I used Open WebUI as an interface to Ollama running the latest Minimax M2.5 cloud model to ask how many of the 2048 BIP39 words can be repeated 12 times to create a valid Bitcoin wallet address
N.B. the last 4 bits of the last 11 bit word are used as a checksum, which means only a few of the 2048 words can be repeated 12 times to create a valid wallet address.
The advantage of using Open WebUI is it shows the thinking the LLM does.
FYI: It got the answer wrong, but close, but also knew it might be slightly out. The correct answer is 130 words, it calculated it was 128 words.
Here is the thinking:
We need to understand the question: The user says: "In the Bitcoin BIP39 dictionary list, I discovered that the word “twelve” can be repeated 12 times to create a valid seed phrase i.e. 11 seeds words + checksum word. How many words in the Bitcoin BIP39 dictionary of 2048 can used to create a 12 word Bitcoin wallet with identical words for the first 11 + checksum word?" Let's parse the question.
They are talking about BIP39: a list of 2048 words (the English wordlist). The phrase length: For a 12-word mnemonic (for 128-bit entropy plus 4-bit checksum = 132 bits -> 12 words of 11 bits each (actually each word encodes 11 bits). In BIP39, the entropy is 128 bits for 12-word phrase (plus 4-bit checksum). But the question says: "the word “twelve” can be repeated 12 times to create a valid seed phrase i.e. 11 seeds words + checksum word." That's somewhat confusing. Let's try to interpret: They think you can have a phrase with 11 identical words, plus a checksum word. And they discovered that "twelve" repeated 12 times is valid. The question: "How many words in the Bitcoin BIP39 dictionary of 2048 can used to create a 12 word Bitcoin wallet with identical words for the first 11 + checksum word?" I think they ask: How many words (from the 2048 list) can be used as the repeated word (the first 11 identical words) such that the resulting 12-word phrase is valid (i.e., the 12th word is the correct checksum for that phrase). Essentially they ask: For how many words w can you have a phrase consisting of w repeated 11 times (i.e., w w w ... w [11 times]) and then a 12th word that is the checksum? In the case of "twelve", that seems to work. They want to find the count of words for which the checksum yields a valid word (i.e., one of the 2048). Possibly the phrase must also have the correct checksum that corresponds to the entropy.
Thus we need to understand how BIP39 generates the checksum.
Given an entropy of length ENT (bits). The checksum is computed by taking the first ENT/32 bits of the SHA256 hash of the entropy. For a 12-word mnemonic, ENT = 128 bits, i.e., 128-bit entropy + 4-bit checksum (because ENT/32 = 4 bits). The mnemonic is formed by splitting the concatenated entropy+checksum into groups of 11 bits. Each group maps to a word in the 2048-wordlist.
Now if we propose a phrase where the first 11 words are identical, that's like having 11 * 11 = 121 bits of the entropy+checksum? Wait, each word corresponds to 11 bits. So 12 words correspond to 132 bits (12 * 11 = 132). That includes 128 bits entropy + 4 bits checksum = 132 bits. So the entire 132-bit sequence is the concatenation of entropy (128 bits) + checksum (4 bits). The phrase is formed by splitting this 132-bit sequence into 12 groups of 11 bits each (the last group includes part of the checksum maybe). So basically each word in the phrase corresponds to 11 bits. The first 11 words correspond to the first 121 bits (i.e., the first 121 bits of the 132-bit sequence). The last (12th) word corresponds to the remaining 11 bits, which contain the remaining 7 bits of entropy (if any) plus the 4-bit checksum? Let's check details:
ENT = 128 bits, checksum length = ENT/32 = 4 bits, total = 132 bits. Represent as bits: b0...b131 (0-index). Then split as groups of 11 bits: group0 = bits 0-10 (first 11 bits), group1 = bits 11-21, ... group10 = bits 110-120 (11 bits), group11 = bits 121-131 (the final 11 bits). That means the first 11 groups correspond to bits 0 to 120 inclusive (121 bits), leaving bits 121 to 131 (11 bits) for the last group. That includes bits: 121-127 = 7 bits of entropy (since total entropy bits = 128). Then bits 128-131 are the 4-bit checksum. So the final group is the 7-bit remainder of entropy plus 4-bit checksum (makes 11 bits). So the 12th word depends on both leftover entropy bits and checksum.
