FormalConjectures.OEIS.«185895»

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import FormalConjecturesUtil

Coefficients of $\prod_{k>0} (1 - x^k/k!)$

The sequence $a(n)$ has exponential generating function $$E(x) = \prod_{k=1}^\infty \left(1 - \frac{x^k}{k!}\right),$$ so that $a(n) = n! [x^n] \prod_{k=1}^n \left(1 - \frac{x^k}{k!}\right)$.

References:

A185895

open Polynomialnamespace OeisA185895

The finite polynomial approximation $\prod_{k=1}^n (1 - X^k / k!)$.

noncomputable def P (n : ℕ) : Polynomial ℚ :=
  ∏ k ∈ Finset.Icc 1 n, (1 - C (1 / (k.factorial : ℚ)) * X ^ k)

The sequence $a(n) = n! [x^n] \prod_{k=1}^n (1 - x^k / k!)$.

noncomputable def a (n : ℕ) : ℤ :=
  if n = 0 then 1
  else (coeff (P n) n * (n.factorial : ℚ)).floor

A natural number $n$ is triangular if $n = k(k+1)/2$ for some $k \in \mathbb{N}$.

def IsTriangular (n : ℕ) : Prop := ∃ k : ℕ, n = k * (k + 1) / 2



@[category API, AMS 11]
lemma C_mul_X_pow_mul (a b : ℚ) (n m : ℕ) :
    (C a * X ^ n) * (C b * X ^ m) = C (a * b) * X ^ (n + m) := bya:ℚb:ℚn:ℕm:ℕ⊢ C a * X ^ n * (C b * X ^ m) = C (a * b) * X ^ (n + m)
  rw [C_mul,a:ℚb:ℚn:ℕm:ℕ⊢ C a * X ^ n * (C b * X ^ m) = C a * C b * X ^ (n + m)a:ℚb:ℚn:ℕm:ℕ⊢ C a * X ^ n * (C b * X ^ m) = C a * C b * (X ^ n * X ^ m) pow_adda:ℚb:ℚn:ℕm:ℕ⊢ C a * X ^ n * (C b * X ^ m) = C a * C b * (X ^ n * X ^ m)a:ℚb:ℚn:ℕm:ℕ⊢ C a * X ^ n * (C b * X ^ m) = C a * C b * (X ^ n * X ^ m)]a:ℚb:ℚn:ℕm:ℕ⊢ C a * X ^ n * (C b * X ^ m) = C a * C b * (X ^ n * X ^ m)
  ringAll goals completed! 🐙

Value of the sequence a at 0.

@[category test, AMS 11]
theorem a_0 : a 0 = 1 := by⊢ a 0 = 1 rflAll goals completed! 🐙

Value of the sequence a at 1.

@[category test, AMS 11]
theorem a_1 : a 1 = -1 := by⊢ a 1 = -1
  dsimp [a, P]⊢ ((∏ k ∈ Finset.Icc 1 1, (1 - C (1 / ↑k.factorial) * X ^ k)).coeff 1 * 1).floor = -1
  simp [coeff_one]⊢ (-1).floor = -1
  rflAll goals completed! 🐙

Value of the sequence a at 2.


