Sierpiński and Riesel numbers

TheoremComposite numbers from Mersenne shifts of an arithmetic progression#●ready to prove

For a natural number \(j\) let \(M_j = 2^j - 1\) be the \(j\)-th Mersenne number (so \(M_0 = 0\)). For every natural number \(t\) there are natural numbers \(a\) and \(d > 0\) such that, for every natural number \(r\), every natural number \(j \le t\) and every natural number \(n\), the number \((a + rd + M_j) \cdot 2^n + 1\) is composite.

theorem_wanted exists_arithProg_sierpinski_add_mersenne
    (t : ℕ) :
    ∃ a d : ℕ, 0 < d ∧ ∀ r : ℕ, ∀ j ≤ t, ∀ n : ℕ,
      Nat.IsComposite ((a + r * d + mersenne j) * 2 ^ n + 1)

Sources

• Carrie E. Finch-Smith and R. Scottfield Groth, "Arbitrarily Long Sequences of Sierpiński Numbers that are the Sum of a Sierpiński Number and a Mersenne Number", Theorem 1.

Uses 1

Statement usesComposite numbers

DefinitionSierpiński number#✓fully proved

A natural number \(k\) is a Sierpiński number if \(k\) is odd and, for every integer \(n \ge 1\), the number \(k \cdot 2^n + 1\) is not prime.

public def IsSierpinskiNumber (k : ℕ) : Prop :=
  Odd k ∧ ∀ n : ℕ, 1 ≤ n → ¬ (k * 2 ^ n + 1).Prime

Sources

• Bispels, Cohen, Harrington, Lowrance, Pontes, Schaumann and Wong, On Sierpiński and Riesel Repdigits and Repintegers, INTEGERS 26 (2026), Article A12, abstract and Section 1.

Used by 9

Used byInfinitely many Narayana numbers are Sierpiński numbers · Infinitely many Sierpiński repdigits in every base \(b > 78557\) · Covering-produced Sierpiński numbers give infinitely many Sierpiński repintegers · Infinitely many Sierpiński repunits in bases \(b \equiv 1, 147, 265, 378 \pmod{641}\) with \(b \not\equiv 1 \pmod 5\) · Infinitely many Sierpiński repunits from primitive divisors of \(2^{2^j} - 1\) · Binary repintegers of 18107 that are Sierpiński numbers · Repdigits with digit 41 that are Sierpiński numbers · Binary repstrings of 659 that are Sierpiński numbers · \(1\) is not a Sierpiński number

DefinitionRiesel number#✓fully proved

A natural number \(k\) is a Riesel number if \(k\) is odd and, for every integer \(n \ge 1\), the number \(k \cdot 2^n - 1\) is not prime.

public def IsRieselNumber (k : ℕ) : Prop :=
  Odd k ∧ ∀ n : ℕ, 1 ≤ n → ¬ (k * 2 ^ n - 1).Prime

Sources

• Bispels, Cohen, Harrington, Lowrance, Pontes, Schaumann and Wong, On Sierpiński and Riesel Repdigits and Repintegers, INTEGERS 26 (2026), Article A12, abstract and Section 1.

Used by 9

Used byInfinitely many Riesel repdigits in every base \(b > 509203\) · Covering-produced Riesel numbers give infinitely many Riesel repintegers · Infinitely many Riesel repunits in bases \(b \equiv 16518444216571 \pmod{2^{64} - 1}\) · Infinitely many Riesel repunits from primitive divisors of \(2^{2^j} - 1\) · Narayana numbers that are Riesel numbers · Riesel repdigits with digit \(101\) in bases \(b \equiv 180 \pmod{11184810}\) · Binary repintegers of 18107 that are Riesel numbers · Binary repstrings of 659 that are Riesel numbers · One is not a Riesel number

Theorem\(1\) is not a Sierpiński number#✓fully proved

The number \(1\) is not a Sierpiński number, since \(1 \cdot 2 + 1 = 3\) is prime.

public theorem not_isSierpinskiNumber_one : ¬ IsSierpinskiNumber 1

Sources

• Bispels, Cohen, Harrington, Lowrance, Pontes, Schaumann and Wong, On Sierpiński and Riesel Repdigits and Repintegers, INTEGERS 26 (2026), Article A12, Section 1.

Uses 1

Statement usesSierpiński number

TheoremOne is not a Riesel number#✓fully proved

The number \(1\) is not a Riesel number, since \(1 \cdot 2^2 - 1 = 3\) is prime.

public theorem not_isRieselNumber_one : ¬ IsRieselNumber 1

Sources

• Bispels, Cohen, Harrington, Lowrance, Pontes, Schaumann and Wong, On Sierpiński and Riesel Repdigits and Repintegers, INTEGERS 26 (2026), Article A12, Section 1.

