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module
public import FormalConjecturesUtilExpansion of $(1 - x)/(1 - 2 x + 3 x^2)$
This sequence is the expansion of $(1 - x)/(1 - 2 x + 3 x^2)$ in powers of $x$. It satisfies the linear recurrence relation $a(n) = 2 a(n-1) - 3 a(n-2)$ for $n \ge 2$, with initial values $a(0)=1$ and $a(1)=1$.
References:
@[expose] public sectionnamespace OeisA87455
The primary defining sequence a, satisfying $a(n) = 2 a(n-1) - 3 a(n-2)$ for $n \ge 2$,
with initial values $a(0)=1$ and $a(1)=1$.
def a : ℕ → ℤ
| 0 => 1
| 1 => 1
| n + 2 => 2 * a (n + 1) - 3 * a n
The decimal significand of an integer x: the unique real number $S(x) \in [1, 10)$ such that
$|x| = S(x) \cdot 10^k$ for some integer $k$, with the convention $S(0) = 0$.
noncomputable def significand (x : ℤ) : ℝ :=
(x.natAbs : ℝ) / 10 ^ ((Nat.digits 10 x.natAbs).length - 1)A sequence of integers satisfies Benford's law if, for every $t \in [1, 10)$, the asymptotic relative frequency of terms whose decimal significand is at most $t$ equals $\log_{10} t$ (Berger-Hill). This is strictly stronger than the law for the leading digit alone.
def SatisfiesBenford (s : ℕ → ℤ) : Prop :=
∀ t ∈ Set.Ico (1 : ℝ) 10,
Filter.Tendsto
(fun N : ℕ =>
((Finset.range N).filter (fun n => significand (s n) ≤ t)).card / (N : ℝ))
Filter.atTop
(nhds (Real.logb 10 t))
Value of the sequence a at 0.
@[category test, AMS 11]
theorem a_0 : a 0 = 1 := ⊢ a 0 = 1 All goals completed! 🐙
Value of the sequence a at 1.
@[category test, AMS 11]
theorem a_1 : a 1 = 1 := ⊢ a 1 = 1 All goals completed! 🐙
Value of the sequence a at 2.
@[category test, AMS 11]
theorem a_2 : a 2 = -1 := ⊢ a 2 = -1 All goals completed! 🐙
Value of the sequence a at 3.
@[category test, AMS 11]
theorem a_3 : a 3 = -5 := ⊢ a 3 = -5 All goals completed! 🐙
Value of the sequence a at 4.
@[category test, AMS 11]
theorem a_4 : a 4 = -7 := ⊢ a 4 = -7 All goals completed! 🐙It is an open question whether or not this sequence satisfies Benford's law [Berger-Hill, 2017; Arno Berger, email, Jan 06 2017]. - N. J. A. Sloane, Feb 08 2017
@[category research open, AMS 11 60]
theorem conjecture : answer(sorry) ↔ SatisfiesBenford a := ⊢ True ↔ SatisfiesBenford a
All goals completed! 🐙end OeisA87455