@[category test, AMS 11]
theorem a_2 : a 2 = -1 := by⊢ a 2 = -1
  dsimp [a, P]⊢ ((∏ k ∈ Finset.Icc 1 2, (1 - C (1 / ↑k.factorial) * X ^ k)).coeff 2 * 2).floor = -1
  have hI : (Finset.Icc 1 2 : Finset ℕ) = {1, 2} := by⊢ a 2 = -1hI:Finset.Icc 1 2 = {1, 2}⊢ ((∏ k ∈ Finset.Icc 1 2, (1 - C (1 / ↑k.factorial) * X ^ k)).coeff 2 * 2).floor = -1 decidehI:Finset.Icc 1 2 = {1, 2}⊢ ((∏ k ∈ Finset.Icc 1 2, (1 - C (1 / ↑k.factorial) * X ^ k)).coeff 2 * 2).floor = -1hI:Finset.Icc 1 2 = {1, 2}⊢ ((∏ k ∈ Finset.Icc 1 2, (1 - C (1 / ↑k.factorial) * X ^ k)).coeff 2 * 2).floor = -1
  rw [hI,hI:Finset.Icc 1 2 = {1, 2}⊢ ((∏ k ∈ {1, 2}, (1 - C (1 / ↑k.factorial) * X ^ k)).coeff 2 * 2).floor = -1hI:Finset.Icc 1 2 = {1, 2}⊢ (((1 - C (1 / ↑(Nat.factorial 1)) * X ^ 1) * (1 - C (1 / ↑(Nat.factorial 2)) * X ^ 2)).coeff 2 * 2).floor = -1 Finset.prod_insert (byhI:Finset.Icc 1 2 = {1, 2}⊢ 1 ∉ {2}hI:Finset.Icc 1 2 = {1, 2}⊢ (((1 - C (1 / ↑(Nat.factorial 1)) * X ^ 1) * (1 - C (1 / ↑(Nat.factorial 2)) * X ^ 2)).coeff 2 * 2).floor = -1 decideAll goals completed! 🐙hI:Finset.Icc 1 2 = {1, 2}⊢ (((1 - C (1 / ↑(Nat.factorial 1)) * X ^ 1) * (1 - C (1 / ↑(Nat.factorial 2)) * X ^ 2)).coeff 2 * 2).floor = -1), Finset.prod_singletonhI:Finset.Icc 1 2 = {1, 2}⊢ (((1 - C (1 / ↑(Nat.factorial 1)) * X ^ 1) * (1 - C (1 / ↑(Nat.factorial 2)) * X ^ 2)).coeff 2 * 2).floor = -1hI:Finset.Icc 1 2 = {1, 2}⊢ (((1 - C (1 / ↑(Nat.factorial 1)) * X ^ 1) * (1 - C (1 / ↑(Nat.factorial 2)) * X ^ 2)).coeff 2 * 2).floor = -1]hI:Finset.Icc 1 2 = {1, 2}⊢ (((1 - C (1 / ↑(Nat.factorial 1)) * X ^ 1) * (1 - C (1 / ↑(Nat.factorial 2)) * X ^ 2)).coeff 2 * 2).floor = -1
  simp only [mul_sub, sub_mul, one_mul, mul_one, C_mul_X_pow_mul,
    coeff_sub, coeff_one, coeff_C_mul_X_pow]hI:Finset.Icc 1 2 = {1, 2}⊢ ((if 2 = 0 then 1 else 0) * 2 - (if 2 = 1 then 1 / ↑(Nat.factorial 1) else 0) * 2 -
      ((if True then 1 / ↑(Nat.factorial 2) else 0) * 2 -
        (if 2 = 1 + 2 then 1 / ↑(Nat.factorial 1) * (1 / ↑(Nat.factorial 2)) else 0) * 2)).floor =
  -1
  decide +nativeAll goals completed! 🐙

Value of the sequence a at 3.