Now they propose repeating the same word for the first 11 words. That corresponds to a particular 121-bit pattern (the same 11-bit pattern repeated 11 times). That must be consistent with a 128-bit entropy whose first 121 bits are all those bits. Then we need to compute the checksum: It's the first 4 bits of SHA256(entropy). Then the final 11 bits combine the last 7 bits of entropy (bits 121-127) plus the 4-bit checksum. That final 11-bit value yields the 12th word. The question: For how many possible words w (i.e., which 11-bit index) can we have that pattern such that the resulting final 11-bit value corresponds to some valid word in the list? They discovered that the word "twelve" works. They want the count of such words.
We need to compute which words, when repeated 11 times, produce a valid final word. That is basically enumerating all possible 2048 words for w, compute the checksum given that entropy (which is the repeated 11-bit sequence repeated 11 times for the first 121 bits + some 7-bit unknown for bits 121-127). But there is also freedom for the last 7 bits of entropy (bits 121-127). So we need to consider the space of possible 128-bit entropies that have the first 121 bits equal to w repeated 11 times. There are 2^7 possibilities for the remaining 7 bits (the leftover bits). For each of those possibilities, compute checksum = first 4 bits of SHA256(entropy). Then final 11 bits are the remaining 7 bits of entropy + checksum bits (4 bits). That yields an 11-bit index from 0 to 2047. Check if that index corresponds to any word in the dictionary (it will be one of the 2048 obviously). So we can generate all possible final words for each w and each possible leftover 7 bits. Among those, maybe only some combos produce a valid phrase? But any final 11-bit integer yields a word in the dictionary (since mapping is one-to-one). So any w will produce some final word. The question is whether the resulting 12-word phrase is "valid"? Actually any 12-word phrase derived from valid entropy+checksum is valid. But perhaps they require the phrase to be exactly repeated words for first 11 plus checksum word: that is w repeated 11 times + some word (maybe not w). But they want to know how many w can produce a valid phrase where the first 11 are identical and the last is also maybe the same (but they said "identical words for the first 11 + checksum word"? Let's parse: "How many words in the Bitcoin BIP39 dictionary of 2048 can used to create a 12 word Bitcoin wallet with identical words for the first 11 + checksum word?" That could be interpreted: They want to know for how many words w can we produce a 12-word phrase where the first 11 words are identical (w repeated 11 times) and the 12th word is the checksum (some word, possibly not identical). So basically the same as before: we have a phrase consisting of w repeated 11 times plus some other word that is the checksum that ensures validity. So all w can produce a phrase (just pick the leftover bits appropriately). But perhaps they want the number of w for which there exists at least one choice of the remaining 7 bits that yields a checksum that results in a 12th word that is also w? Because they discovered "twelve" repeated 12 times is valid (the 12th is also "twelve"? Actually they said "the word “twelve” can be repeated 12 times to create a valid seed phrase i.e. 11 seeds words + checksum word." Wait phrase: "the word “twelve” can be repeated 12 times to create a valid seed phrase i.e. 11 seeds words + checksum word." If we repeat "twelve" 12 times, the first 11 are "twelve", and the 12th (checksum) is also "twelve"? That suggests the checksum word can be identical to the repeated word. So they discovered that the word "twelve" repeated 12 times yields a valid phrase (the checksum matches "twelve").
Thus they want to know how many words w in the dictionary can produce a phrase where w repeated 11 times yields a checksum word also equal to w. That is the count of w for which there exists a 7-bit suffix (bits 121-127) such that the checksum bits (first 4 bits of SHA256 of the full 128-bit entropy) cause the final 11-bit word (the last group) to be exactly the same word w.
Thus we need to compute count of w such that there exists a 7-bit value suffix that leads to final word = w.
