compute for a given index i (which is n in the sequence term a(n)).

Steps for A126071(i):
count = 0
for base in range(2, i+2): # because bases from 2 to i+1 (inclusive)
# Convert n (which is i) to base base
if i == 0:
# then in base base, representation is "0", which is a palindrome?
# but note: the problem says n>=1? and our index i is at least 1?
# so we can skip.
pass
else:
digits = []
num = i
while num:
digits.append(num % base)
num //= base

However, note: the base might be greater than n. In that case, the representation is a single digit.
For example, when base > n, then the representation is the digit n.
But note: our base goes from 2 to i+1. Since i>=1, then base can be as large as i+1, which is greater than i (if i>=1) only when i>=1?
Actually, when base > i, then the representation is a single digit?
But note: if base = i+1, then i < base, so the representation is a single digit (i) and that is a palindrome.

However, note: the representation for base > n is a single digit, but what about base == n?
We already considered: base n yields "10", which is not a palindrome.

However, note: the above code might be inefficient for large i?
The outer loop runs about i times (from 2 to i+1, which is i iterations).
For each base, we do a conversion which takes O(log_base(i)) and then a palindrome check O(log_base(i)).
Since the base is at least 2, the logarithm is at most about log2(i).
So the total time complexity is O(i * log(i))?
But note: when base is large (like base>sqrt(i)), then the number of digits is 2 (or 1) because:
i = a * base + b, with 0<=b<base, and a = i//base.
If base>i, then we break in one step.
If base>sqrt(i), then a = i//base < base, and a is at most base-1, so the representation has at most 2 digits.