@[category test, AMS 11]
theorem a_3 : a 3 = 2 := by⊢ a 3 = 2
  dsimp [a, P]⊢ ((∏ k ∈ Finset.Icc 1 3, (1 - C (1 / ↑k.factorial) * X ^ k)).coeff 3 * ↑(Nat.factorial 3)).floor = 2
  have hI : (Finset.Icc 1 3 : Finset ℕ) = {1, 2, 3} := by⊢ a 3 = 2hI:Finset.Icc 1 3 = {1, 2, 3}⊢ ((∏ k ∈ Finset.Icc 1 3, (1 - C (1 / ↑k.factorial) * X ^ k)).coeff 3 * ↑(Nat.factorial 3)).floor = 2 decidehI:Finset.Icc 1 3 = {1, 2, 3}⊢ ((∏ k ∈ Finset.Icc 1 3, (1 - C (1 / ↑k.factorial) * X ^ k)).coeff 3 * ↑(Nat.factorial 3)).floor = 2hI:Finset.Icc 1 3 = {1, 2, 3}⊢ ((∏ k ∈ Finset.Icc 1 3, (1 - C (1 / ↑k.factorial) * X ^ k)).coeff 3 * ↑(Nat.factorial 3)).floor = 2
  rw [hI,hI:Finset.Icc 1 3 = {1, 2, 3}⊢ ((∏ k ∈ {1, 2, 3}, (1 - C (1 / ↑k.factorial) * X ^ k)).coeff 3 * ↑(Nat.factorial 3)).floor = 2hI:Finset.Icc 1 3 = {1, 2, 3}⊢ (((1 - C (1 / ↑(Nat.factorial 1)) * X ^ 1) *
            ((1 - C (1 / ↑(Nat.factorial 2)) * X ^ 2) * (1 - C (1 / ↑(Nat.factorial 3)) * X ^ 3))).coeff
        3 *
      ↑(Nat.factorial 3)).floor =
  2 Finset.prod_insert (byhI:Finset.Icc 1 3 = {1, 2, 3}⊢ 1 ∉ {2, 3}hI:Finset.Icc 1 3 = {1, 2, 3}⊢ (((1 - C (1 / ↑(Nat.factorial 1)) * X ^ 1) *
            ((1 - C (1 / ↑(Nat.factorial 2)) * X ^ 2) * (1 - C (1 / ↑(Nat.factorial 3)) * X ^ 3))).coeff
        3 *
      ↑(Nat.factorial 3)).floor =
  2 decideAll goals completed! 🐙hI:Finset.Icc 1 3 = {1, 2, 3}⊢ (((1 - C (1 / ↑(Nat.factorial 1)) * X ^ 1) *
            ((1 - C (1 / ↑(Nat.factorial 2)) * X ^ 2) * (1 - C (1 / ↑(Nat.factorial 3)) * X ^ 3))).coeff
        3 *
      ↑(Nat.factorial 3)).floor =
  2), Finset.prod_insert (byhI:Finset.Icc 1 3 = {1, 2, 3}⊢ 2 ∉ {3}hI:Finset.Icc 1 3 = {1, 2, 3}⊢ (((1 - C (1 / ↑(Nat.factorial 1)) * X ^ 1) *
            ((1 - C (1 / ↑(Nat.factorial 2)) * X ^ 2) * (1 - C (1 / ↑(Nat.factorial 3)) * X ^ 3))).coeff
        3 *
      ↑(Nat.factorial 3)).floor =
  2 decideAll goals completed! 🐙hI:Finset.Icc 1 3 = {1, 2, 3}⊢ (((1 - C (1 / ↑(Nat.factorial 1)) * X ^ 1) *
            ((1 - C (1 / ↑(Nat.factorial 2)) * X ^ 2) * (1 - C (1 / ↑(Nat.factorial 3)) * X ^ 3))).coeff
        3 *
      ↑(Nat.factorial 3)).floor =
  2), Finset.prod_singletonhI:Finset.Icc 1 3 = {1, 2, 3}⊢ (((1 - C (1 / ↑(Nat.factorial 1)) * X ^ 1) *
            ((1 - C (1 / ↑(Nat.factorial 2)) * X ^ 2) * (1 - C (1 / ↑(Nat.factorial 3)) * X ^ 3))).coeff
        3 *
      ↑(Nat.factorial 3)).floor =
  2hI:Finset.Icc 1 3 = {1, 2, 3}⊢ (((1 - C (1 / ↑(Nat.factorial 1)) * X ^ 1) *
            ((1 - C (1 / ↑(Nat.factorial 2)) * X ^ 2) * (1 - C (1 / ↑(Nat.factorial 3)) * X ^ 3))).coeff
        3 *
      ↑(Nat.factorial 3)).floor =
  2]hI:Finset.Icc 1 3 = {1, 2, 3}⊢ (((1 - C (1 / ↑(Nat.factorial 1)) * X ^ 1) *
            ((1 - C (1 / ↑(Nat.factorial 2)) * X ^ 2) * (1 - C (1 / ↑(Nat.factorial 3)) * X ^ 3))).coeff
        3 *
      ↑(Nat.factorial 3)).floor =
  2
  simp only [mul_sub, sub_mul, one_mul, mul_one, C_mul_X_pow_mul,
    coeff_sub, coeff_one, coeff_C_mul_X_pow]hI:Finset.Icc 1 3 = {1, 2, 3}⊢ ((if 3 = 0 then 1 else 0) * ↑(Nat.factorial 3) - (if 3 = 1 then 1 / ↑(Nat.factorial 1) else 0) * ↑(Nat.factorial 3) -
        ((if 3 = 2 then 1 / ↑(Nat.factorial 2) else 0) * ↑(Nat.factorial 3) -
          (if True then 1 / ↑(Nat.factorial 1) * (1 / ↑(Nat.factorial 2)) else 0) * ↑(Nat.factorial 3)) -
      ((if True then 1 / ↑(Nat.factorial 3) else 0) * ↑(Nat.factorial 3) -
          (if 3 = 1 + 3 then 1 / ↑(Nat.factorial 1) * (1 / ↑(Nat.factorial 3)) else 0) * ↑(Nat.factorial 3) -
        ((if 3 = 2 + 3 then 1 / ↑(Nat.factorial 2) * (1 / ↑(Nat.factorial 3)) else 0) * ↑(Nat.factorial 3) -
          (if 3 = 1 + (2 + 3) then 1 / ↑(Nat.factorial 1) * (1 / ↑(Nat.factorial 2) * (1 / ↑(Nat.factorial 3)))
            else 0) *
            ↑(Nat.factorial 3)))).floor =
  2
  decide +nativeAll goals completed! 🐙

Value of the sequence a at 4.