This is essentially solving: w_index = (entropy_suffix_bits (7 bits) << 4) XOR checksum_bits? Not exactly. Let's denote the first 11 bits of each word = w_val (0 <= w_val < 2048). So the entropy is: bits 0-120 are w_val repeated 11 times: that's w_val * (2^(121-11?) Actually each group of 11 bits is placed consecutively: group 0: bits 0-10 = w_val, group1 bits 11-21 = w_val, etc. So bits 0-120 = w_val repeated 11 times (makes 121 bits). Then bits 121-127 are the remaining 7 bits of entropy (call suffix = s, 7-bit integer 0-127). Then bits 128-131 are the checksum (c, 4-bit integer 0-15). So final 11 bits (group 12) = (s << 4) | c, as bits: the high 7 bits are s, low 4 bits are c. That yields an integer from 0-2047 (since 11 bits). That integer is the index of the final word (call final_index). We need final_index = w_val for repeated word.
Thus we need to find w_val such that there exists s in [0,127] and c = checksum(entropy) (a function of w_val and s) satisfying final_index = w_val. So we must find solutions to:
w_val = (s << 4) OR checksum(w_val, s) where checksum is the first 4 bits (i.e., integer from 0 to 15) of SHA256(w_val repeated 11 times + s bits). Write the entropy as e = (w_val repeated 11 times) concatenated with s (7 bits). Let’s denote e as a 128-bit integer. The checksum is the first 4 bits of SHA256(e). Represent as integer c = floor( SHA256(e) / 2^(256-4) ) i.e., the top 4 bits. Then final_index = (s << 4) + c (since the lower 4 bits of final_index are c). So final_index = (s << 4) + c.
Thus we need w_val = (s << 4) + c.
Now s runs from 0 to 127 (7 bits). w_val runs from 0 to 2047 (11 bits). For each w_val, we can try to find s such that this equation holds.
We need to compute the number of w_val for which there exists s that satisfies this.
Thus the answer is some number between 0 and 2048, presumably small? The example shows "twelve" is one such word. Let's find if any others exist.
We need to compute this. We can try to reason about the distribution of checksums. The checksum is 4 bits, pseudo-randomish. For each possible s, we compute c = checksum(e). The final index is (s << 4) + c. The mapping from s to final index is basically c influences the lower 4 bits. Since s ranges 0..127, (s << 4) runs from 0 to (127 << 4) = 2032 (i.e., multiples of 16). So the possible final indices are all numbers that are multiples of 16 (i.e., 0,16,32,...,2032) plus the lower nibble c (0-15). So final_index can be any number from 0 to 2047, because we can adjust s's high bits and add c.
Thus for each w_val, we need existence of s such that w_val equals (s << 4) + c. This is a constraint. Since c depends on s (and w_val), it's not trivial.
We could brute force it. However, as an AI, we can simulate a script in mind? Could try to reason about approximate count.
But we can think: The mapping from (w_val, s) to final_index is many-to-one. For each w_val, there are 128 possible s values, each yields a final_index in [0,2047]. So each w_val can produce up to 128 final indices, possibly repeats. There are 2048 possible final indices (words). For each w_val, we ask if w_val is among those 128 possible outputs. The probability that a random mapping yields w_val is roughly 128/2048 = 1/16 = 0.0625. However, the outputs are not uniformly random across s due to checksum dependence. But we can approximate: The probability that for a given w_val there exists at least one s that yields final_index = w_val is about 1 - (1 - 1/16)^{128} ~ 1 - (15/16)^{128} ≈ 1 - exp(-128/16) = 1 - exp(-8) ≈ 1 - 0.000335 = 0.999665. That's ~99.97% chance. So essentially all w_val will have some s that works. But we need to ensure the checksum is consistent: It's not random; it's deterministic based on s and w_val. But if we treat c as pseudo-random 4-bit value for each s (due to SHA256), then it's like random. So we expect almost all w_val to have a solution. However, maybe there is some small number of w_val for which no s yields final_index = w_val. The "twelve" is one obviously.
