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[March for reflection |Maynard] staring and factor

已有 466 次阅读 2022-7-16 19:44 |个人分类:牛津大学|系统分类:科研笔记

[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.

High paper, high view, with needed cornerstones.

      ♘   7         5

 

      ♗   2         3

Story - The first problem is to verify the coverage of power...

 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) := (κ + o(1)) (b - 1)^k / klogbas k --> through the integers, where κ = b/(b-1) or b(b'-1)/ b'(b-1).

Note: b':=φ(b), but φ appears not defined in the paper... Wait, φ is called Euler's phi function*. (b' = 4 for b = 10).

---- For a bigger b^k testing.

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

---- #m(2, 10^7) ≈ (10/9 + 0)·9^7/ 7·log10 ≈ 329716.99

---- The exact number of M(2) is of 352155.

---- The error is of  -6.37%.

---- By replacing o(1) with 0.075615, the estimation reaches to the exact value.

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Comments: the estimation is powful for large b^k, with fine accuracy.

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3. Sum of digits of primes

Any number n whose sum of digits s10(n) in base 10 is a multiple of 3 must itself be a multiple of 3.

---- sum of digits of a number may have common factor with that number.

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More generally, in base b, n = ∑nib^i ≡ ∑ni = sb(n)    (mod b - 1), so any integer whose sum of digits in base b has a common factor h with b - 1 must itself be a multiple of h.

---- Take n= 1234, one has 1234 ≡ 10 = sb(1234)   (mod 10 - 1), with h = 1...

---- I see, 1*10^3 + 2*10^2 + 3*10 + 4 = 1 + 2 + 3 + 4 = 10   (mod 10 - 1)

---- Take n = 12,    one has  12    ≡   3 = sb(12)       (mod  10 - 1), with h = 3...

---- I see, 1*10+2*1 = 1 + 2 = 3  mod (10 - 1).

---- The essense is of nib^i = ni   mod (b - 1).

---- nib^i = b^i + b^i + ... + b^i   (there are ni terms of b^i).

---- For each b^i, one has b^i = 1   mod (b - 1).

---- So, nib^i = 1 + 1 + ... + 1 (there are ni terms of 1) = ni   (mod  b - 1).

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Comments: the operation of mod is linear (therefore addicative). I perhaps met the operation of "mod" in middle school (or earlier) but didn't grasp too much of the spirit. It appears to me to use the mod as a ruler, to trim off the maximum multiple (of mod) yet smaller than the operant number, to gain the remainder which is treated to be equivalent to the operant number. In short, the operant number is viewed equivalent to the remainder under the mod: operant ~ remainder  (mod  m).

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Note: Consider n ~ u  (mod  b-1). That is to say, n = q·(b - 1) + u. Take z' the complementary factor of z for some factor of z. One has n = q·(b - 1)'·h + u'·h = [q(b - 1)' + u']·h. That is to say, the common factor of u and (b - 1) is also a factor of n.

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Memo: n ~ u  (mod  b-1) ==> n'h ~ u'h   (mod  (b-1)'h).  |  Betterstarings share factors.

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In particular, the sum of digits sb(p) in base b of a prime number p > b - 1 must necessarily be coprime to b - 1.

---- Let p = q·(b - 1) + sb(p).

---- For one thing, p > b - 1 means p can not be a factor of b - 1.

---- For another, p has no factor to share with b - 1.

---- So, (p, b - 1) = 1.

---- Clearly, b - 1 and sb(p) can not have a common factor greater than 1. (Or, that factor would be shared with p, a contradiction).

---- As p > sb(p), p is also coprime to sb(p).

---- That is to say, p, b - 1 and sb(p) are coprime to each other.

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|| After a moment's thought, it appears difficult to think of any other property linking the sum of digits function with the primes.

---- clear.

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This might encourage us to guess that if we look at large numbers, the properties `being prime' and `having sum of digits equal to a' are roughly independent, apart from the property that a must be coprime with b - 1.

---- guess: n being a prime is not related to sb(n), except for sb(n) and b - 1 being coprime.

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In particular, we might guess that asymptotically 50% of prime numbers have an even sum of digits, and 50% have an odd sum of digits...

---- Sum of digits of primes might be equaly grouped by even and odd.

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... or that 50% of primes have sum of digits which is 1 (mod 3) and 50% a sum of digits which is 2 (mod 3) (since 3 is the only prime with sum of digits 0 (mod 3)).

---- Another way to view the 50% vs. 50% division of the sums of digits of primes.

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Theorem 2.1 is confirming this guess.

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Comments: this second paragraph offers an qualitative analysis concerning the 50% even-odd distribution of the sum of digits of primes, referring to Th.2.1. The first paragraph offers the key property of the sum of digits of primes.

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Summary: in this section Maynard shared insights concerning Th.2.1.

 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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