Uses 1

Statement usesRiesel number

DefinitionRepstrings#✓fully proved

For natural numbers \(b\), \(k\), \(z\) and \(t\), let \(\ell\) be the number of base-\(b\) digits of \(k\). The \(b\)-repstring is

\[k_b^{(z;t)} = \left\lfloor \frac{k\,(b^{(z+\ell)t} - 1)}{b^{z+\ell} - 1} \right\rfloor,\]

where subtraction is truncated at \(0\) and the integer division gives \(0\) when the denominator is \(0\). For \(z = 0\) it is the \(b\)-repinteger \(k_b^{(t)}\).

For \(b \ge 2\) and \(k \ge 1\) the division is exact, and \(k_b^{(z;t)}\) is the number whose base-\(b\) expansion is \(k\) written \(t\) times with \(z\) zeros between consecutive copies.

public def repstring (b k z t : ℕ) : ℕ :=
  k * (b ^ ((z + (Nat.digits b k).length) * t) - 1) / (b ^ (z + (Nat.digits b k).length) - 1)

Sources

• Bispels, Cohen, Harrington, Lowrance, Pontes, Schaumann and Wong, On Sierpiński and Riesel Repdigits and Repintegers, INTEGERS 26 (2026), Article A12, Sections 1 and 4.

Used by 17

Used byInfinitely many Riesel repdigits in every base \(b > 509203\) · Covering-produced Riesel numbers give infinitely many Riesel repintegers · Infinitely many Riesel repunits in bases \(b \equiv 16518444216571 \pmod{2^{64} - 1}\) · Infinitely many Riesel repunits from primitive divisors of \(2^{2^j} - 1\) · Infinitely many Sierpiński repdigits in every base \(b > 78557\) · Covering-produced Sierpiński numbers give infinitely many Sierpiński repintegers · Infinitely many Sierpiński repunits in bases \(b \equiv 1, 147, 265, 378 \pmod{641}\) with \(b \not\equiv 1 \pmod 5\) · Infinitely many Sierpiński repunits from primitive divisors of \(2^{2^j} - 1\) · Riesel repdigits with digit \(101\) in bases \(b \equiv 180 \pmod{11184810}\) · Binary repintegers of 18107 that are Riesel numbers · Binary repstrings of 659 that are Riesel numbers · Binary repintegers of 18107 that are Sierpiński numbers · Repdigits with digit 41 that are Sierpiński numbers · Binary repstrings of 659 that are Sierpiński numbers · A \(10\)-repstring · The three-digit \(2\)-repunit · Theorem 4.6 (Sierpiński and Riesel repstring coverings correspond)

TheoremA \(10\)-repstring#✓fully proved

The \(10\)-repstring with \(k = 1\), \(z = 2\) and \(t = 3\) is \(1_{10}^{(2;3)} = 1001001\).

public theorem repstring_ten_one_two_three : repstring 10 1 2 3 = 1001001

Sources

• Bispels, Cohen, Harrington, Lowrance, Pontes, Schaumann and Wong, On Sierpiński and Riesel Repdigits and Repintegers, INTEGERS 26 (2026), Article A12, Section 4.

Uses 1

Statement usesRepstrings

TheoremThe three-digit \(2\)-repunit#✓fully proved

The \(2\)-repstring with \(k = 1\), \(z = 0\) and \(t = 3\), that is, the \(2\)-repunit \(1_2^{(3)}\) with three digits, equals \(7\).

public theorem repstring_two_one_zero_three : repstring 2 1 0 3 = 7

Sources

• Bispels, Cohen, Harrington, Lowrance, Pontes, Schaumann and Wong, On Sierpiński and Riesel Repdigits and Repintegers, INTEGERS 26 (2026), Article A12, Section 1.

Uses 1

Statement usesRepstrings

DefinitionSierpiński covering data#✓fully proved

Let \(k\) be a natural number and \(S\) a finite set of triples \((r, m, p)\) of natural numbers. Then \(S\) is Sierpiński covering data for \(k\) if every integer is congruent to \(r\) modulo \(m\) for some \((r, m, p) \in S\), distinct triples in \(S\) have distinct third entries, and every \((r, m, p) \in S\) has \(m \ge 1\), \(p\) prime, \(p \mid 2^m - 1\) and \(p \mid k \cdot 2^r + 1\).

This is the input of the covering system method for Sierpiński numbers; the last condition says that \(k \equiv -2^{-r} \pmod{p}\).

public def IsSierpinskiCoveringData (k : ℕ) (S : Finset (ℕ × ℕ × ℕ)) : Prop :=
  Int.IsCoveringSystem (S.image fun s => ((s.1 : ℤ), s.2.1)) ∧
    Set.InjOn (fun s : ℕ × ℕ × ℕ => s.2.2) (S : Set (ℕ × ℕ × ℕ)) ∧
    ∀ s ∈ S, 1 ≤ s.2.1 ∧ s.2.2.Prime ∧ s.2.2 ∣ 2 ^ s.2.1 - 1 ∧ s.2.2 ∣ k * 2 ^ s.1 + 1

Sources

• Bispels, Cohen, Harrington, Lowrance, Pontes, Schaumann and Wong, On Sierpiński and Riesel Repdigits and Repintegers, INTEGERS 26 (2026), Article A12, Section 2, Equation (2).