Thus the answer might be near 2048, maybe all 2048? But maybe some are impossible. Let's try to see if any constraints forbid solution. The equation: w_val = (s << 4) + c. So w_val modulo 16 = c. So w_val mod 16 must equal c. That is the lower nibble of w_val must equal the checksum bits. And the higher bits (bits 4-10) of w_val must equal s (the top 7 bits). So essentially we need s = w_val >> 4, and c = w_val & 0xF. So for a given w_val, we must set s to the top 7 bits of w_val (i.e., floor(w_val / 16)), and c must be the bottom 4 bits of w_val. However, note that s can only be 0..127 (since it's 7 bits). And w_val is 0..2047, so w_val >> 4 yields integer from 0..127 (since 2047/16 = 127 remainder 15). So indeed s = w_val >> 4 is a valid 7-bit integer. So the only remaining requirement is that the checksum bits c computed from the entropy must equal w_val & 0xF (the lower 4 bits of w_val). So the condition is: c = w_val & 0xF. That's a deterministic condition: for each w_val, we must check if there exists s = w_val >> 4 (which is determined) such that the checksum of the 128-bit entropy formed by w repeated 11 times and s yields the lower nibble equal to w_val & 0xF.
Thus we only need to check a single s per w_val, not 128 possibilities. Because if we want final_index = w_val, we must set s to w_val >> 4. The final 11 bits are s (bits 4-10) plus c (bits 0-3). That matches w_val exactly. So we must pick s as the high bits of w_val. Then final_index = w_val automatically, regardless of c? Wait we must ensure that final_index = (s << 4) + c = w_val. Since s = w_val >> 4, (s << 4) = w_val - (w_val & 0xF). Then w_val = (s << 4) + (w_val & 0xF). So final_index = (s << 4) + c = w_val if and only if c = w_val & 0xF. So we must set c = lower nibble.
Thus the problem reduces to: For each w_val (0 to 2047), compute s = floor(w_val/16). Then compute the checksum c for the entropy consisting of w_val repeated 11 times (bits 0-120) plus s bits bits 121-127. Then see if c == w_val & 0xF. That's it.
Thus the question: How many w satisfy this? Let's denote w_val index i. There are 2048 possibilities. We can brute force mentally? Not realistic but we can approximate. However, we can possibly deduce that it's around 1/16 of them? Actually each w's lower nibble must match checksum. The checksum is 4 bits, essentially uniformly distributed (if entropy is random). But w's lower nibble is fixed for each w. For each w, there is a specific s. So probability that c equals that specific nibble is 1/16. So we would expect roughly 2048/16 = 128 solutions. That's the expected count if the checksum is independent of the lower bits of w and s (which is essentially random). So we expect roughly 128 words that satisfy that condition.
Thus the answer likely: Approximately 128 words (i.e., about 1/16 of the dictionary). But we need to verify whether any further constraints could slightly change that. But the distribution might not be exactly uniform; however, for random-looking functions, we expect ~128. Also, the example "twelve" is one of those. The user says "the word 'twelve' can be repeated 12 times". We can test "twelve" in the BIP39 wordlist: "twelve" is the word at index? Let's locate "twelve" in the BIP39 English wordlist. The list is: index 0 = "abandon", 1 = "ability", ... we need to find "twelve". Let's quickly recall typical list: It includes "test", "testing"... Not sure. But we can search memory: "twelve" appears at index something like 1795? Actually not sure. Let's check approximate: The BIP39 wordlist is alphabetically sorted. Words: "ticket", "tide", "tiger", "tiny", "tip", "title", "toast", "today", "told", "tomato", "tomorrow", "tone", "tonight", "too", "tool", "tooth", "top", "torch", "tornado", "toss", "total", "tourist", "toward", "tower", "track", "trade", "traffic", "train", "transform", "travel", "treat", "tree", "trend", "trial", "trip", "trove", "truck", "truly", "trust", "truth", "try", "tube", "turn", "turtle", "twelve", "twenty", "two", ... Actually "twelve" appears near the end of the list? Let's see order: The list goes up to index 2047. "twenty" maybe near 1870ish. But anyway.
But we can compute the lower nibble of its index: If "twelve" is index 1795? Let's check. 1795 decimal in hex is 0x703? Actually 1795 decimal = 0x703? Let's compute: 0x700 = 1792, plus 3 = 1795, so 0x703. Lower nibble = 3. So lower nibble is 3. If "twelve" works, then its lower nibble should equal its checksum. So checksum for its entropy must be 3.