@[category test, AMS 11]
theorem a_4 : a 4 = 3 := by⊢ a 4 = 3
  dsimp [a, P]⊢ ((∏ k ∈ Finset.Icc 1 4, (1 - C (1 / ↑k.factorial) * X ^ k)).coeff 4 * ↑(Nat.factorial 4)).floor = 3
  have hI : (Finset.Icc 1 4 : Finset ℕ) = {1, 2, 3, 4} := by⊢ a 4 = 3hI:Finset.Icc 1 4 = {1, 2, 3, 4}⊢ ((∏ k ∈ Finset.Icc 1 4, (1 - C (1 / ↑k.factorial) * X ^ k)).coeff 4 * ↑(Nat.factorial 4)).floor = 3 decidehI:Finset.Icc 1 4 = {1, 2, 3, 4}⊢ ((∏ k ∈ Finset.Icc 1 4, (1 - C (1 / ↑k.factorial) * X ^ k)).coeff 4 * ↑(Nat.factorial 4)).floor = 3hI:Finset.Icc 1 4 = {1, 2, 3, 4}⊢ ((∏ k ∈ Finset.Icc 1 4, (1 - C (1 / ↑k.factorial) * X ^ k)).coeff 4 * ↑(Nat.factorial 4)).floor = 3
  rw [hI,hI:Finset.Icc 1 4 = {1, 2, 3, 4}⊢ ((∏ k ∈ {1, 2, 3, 4}, (1 - C (1 / ↑k.factorial) * X ^ k)).coeff 4 * ↑(Nat.factorial 4)).floor = 3hI:Finset.Icc 1 4 = {1, 2, 3, 4}⊢ (((1 - C (1 / ↑(Nat.factorial 1)) * X ^ 1) *
            ((1 - C (1 / ↑(Nat.factorial 2)) * X ^ 2) *
              ((1 - C (1 / ↑(Nat.factorial 3)) * X ^ 3) * (1 - C (1 / ↑(Nat.factorial 4)) * X ^ 4)))).coeff
        4 *
      ↑(Nat.factorial 4)).floor =
  3 Finset.prod_insert (byhI:Finset.Icc 1 4 = {1, 2, 3, 4}⊢ 1 ∉ {2, 3, 4}hI:Finset.Icc 1 4 = {1, 2, 3, 4}⊢ (((1 - C (1 / ↑(Nat.factorial 1)) * X ^ 1) *
            ((1 - C (1 / ↑(Nat.factorial 2)) * X ^ 2) *
              ((1 - C (1 / ↑(Nat.factorial 3)) * X ^ 3) * (1 - C (1 / ↑(Nat.factorial 4)) * X ^ 4)))).coeff
        4 *
      ↑(Nat.factorial 4)).floor =
  3 decideAll goals completed! 🐙hI:Finset.Icc 1 4 = {1, 2, 3, 4}⊢ (((1 - C (1 / ↑(Nat.factorial 1)) * X ^ 1) *
            ((1 - C (1 / ↑(Nat.factorial 2)) * X ^ 2) *
              ((1 - C (1 / ↑(Nat.factorial 3)) * X ^ 3) * (1 - C (1 / ↑(Nat.factorial 4)) * X ^ 4)))).coeff
        4 *
      ↑(Nat.factorial 4)).floor =
  3), Finset.prod_insert (byhI:Finset.Icc 1 4 = {1, 2, 3, 4}⊢ 2 ∉ {3, 4}hI:Finset.Icc 1 4 = {1, 2, 3, 4}⊢ (((1 - C (1 / ↑(Nat.factorial 1)) * X ^ 1) *
            ((1 - C (1 / ↑(Nat.factorial 2)) * X ^ 2) *
              ((1 - C (1 / ↑(Nat.factorial 3)) * X ^ 3) * (1 - C (1 / ↑(Nat.factorial 4)) * X ^ 4)))).coeff
        4 *
      ↑(Nat.factorial 4)).floor =
  3 decideAll goals completed! 🐙hI:Finset.Icc 1 4 = {1, 2, 3, 4}⊢ (((1 - C (1 / ↑(Nat.factorial 1)) * X ^ 1) *
            ((1 - C (1 / ↑(Nat.factorial 2)) * X ^ 2) *
              ((1 - C (1 / ↑(Nat.factorial 3)) * X ^ 3) * (1 - C (1 / ↑(Nat.factorial 4)) * X ^ 4)))).coeff
        4 *
      ↑(Nat.factorial 4)).floor =
  3),
      Finset.prod_insert (byhI:Finset.Icc 1 4 = {1, 2, 3, 4}⊢ 3 ∉ {4}hI:Finset.Icc 1 4 = {1, 2, 3, 4}⊢ (((1 - C (1 / ↑(Nat.factorial 1)) * X ^ 1) *
            ((1 - C (1 / ↑(Nat.factorial 2)) * X ^ 2) *
              ((1 - C (1 / ↑(Nat.factorial 3)) * X ^ 3) * (1 - C (1 / ↑(Nat.factorial 4)) * X ^ 4)))).coeff
        4 *
      ↑(Nat.factorial 4)).floor =
  3 decideAll goals completed! 🐙hI:Finset.Icc 1 4 = {1, 2, 3, 4}⊢ (((1 - C (1 / ↑(Nat.factorial 1)) * X ^ 1) *
            ((1 - C (1 / ↑(Nat.factorial 2)) * X ^ 2) *
              ((1 - C (1 / ↑(Nat.factorial 3)) * X ^ 3) * (1 - C (1 / ↑(Nat.factorial 4)) * X ^ 4)))).coeff
