# [March for reflection |Maynard] digit zero differs

[注：下文是群邮件的内容，标题是原有的。内容是学习一篇数学文章的笔记。]

["Terms of awareness /use" folded below] On going is to read a paper of primes to increase generic understanding on mathematics.

It is high papers that make textbooks meaningful, not on the contrary.

♘   7         5

♗   2         3

Story - Bishop assigned a job to check the cornerstones...

ℂ ℍ ℕ ℙ ℚ ℝ ℤ ℭ ℜ I|φ∪∩∈ ⊆ ⊂ ⊇ ⊃ ⊄ ⊅ ≤ ≥ Γ Θ Λ α Δ δ μ ≠ ⌊ ⌋ ∨∧∞Φ⁻⁰ 1

Review: #m(a0, b^k) := (κi + o(1)) (b - 1)^k / klogbas k --> through the integers, where κ2 = b/(b-1), κ1 = b(b'-1)/ b'(b-1).

Note: b':=φ(b), where φ is called Euler's phi function*. (b' = 4 for b = 10).

---- For a "full" test of bigger b^k, taking a0 = 0,...,9 as the missing digit.

---- Set b = 10, k = 7.

---- κ2 = b/(b-1) = 1.111... for a0 = 0, 2, 4, 5, 6, 8. [Note: κ value need a rivision for digit 0, see below].

---- #m(a0, 10^7) = 329716.99 ≈ 329717

---- Eaxct value for a0 = 0 is of 397866 (accuracy: -17.13%);

---- Eaxct value for a0 = 2 is of 352155 (accuracy: -6.37%);

---- Eaxct value for a0 = 4 is of 354123 (accuracy: -6.89%);

---- Eaxct value for a0 = 5 is of 354910 (accuracy: -7.10%);

---- Eaxct value for a0 = 6 is of 355148 (accuracy: -7.16%);

---- Eaxct value for a0 = 8 is of 355805 (accuracy: -7.33%);

Comment: digit 0 is special in primes.

---- κ1 = b(b'-1)/ b'(b-1) =  0.8333... for a0 = 1, 3, 7, 9.

---- #m(a0, 10^7) = 247287.74 ≈ 247288

---- Eaxct value for a0 = 1 is of 262549 (accuracy: -5.81%);

---- Eaxct value for a0 = 3 is of 265015 (accuracy: -6.69%);

---- Eaxct value for a0 = 7 is of 266823 (accuracy: -7.32%);

---- Eaxct value for a0 = 9 is of 267486 (accuracy: -7.55%);

Comment: digits 1, 3, 7, 9 are greater weighted in primes than the rest digits. Apparently, these four digits may appear as the ending digit of all primes greater than 10, while other digits may not.

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Observation: roughly, the number of primes missing digit 0 is of 4x10^5 greater than other cases of the same type. In particular, if one calculates the difference of the two estimations, namely 329717 -  247288 = 82429, one may want to divide this difference by two and add it onto the estimation for the first type, to gain a corrected estimation of 329717 + 41215 = 370932 (accuracy: -6.77%) for the digit 0. This is reasonable as the digit 0 does not have the chance to be the leading digit for any prime, while all other digits have. So, one can guess (in the case of base 10) for large b^k ——

#m(0, 10^k) = #m(2, 10^k) + #Δ,  where #Δ = [#m(2, 10^k) - #m(1, 10^k) ]/2.

That is to say, for digit 0, one takes κ0 = κ2 + (κ2 - κ1)/2.  For both types of the digits, except for 0, the esitmaition is slightly more accurate for the smaller digits. It's clear that digit 1 is weighted number one by presence in primes for x < 10^7, possibly the case for any upper bound.

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How many primes are there in the case of P(2, 3, 5, 7) where the digits 0, 1, 4, 6, 8, 9 are missing? The exact number is 1903 for x < 10^7. That is to say, P(2, 3, 5, 7) is very sparse in [1, x] for sufficiently large x. By homemade experiments, I got #P(2,3,5,7) = (1 + o(1))·2^3·3^(k-2), verified for 10^k with 3 ≤ k ≤ 9. (I expect to check later for the estimation of multiple missing digits in a major paper by Maynard).

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Review (halfway modulo): n = ∑nib^i  ∑ni = sb(n)    (mod  b - 1) ...... （%)

In the last note, I made an amussing but partial understanding to this formula.

---- The equation n = q·(b - 1) + sb(n) appears to hold, though I need to check it closely.

---- The partial understanding was identified accidentally when checking a case of missing digit 0 in the above experiment.

---- The case is of 9817141, a usual example of primes missing digit 0.

---- Apparently, 9817141 = 1090793·(10 -1) + 4.

---- At the same time, sb(9817141) = 31 = 3·(10 - 1) + 4.

---- That is, 9817141 = 4  (mod 10 - 1), and

----  ......................31 = 4  (mod 10 - 1).

---- The fomula (%) refers to 4 ≡ 4, not 9817141 = sb(9817141)   (mod b - 1).

---- But, the equation n = q·(b - 1) + sb(n) may hold.

---- One can rewrite 9817141 = 1090790·(10 - 1) + s10(9817141), though digit 9 is not coprime to 10 - 1.

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The arguments in the last note only hold partially (i.e. for the coprime case) ——

---- For ni coprime to b - 1, one indeed has ni·b^i = ni   (mod  b - 1).

---- Under this coprime assumption, one indeed has n = sb(n)    (mod  b - 1)  at some stage of modulus.

---- That is, n = q·(b - 1) + sb(n), generally sb(n) > b - 1. (Note: I referred to "n", "b - 1" and sb(n) as "starings" who apparently share factors, if any).

---- For this reason, to the end, n and sb(n) would have the same remainder under modulus b - 1.

---- For example, taking n = 124578, one has sb(124578) = 27 > 10 - 1.

---- Further, 124578 = 13842·(10 - 1) + 0.

---- One can rewrite n = 124578 = 13839·(10 - 1) + 27.

---- That is, 124578 = 27  (mod 10 - 1), which I call halfway modulo.

---- Or, 124578 = sb(124578)   (mod 10 - 1).

---- So comes the form n = q·(b - 1) +  sb(n) for the present case.

---- Or, n = sb(n)   (mod b - 1).

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For the case of ni not coprime to b - 1, one still has ni·b^i = ni   (mod  b - 1), except for ni = b - 1.

---- The inequality ni ≤ b - 1 aways holds.

---- For the case of ni < b - 1, one still has ni·b^i = ni   (mod  b - 1).

---- For the case of ni = b - 1, it is better to check an example.

---- Say, n = 98765, with sb(n) = 35.

---- 9·10^4 = 10^4·(10 - 1) + 0.

---- That is, 9·10^4 = 0   (mod 10 - 1)

---- However, one can always separate a 9 from the form of u·9 for any integer u > 9.

---- That is, 9·10^4 = 10^4·(10 - 1) = 9999·(10 - 1) + 9.

---- In the sense of halfway modulo, one has 9·10^4 = 9  (mod 10 - 1).

---- With such a special treatment, one still has ni·b^i = ni  (mod  b - 1) for ni = b - 1.

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In summary, the equation n = q·(b - 1) + sb(n) holds for any (positive) integer. This equation appears more useful than equation (%).

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Question: what if sb(n) be a prime?

ℂ ℍ ℕ ℙ ℚ ℝ ℤ ℭ ℜ I|φ∪∩∈ ⊆ ⊂ ⊇ ⊃ ⊄ ⊅ ≤ ≥ Γ Θ Λ α Δ δ μ ≠ ⌊ ⌋ ∨∧∞Φ⁻⁰ 1

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