Now what is w_val >> 4 for "twelve"? If index = 1795, then s = floor(1795/16) = 112 (since 112*16 = 1792). So s = 112. So the entropy is w repeated 11 times + bits for s = 112 (binary 1110000?). Then compute checksum, must be lower nibble = 3. That matches.
Thus there are about 128 such words. The user asks "How many words in the Bitcoin BIP39 dictionary of 2048 can used to create a 12 word Bitcoin wallet with identical words for the first 11 + checksum word?" The phrase "identical words for the first 11 + checksum word" could be interpreted as they want the 12th word to be identical as well (like "twelve" repeated 12 times). But it's ambiguous: maybe they want any word such that you can have the 11 identical words plus the checksum word (the 12th). It could also be they want the checksum word itself be identical to the first 11 (so the whole phrase is all same word). But the phrase "identical words for the first 11 + checksum word" suggests first 11 identical, plus checksum word (which might be different). But they also gave example "twelve" repeated 12 times (the checksum word is also "twelve"). But maybe that is just the example they found: they thought it's 11 seeds + checksum, but realized it's all 12 identical. It's ambiguous.
But likely they want the count of words that can generate a valid phrase where the first 11 words are identical and the 12th word is the checksum, possibly identical to the first if that works. So the answer is about 128 words (i.e., roughly 1/16 of the wordlist). Let's produce a precise answer: exactly 128? Might be slightly different due to distribution, but expected 128.
We can even provide explanation: Because the last 4 bits of the word index must match the checksum, and checksum is 4 bits uniformly random. Probability 1/16. So expected 2048/16 = 128.
We can also note that "twelve" is one of them. So answer: Approximately 128 words (maybe exactly 128? Could be slightly off if some words are impossible due to constraints). Could we compute exact number? Possibly it's exactly 128 if the mapping is uniform. However, uniform distribution of checksum across all possible entropies may not be exactly uniform for this specific subset. But we can assume near uniform. So answer is around 128.
Thus we can respond: Around 128 words (i.e., about one-sixteenth of the BIP39 wordlist) can be used to produce a 12-word phrase consisting of the same word repeated 11 times and a checksum that yields that same word as the 12th word.
But we can also mention that the set includes any word whose index modulo 16 equals the checksum computed from the entropy formed by the repeated word and its high-order bits. This yields about 128 words.
But the question: "How many words in the Bitcoin BIP39 dictionary of 2048 can used to create a 12 word Bitcoin wallet with identical words for the first 11 + checksum word?" The answer: about 128.
Let's provide explanation: The checksum is 4 bits, so probability 1/16 that it matches the lower 4 bits of the word index; thus 2048/16 = 128. Provide perhaps more rigorous: For any given word index i, the required entropy is uniquely defined (the 7-bit suffix = i>>4). The checksum is deterministic, but as a pseudo-random function, it yields each of the 16 possible values roughly equally. So roughly 128 indices satisfy the condition.
If we want exact number: we could run a small script but can't here. But answer is approximate.
Thus answer: Approximately 128 words (i.e., roughly 6.25% of the dictionary). The exact count may vary but expected 128.
Also note that "twelve" is one of them.
Now let's also consider that some words may have multiple s values that satisfy. But we just need existence. The count remains near 128.
Thus answer: About 128.
Alternatively, if they require the 12th word also identical to the first 11, then the number is roughly 128 (the same count). If they allow the checksum to be any word, then it's likely all 2048 words can produce a valid phrase (since you can choose appropriate suffix bits). But likely they ask about identical phrase (i.e., all 12 words identical). They specifically mention "twelve repeated 12 times". That's a phrase with all identical words. So they ask: "How many words ... can be used to create a 12 word Bitcoin wallet with identical words for the first 11 + checksum word?" Could be they want number of words that can form such phrase where the final word can be anything, but first 11 identical. But they also gave example of all identical. I think they'd like to know number of words that can be repeated 12 times (the first 11 identical plus checksum yields the same word). That is exactly the scenario where the checksum matches the same word.
Thus answer: about 128.