        4 *
      ↑(Nat.factorial 4)).floor =
  3), Finset.prod_singletonhI:Finset.Icc 1 4 = {1, 2, 3, 4}⊢ (((1 - C (1 / ↑(Nat.factorial 1)) * X ^ 1) *
            ((1 - C (1 / ↑(Nat.factorial 2)) * X ^ 2) *
              ((1 - C (1 / ↑(Nat.factorial 3)) * X ^ 3) * (1 - C (1 / ↑(Nat.factorial 4)) * X ^ 4)))).coeff
        4 *
      ↑(Nat.factorial 4)).floor =
  3hI:Finset.Icc 1 4 = {1, 2, 3, 4}⊢ (((1 - C (1 / ↑(Nat.factorial 1)) * X ^ 1) *
            ((1 - C (1 / ↑(Nat.factorial 2)) * X ^ 2) *
              ((1 - C (1 / ↑(Nat.factorial 3)) * X ^ 3) * (1 - C (1 / ↑(Nat.factorial 4)) * X ^ 4)))).coeff
        4 *
      ↑(Nat.factorial 4)).floor =
  3]hI:Finset.Icc 1 4 = {1, 2, 3, 4}⊢ (((1 - C (1 / ↑(Nat.factorial 1)) * X ^ 1) *
            ((1 - C (1 / ↑(Nat.factorial 2)) * X ^ 2) *
              ((1 - C (1 / ↑(Nat.factorial 3)) * X ^ 3) * (1 - C (1 / ↑(Nat.factorial 4)) * X ^ 4)))).coeff
        4 *
      ↑(Nat.factorial 4)).floor =
  3
  simp only [mul_sub, sub_mul, one_mul, mul_one, C_mul_X_pow_mul,
    coeff_sub, coeff_one, coeff_C_mul_X_pow]hI:Finset.Icc 1 4 = {1, 2, 3, 4}⊢ ((if 4 = 0 then 1 else 0) * ↑(Nat.factorial 4) - (if 4 = 1 then 1 / ↑(Nat.factorial 1) else 0) * ↑(Nat.factorial 4) -
          ((if 4 = 2 then 1 / ↑(Nat.factorial 2) else 0) * ↑(Nat.factorial 4) -
            (if 4 = 1 + 2 then 1 / ↑(Nat.factorial 1) * (1 / ↑(Nat.factorial 2)) else 0) * ↑(Nat.factorial 4)) -
        ((if 4 = 3 then 1 / ↑(Nat.factorial 3) else 0) * ↑(Nat.factorial 4) -
            (if True then 1 / ↑(Nat.factorial 1) * (1 / ↑(Nat.factorial 3)) else 0) * ↑(Nat.factorial 4) -
          ((if 4 = 2 + 3 then 1 / ↑(Nat.factorial 2) * (1 / ↑(Nat.factorial 3)) else 0) * ↑(Nat.factorial 4) -
            (if 4 = 1 + (2 + 3) then 1 / ↑(Nat.factorial 1) * (1 / ↑(Nat.factorial 2) * (1 / ↑(Nat.factorial 3)))
              else 0) *
              ↑(Nat.factorial 4))) -
      ((if True then 1 / ↑(Nat.factorial 4) else 0) * ↑(Nat.factorial 4) -
            (if 4 = 1 + 4 then 1 / ↑(Nat.factorial 1) * (1 / ↑(Nat.factorial 4)) else 0) * ↑(Nat.factorial 4) -
          ((if 4 = 2 + 4 then 1 / ↑(Nat.factorial 2) * (1 / ↑(Nat.factorial 4)) else 0) * ↑(Nat.factorial 4) -
            (if 4 = 1 + (2 + 4) then 1 / ↑(Nat.factorial 1) * (1 / ↑(Nat.factorial 2) * (1 / ↑(Nat.factorial 4)))
              else 0) *
              ↑(Nat.factorial 4)) -
        ((if 4 = 3 + 4 then 1 / ↑(Nat.factorial 3) * (1 / ↑(Nat.factorial 4)) else 0) * ↑(Nat.factorial 4) -
            (if 4 = 1 + (3 + 4) then 1 / ↑(Nat.factorial 1) * (1 / ↑(Nat.factorial 3) * (1 / ↑(Nat.factorial 4)))
              else 0) *
              ↑(Nat.factorial 4) -
          ((if 4 = 2 + (3 + 4) then 1 / ↑(Nat.factorial 2) * (1 / ↑(Nat.factorial 3) * (1 / ↑(Nat.factorial 4)))
              else 0) *
              ↑(Nat.factorial 4) -
            (if 4 = 1 + (2 + (3 + 4)) then
                1 / ↑(Nat.factorial 1) * (1 / ↑(Nat.factorial 2) * (1 / ↑(Nat.factorial 3) * (1 / ↑(Nat.factorial 4))))
              else 0) *
              ↑(Nat.factorial 4))))).floor =
  3
  decide +nativeAll goals completed! 🐙

