Paperfolding series for odd zeta values

TheoremHurwitz zeta at \(3/4\) and odd arguments#●ready to prove

For every integer \(k \ge 1\),

\[ \zeta(2k+1, \tfrac34) = -\frac{2^{2k-1} \pi^{2k+1} |E_{2k}|}{(2k)!} + 2^{2k} (2^{2k+1} - 1) \zeta(2k+1), \]

where \(\zeta(s, a)\) is the Hurwitz zeta function (for \(\operatorname{Re} s > 1\), \(\zeta(s, a) = \sum_{n \ge 0} (n + a)^{-s}\)), \(\zeta(s)\) is the Riemann zeta function and \(E_{2k}\) is the even-index Euler number.

The even-index Euler numbers are defined by a binomial recurrence equivalent to the generating function \(1 / \cosh t = \sum_n E_n t^n / n!\). The source writes \(k \in \mathbb{N}\); \(k = 0\) is excluded since \(s = 1\) is the pole of both zeta functions.

public theorem_wanted hurwitzZeta_three_quarters_two_mul_add_one (k : ℕ) (hk : 1 ≤ k) :
    HurwitzZeta.hurwitzZeta ((3 / 4 : ℝ) : UnitAddCircle) (2 * k + 1) =
      -((2 : ℂ) ^ (2 * k - 1) / (2 * k).factorial) * (Real.pi : ℂ) ^ (2 * k + 1) *
          ((|chapter5EulerEven k| : ℤ) : ℂ) +
        2 ^ (2 * k) * (2 ^ (2 * k + 1) - 1) * riemannZeta (2 * k + 1)

Sources

• László Tóth, Evaluating ζ(s) at Odd Positive Integers Using Automatic Dirichlet Series, Journal of Integer Sequences 28 (2025), Article 25.4.4, Lemma 3 (jis-2025-25.4.4-lem-3).

Uses 1

Statement usesEven-index Euler numbers

TheoremTóth's paperfolding and Thue–Morse series for \(\zeta(2k+1)\)#●ready to prove

Let \((f_n)_{n \ge 0}\) be the regular paperfolding sequence and \((t_n)_{n \ge 0}\) the Thue–Morse sequence, both indexed from \(0\), and let \(E_{2k}\) be the even-index Euler number. For every integer \(k \ge 1\),

\[ \zeta(2k+1) - \frac{2^{2k-1} |E_{2k}| \pi^{2k+1}}{(2k)!} = \sum_{n \ge 0} \frac{N_k(n)}{(n+1)^{2k+1}}, \]

where

\[ N_k(n) = (2^{4k+1} - 2^{2k})(2 f_n - 1) + 2^{-(2k+1)} \bigl((2^{2k+1} + 1) t_n + (2^{2k+1} - 1) t_{n+1}\bigr). \]

In the source's indexing \(b_n = f_{n-1}\) for \(n \ge 1\) (so \(b_{2n} = b_n\), \(b_{4n+1} = 0\) and \(b_{4n+3} = 1\)), the right side is \(\sum_{n \ge 1} N(n;k)/n^{2k+1}\) with \(N(n;k) = (2^{4k+1} - 2^{2k})(2b_n - 1) + 2^{-(2k+1)}((2^{2k+1} + 1) t_{n-1} + (2^{2k+1} - 1) t_n)\).

public theorem_wanted riemannZeta_two_mul_add_one_sub_eq_tsum_paperfolding_thueMorse (k : ℕ)
    (hk : 1 ≤ k) :
    riemannZeta (2 * k + 1) - (2 : ℂ) ^ (2 * k - 1) / (2 * k).factorial *
        ((|chapter5EulerEven k| : ℤ) : ℂ) * (Real.pi : ℂ) ^ (2 * k + 1) =
      ((∑' n : ℕ, (((2 : ℝ) ^ (4 * k + 1) - 2 ^ (2 * k)) *
            (2 * ((paperfoldingSeq n).toNat : ℝ) - 1) +
          ((2 : ℝ) ^ (2 * k + 1))⁻¹ * (((2 : ℝ) ^ (2 * k + 1) + 1) * ((thueMorse n).toNat : ℝ) +
            ((2 : ℝ) ^ (2 * k + 1) - 1) * ((thueMorse (n + 1)).toNat : ℝ))) /
          ((n + 1 : ℕ) : ℝ) ^ (2 * k + 1) : ℝ) : ℂ)

Sources

• László Tóth, Evaluating ζ(s) at Odd Positive Integers Using Automatic Dirichlet Series, Journal of Integer Sequences 28 (2025), Article 25.4.4, Theorem 2 (jis-2025-25.4.4-thm-2); bₙ, tₙ read from arXiv:2508.04151v1.