Let's double-check the logic: The phrase comprises 12 words. The first 11 correspond to bits 0-120 of the entropy+checksum. The 12th is bits 121-131: the top 7 bits are the remaining entropy bits (bits 121-127), the lower 4 bits are checksum. So if we set the first 11 words to be identical with index i, the entropy bits are i repeated 11 times (i.e., bits 0-120). The remaining bits (121-127) must be i>>4 (i.e., the high 7 bits of i). So s = i>>4. Then compute checksum c = first 4 bits of SHA256(e). Then final word index is (s << 4) + c = i if and only if c = i & 0xF. So we need that checksum for that specific e yields the lower nibble. That's 1/16 chance. So around 1/16 of the words satisfy.
Thus answer: approx 128.
We can also note that due to the deterministic nature, some might be impossible, but it's extremely close to 128. Could be exactly 128 for typical wordlist, but need to verify by enumeration. Could be that there are 127 or 129, but it's likely exactly 128 due to uniform distribution of SHA256. Actually the function mapping from i to checksum bits is deterministic, but for different i you get different checksums. There's no guarantee that exactly 1/16 of them will match. However, it's plausible. We could note that in practice you'd find about 128 words. To be precise you'd need to run a script. But we can answer as approximate.
Thus final answer: roughly 128 words.
Now respond accordingly, with explanation and some context.
Also mention that if you consider the more general case where the final checksum word can be any word (not necessarily identical), any of the 2048 words can be used as the repeated first 11 words; you can always find a suffix to make the checksum produce any desired final word. But if you require final word identical, it's about 128.
Thus answer: roughly 128.
I will produce final answer: Approximately one-sixteenth of the 2048-word BIP39 list, i.e., about 128 words. Provide explanation. Also perhaps include the example of "twelve". And note that if you allow any checksum word, any word works.
I'll include all that.
Short answer: ≈ 128 words (about 1⁄16 of the 2 048‑word BIP‑39 list).
Why ≈ 128?
A 12‑word BIP‑39 seed is built from 128 bits of entropy + 4 bits of checksum = 132 bits.
If we force the first 11 words to be the same word w (index i, 0 ≤ i < 2048), the bit pattern of the entropy is completely fixed:
BITS
CONTENT
0‑120
i repeated 11 times (the first 11 words)
121‑127
the 7 high‑order bits of i ( i >> 4 )
128‑131
the 4‑bit checksum (first 4 bits of SHA‑256 of the 128‑bit entropy)
The 12‑th word therefore becomes
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final_index = ( (i >> 4) << 4 ) + checksum
= (i & ~0xF) + checksum
For the 12‑th word to be exactly the same as the first 11, we need
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checksum = i & 0xF (the lower 4 bits of i)
The checksum is a deterministic 4‑bit value produced by SHA‑256, but for a random‑looking function each of the 16 possible 4‑bit values is roughly equally likely.
Thus, for a given index i there is a 1‑in‑16 chance that the required checksum matches the lower nibble of i.
With 2 048 possible indices, the expected number that satisfy the condition is
16
2048
=128.
In practice the exact number may be a few off (e.g., 127 or 129) because the SHA‑256 output isn’t perfectly uniform, but it will be very close to 128.
What about the example “twelve”?
“twelve” is simply one of those ≈ 128 words. Its index (let’s call it iₜ) has the property that the checksum computed from the entropy formed by “twelve” repeated 11 times plus the 7 high‑order bits of iₜ equals the lower 4 bits of iₜ, so the 12th word also happens to be “twelve”.
General case (any checksum word)
If you don’t require the 12th word to be identical to the first 11, you can pick any of the 2 048 words as the repeated word. For any chosen word you can freely choose the 7‑bit suffix (bits 121‑127) and obtain a checksum that makes the final 11‑bit word any value you like. In that sense all 2 048 words work.
Bottom line
Identical 12‑word phrase (all words the same): ≈ 128 words, i.e. about one‑sixteenth of the BIP‑39 dictionary.
Any 12‑word phrase with the first 11 identical (checksum can be different): any of the 2 048 words works.
So the answer to the specific question (“identical words for the first 11 + checksum word”) is ≈ 128.
