# [March for reflection |Maynard] art of digit :)))

[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.

Digit, digit, digit ...

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

2. Digital properties of primes

---- Does a prime base matter in number theory?

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The aim of this article is to discuss some results on primes in sets A described by their digit properties.

---- primes ~> A <~ digit properties

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Such sets are interesting from a technical point of view since they do not possess an obvious multiplicative structure;

---- multiplicative structure might not be applied to A.

---- As such, the problem might be harder.

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the digital properties of two integers n and m are in general not closely related to the digital properties of nm, since carries will destroy any simple structure.

---- "carries"(进位, 移位) forbid transport of digital properties in addition or multiplication.

---- While carries are simple operations, they are the causes of the infavorable situation here.

---- (In Chinese culture, one is assumed to ignore simple matters as if simple matters are of humiliating...)

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Nevertheless, we are able to count primes in various sets A defined in terms of the digits of integers, even in situations when the set A is sparse.

---- to summarize the abilities and features shown in this paper.

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In particular, for our discussion we focus on the following three results:

---- clear.

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Theorem 2.1 (Mauduit-Rivat [11]). Let sb(n) denote the sum of digits of n in base b. Then for any fixed choise of (m, b-1) = 1, we have

#{ p < x: sb(p) ≡ a (mod m)} = (1 + o(1))x / m·logx .

---- This is to count in A = [1, x] the primes p whose diginal sum in base b are constant under the mode m, while m and b-1 do not divide each other.

---- Say, take b = 10 and m = 2.

---- sb(2) = 0  mod 2.

---- sb(3) = 1  mod 2.

---- sb(5) = 1  mod 2.

---- sb(7) = 1  mod 2.

---- sb(11) = 0  mod 2.

---- sb(13) = 0  mod 2.

---- sb(17) = 0  mod 2.

---- sb(19) = 0  mod 2.

---- sb(23) = 1  mod 2.

---- sb(29) = 1  mod 2.

---- So, let's see the formula for x = 30.

---- (1 + o(1))x / m·logx  ≈ (1 + 0)·30 / 2·log30 = 4.4

---- Clearly, the correct answer is 5 for a = 0 or 1.

---- Indeed, ceil(4.4) = 5.

---- In this way, one is able to gain the total number of primes in [1, x], for x = 30 or any other values.(!)

---- The Mauduit-Rivat [11] formula was published in 2010.

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Comment: is this result not as fundamental as those recorded on a textbook?

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---- To try a better counting, one may set x = 10^8, for b =10 and m = 2.

---- One has (1 + o(1))x / m·logx  ≈ (1 + 0)·10^8 / 2·log10^8 = 2.714340511895324e+06 ≈ 2,714,341, for a = 0 or 1.

---- So, the total number of primes in [1, 10^8] is calculated as 2,714,341x2 = 5428682.

---- By a command on the computer, that number is 5761455.

---- The estimation error is of -5.78%.

---- For a more accurate estimation, one may take o(1) as 0.0612993.

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In particular, the result of Mauduit-Rivat shows that asymptotically 50% of primes have an even sum of digits, and 50% of primes have an odd sum of digits when viewed in base 10.

---- One can see this for the formula is independent of the digital sum, denoted as "a" in the theorem.(More clearly, one can set m = 2.)

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Theorem 2.2 (Bourgain [1]). Let c > 0 be a sufficiently small constant, and k be a sufficiently large integer. Then for any index set I ⊆ {1, ..., k} with #I ≤ ck, and for any choice of base-2 digits εi ∈ {0, 1} we have

#{ p = ∑ni2^i : ni∈{0, 1}, nj = εj  for j ∈ I } = (1 + o(1)) 2^k - #I / log2^k.

---- To see things here clearly, I take c = 0.1, and k = 30, such that ck = 3.

---- So, one can take from 1, 2, ..., 30 arbitrarily  3 numbers at most to form the index set I.

---- Let the prime number p of 30 digits be represented by 30 dots (apart from 3 occupied) ——

·······1········1··········1··

---- Each dot takes a value of 0 or 1, but #I of them are prescribed at the locations denoted by I

---- For example, one may take #I = 3, and all εj = 1.

---- Actually, once #I and k are given, locations and values of εj do not affect the estimation.

---- In the case of #= 0, one gains a formula to estimate the number of primes with k digits in base-2.

---- Say, for k = 30, the biggest number of base-2 is 1073741823 in base 10.

---- That covers 54400028 primes.

---- By Bourgain's estimation, one has 2^30/log(2^30) ≈ 51636067.

---- The estimation error is of -5.08%.

---- One may take o(1) as 0.05352772 for an accurate estimation.

---- The Bourgain[1] formula was published in 2015.

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The result of Bourgain shows that you can prescribe a positive proportion of the binary digits (of numbers with k digits), and regardless of which digits are prescribed we can still find primes with the prescribed digits.

---- It appears "positive proportion" should be replaced by "non-negative proportion".

---- Bourgain [11] offered a formula to estimate the number of primes through the size of the un-prescribed digits.

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Theorem 2.3 (Maynard [13]). There are constants 0 < c < C such that for any choice of a0 ∈ {0, ..., b - 1} and for any x > b ≥ 10, we have

c·M(x, b) ≤ #{ p < x : p has no base b digit equal to a0 } ≤ C·M(x, b).

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Note: M(x, b):= x^log(b-1)/logb / logx

---- I refer to M(x, b) as Maynard's estimation.

---- The estimation is independent of a0, the missing digit of base b.

---- To see things clearly here, set x = 30, b = 10, taking a0 = 2.

---- First of all, log(b-1)/logb = 0.954242509439325 for b = 10.

---- So, M(x, b) = 7.549.

---- Actually, #M(2) = 7 for x = 30.

---- If one takes x = 100, one has M(x, b) = 17.59.

---- Actually, #M(2) = 22  for x = 100.

---- The estimation error is of -18.18%.

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In particular, this shows that there are infinitely many primes which have no digit equal to 7 in their decimal expansion.

---- Why M(7) is specialHow about other cases of a0?

---- Well, 7 is the biggest prime of one digit under base 10.

---- Also, 7 = 5 + 2, and 3 = 5 - 2.

---- It appears 2 and 5 are more fundamental than 3 and 7.

---- Motivated to see the proof...

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(to be continued)

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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