Uses 3

Statement usesEven-index Euler numbers · Regular paperfolding sequence · Thue–Morse sequence

TheoremA paperfolding Dirichlet series at odd integers#●ready to prove

Let \((f_n)_{n \ge 0}\) be the regular paperfolding sequence, indexed from \(0\), and let \(E_{2k}\) be the even-index Euler number. For every integer \(k \ge 1\),

\[ (2^{4k+1} - 2^{2k}) \sum_{n \ge 0} \frac{2 f_n - 1}{(n+1)^{2k+1}} = -\frac{2^{2k-1} \pi^{2k+1} |E_{2k}|}{(2k)!}. \]

In the source's indexing \(b_n = f_{n-1}\) for \(n \ge 1\), the sum is \(\sum_{n \ge 1} (2 b_n - 1)/n^{2k+1}\).

public theorem_wanted tsum_paperfolding_two_mul_add_one (k : ℕ) (hk : 1 ≤ k) :
    ((2 : ℝ) ^ (4 * k + 1) - 2 ^ (2 * k)) *
        ∑' n : ℕ, (2 * ((paperfoldingSeq n).toNat : ℝ) - 1) / ((n + 1 : ℕ) : ℝ) ^ (2 * k + 1) =
      -((2 : ℝ) ^ (2 * k - 1) / (2 * k).factorial) * Real.pi ^ (2 * k + 1) *
        ((|chapter5EulerEven k| : ℤ) : ℝ)

Sources

• László Tóth, Evaluating ζ(s) at Odd Positive Integers Using Automatic Dirichlet Series, Journal of Integer Sequences 28 (2025), Article 25.4.4, Lemma 5 (jis-2025-25.4.4-lem-5); bₙ read from arXiv:2508.04151v1.

Uses 2

Statement usesEven-index Euler numbers · Regular paperfolding sequence

TheoremThe \(\pm 1\) paperfolding Dirichlet series at odd integers#●ready to prove

Let \((f_n)_{n \ge 0}\) be the regular paperfolding sequence, indexed from \(0\), and let \(E_{2k}\) be the even-index Euler number. For every integer \(k \ge 1\),

\[ \sum_{n \ge 0} \frac{(-1)^{f_n}}{(n+1)^{2k+1}} = \frac{\pi^{2k+1} |E_{2k}|}{(2^{2k+2} - 2)(2k)!}. \]

In the source's indexing \(b_n = f_{n-1}\) for \(n \ge 1\), this is the sum over \(n \ge 1\) of \((-1)^{b_n}/n^{2k+1}\). The source prints the sum as \(\sum_{n \ge 0} (-1)^{b_n}/(n+1)^{2k+1}\), which uses \(b_0\), left undetermined by the recurrence \(b_{2n} = b_n\); its proof (from Lemma 5) gives the sum over \(n \ge 1\) stated here.

public theorem_wanted tsum_neg_one_pow_paperfolding (k : ℕ) (hk : 1 ≤ k) :
    ∑' n : ℕ, (-1 : ℝ) ^ (paperfoldingSeq n).toNat / ((n + 1 : ℕ) : ℝ) ^ (2 * k + 1) =
      Real.pi ^ (2 * k + 1) * ((|chapter5EulerEven k| : ℤ) : ℝ) /
        (((2 : ℝ) ^ (2 * k + 2) - 2) * (2 * k).factorial)

Sources

• László Tóth, Evaluating ζ(s) at Odd Positive Integers Using Automatic Dirichlet Series, Journal of Integer Sequences 28 (2025), Article 25.4.4, Corollary 7 (jis-2025-25.4.4-cor-7); bₙ read from arXiv:2508.04151v1.

Uses 2

Statement usesEven-index Euler numbers · Regular paperfolding sequence

4 ready to prove

The dependency explorer needs JavaScript and its graph data.

Download the complete graph data.