The $n$-th coefficient of the square of the ordinary generating function $A(x)^2$.

noncomputable def c (n : ℕ) : ℤ :=
  ∑ k ∈ Finset.range (n + 1), a k * a (n - k)

$a(n)$ differs in sign from $a(n-1)$ if and only if $n$ is a triangular number (checked up to $n = 1225 = (50 \cdot 51)/2$).

Peter Bala, Mar 17 2022

@[category research open, AMS 11]
theorem conjecture1 (n : ℕ) (hn : 0 < n) :
    a n * a (n - 1) < 0 ↔ IsTriangular n := byn:ℕhn:0 < n⊢ a n * a (n - 1) < 0 ↔ IsTriangular n
  sorryAll goals completed! 🐙

The coefficients $c(n)$ of $A(x)^2 = (\sum_{n \ge 0} a(n) x^n)^2$ differ in sign from $c(n-1)$ if and only if $n$ is a triangular number.

Peter Bala, Mar 17 2022

@[category research open, AMS 11]
theorem conjecture2 (n : ℕ) (hn : 0 < n) :
    c n * c (n - 1) < 0 ↔ IsTriangular n := byn:ℕhn:0 < n⊢ c n * c (n - 1) < 0 ↔ IsTriangular n
  sorryAll goals completed! 🐙

The Gauss congruences $a(n \cdot p^k) \equiv a(n \cdot p^{k-1}) \pmod{p^k}$ hold for all primes $p$ and positive integers $n$ and $k$.

Peter Bala, Mar 17 2022

@[category research open, AMS 11]
theorem conjecture3 (p : ℕ) (hp : p.Prime) (n k : ℕ) (hn : 0 < n) (hk : 0 < k) :
    a (n * p ^ k) ≡ a (n * p ^ (k - 1)) [ZMOD (p : ℤ) ^ k] := byp:ℕhp:Nat.Prime pn:ℕk:ℕhn:0 < nhk:0 < k⊢ a (n * p ^ k) ≡ a (n * p ^ (k - 1)) [ZMOD ↑p ^ k]
  sorryAll goals completed! 🐙

end OeisA185895