Sylvester-Schur: binomial and consecutive forms
TheoremLarge prime divisors of a block term divide the binomial coefficient#✓fully proved
Nat.SylvesterSchur.prime_dvd_choose_of_dvd_consecutive
Let \(n\), \(k\) and \(j\) be natural numbers with \(j < n\), and let \(p\) be a prime with \(p > n\). If \(p\) divides \(k + j\), one of the \(n\) consecutive integers \(k, \dots, k+n-1\), then \(p\) divides \(\binom{k+n-1}{n}\).
theorem prime_dvd_choose_of_dvd_consecutive {n k p j : ℕ} (hp : p.Prime) (hp_gt : n < p)
(hj : j < n) (hdvd : p ∣ k + j) :
p ∣ Nat.choose (k + n - 1) n
TheoremPrime divisors of the block binomial coefficient divide a block term#✓fully proved
Nat.SylvesterSchur.exists_dvd_consecutive_of_prime_dvd_choose
Let \(n\) and \(k\) be natural numbers and let \(p\) be a prime. If \(p\) divides \(\binom{k+n-1}{n}\), then \(p\) divides \(k + i\) for some \(i\) with \(0 \le i < n\).
theorem exists_dvd_consecutive_of_prime_dvd_choose (n k p : ℕ) (hp : p.Prime)
(hp_dvd : p ∣ Nat.choose (k + n - 1) n) :
∃ i < n, p ∣ k + i
TheoremFrom a large prime divisor of the binomial coefficient to one in the block#✓fully proved
Nat.SylvesterSchur.consecutive_of_exists_prime_dvd_choose
Let \(n\) and \(k\) be natural numbers. If some prime \(p > n\) divides \(\binom{k+n-1}{n}\), then some prime \(p > n\) divides one of the integers \(k + i\) with \(0 \le i < n\).
theorem consecutive_of_exists_prime_dvd_choose {n k : ℕ}
(hbin : ∃ p : ℕ, p.Prime ∧ n < p ∧ p ∣ Nat.choose (k + n - 1) n) :
∃ i < n, ∃ p : ℕ, p.Prime ∧ n < p ∧ p ∣ k + i
Uses 1 · Used by 7
Used byErdős layer criterion for a large prime in a consecutive block · Erdős layer criterion with a one-third first layer for a large prime in a consecutive block · Half-exponent criterion for a large prime in a consecutive block · Prime-counting criterion for a large prime in a consecutive block · Square-root and primorial criterion for a large prime in a consecutive block · Large starts give a large prime in a consecutive block · Quadratic starts give a large prime in a consecutive block
3 fully proved
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