Used by 4

Used byCovering-produced Sierpiński numbers give infinitely many Sierpiński repintegers · Empty Sierpiński covering data · Theorem 3.2 (no Sierpiński \(2\)-repdigits from coverings) · Theorem 4.6 (Sierpiński and Riesel repstring coverings correspond)

DefinitionRiesel covering data#✓fully proved

Let \(k\) be a natural number and let \(S\) be a finite set of triples \((r, m, p)\) of natural numbers. Then \(S\) is Riesel covering data for \(k\) (data producing \(k\) as a Riesel number by the covering system method) when every integer is congruent to \(r\) modulo \(m\) for some \((r, m, p) \in S\), distinct triples in \(S\) have distinct third entries \(p\), and every \((r, m, p) \in S\) has \(m \ge 1\), \(p\) prime, \(p \mid 2^m - 1\) and \(p \mid k \cdot 2^r - 1\) (for \(k \ge 1\) the last condition says \(k \equiv 2^{-r} \pmod p\)).

Subtraction is in the natural numbers, so for \(k = 0\) the condition \(p \mid k \cdot 2^r - 1\) reads \(p \mid 0\) and holds trivially.

public def IsRieselCoveringData (k : ℕ) (S : Finset (ℕ × ℕ × ℕ)) : Prop :=
  Int.IsCoveringSystem (S.image fun s => ((s.1 : ℤ), s.2.1)) ∧
    Set.InjOn (fun s : ℕ × ℕ × ℕ => s.2.2) (S : Set (ℕ × ℕ × ℕ)) ∧
    ∀ s ∈ S, 1 ≤ s.2.1 ∧ s.2.2.Prime ∧ s.2.2 ∣ 2 ^ s.2.1 - 1 ∧ s.2.2 ∣ k * 2 ^ s.1 - 1

Sources

• Bispels, Cohen, Harrington, Lowrance, Pontes, Schaumann and Wong, On Sierpiński and Riesel Repdigits and Repintegers, INTEGERS 26 (2026), Article A12, Section 2, Equation (3).

Used by 3

Used byCovering-produced Riesel numbers give infinitely many Riesel repintegers · The empty set is not Riesel covering data · Theorem 4.6 (Sierpiński and Riesel repstring coverings correspond)

TheoremEmpty Sierpiński covering data#✓fully proved

For every natural number \(k\), the empty set is not Sierpiński covering data for \(k\): the empty family of residue classes does not cover \(\mathbb{Z}\).

public theorem not_isSierpinskiCoveringData_empty (k : ℕ) : ¬ IsSierpinskiCoveringData k ∅

Sources

• Bispels, Cohen, Harrington, Lowrance, Pontes, Schaumann and Wong, On Sierpiński and Riesel Repdigits and Repintegers, INTEGERS 26 (2026), Article A12, Section 2.

Uses 1

Statement usesSierpiński covering data

TheoremThe empty set is not Riesel covering data#✓fully proved

For every natural number \(k\), the empty set is not Riesel covering data for \(k\), because its residue classes do not cover \(\mathbb{Z}\).

public theorem not_isRieselCoveringData_empty (k : ℕ) : ¬ IsRieselCoveringData k ∅

Sources

• Bispels, Cohen, Harrington, Lowrance, Pontes, Schaumann and Wong, On Sierpiński and Riesel Repdigits and Repintegers, INTEGERS 26 (2026), Article A12, Section 2.

Uses 1

Statement usesRiesel covering data

TheoremTheorem 3.2 (no Sierpiński \(2\)-repdigits from coverings)#●ready to prove

For every natural number \(t \ge 1\), the covering system method cannot produce the \(2\)-repdigit \(2^t - 1\) as a Sierpiński number: no finite set of triples \((r_j, m_j, p_j)\) is Sierpiński covering data for \(2^t - 1\).

public theorem_wanted not_isSierpinskiCoveringData_two_pow_sub_one {t : ℕ} (ht : 1 ≤ t)
    (S : Finset (ℕ × ℕ × ℕ)) : ¬ IsSierpinskiCoveringData (2 ^ t - 1) S

Sources

• Bispels, Cohen, Harrington, Lowrance, Pontes, Schaumann and Wong, On Sierpiński and Riesel Repdigits and Repintegers, INTEGERS 26 (2026), Article A12, Theorem 3.2.

Uses 1

Statement usesSierpiński covering data

TheoremCovering-produced Sierpiński numbers give infinitely many Sierpiński repintegers#●ready to prove

Let \(k\) be a Sierpiński number that has Sierpiński covering data, so that \(k\) is produced as a Sierpiński number by the covering system method. Then for every natural number \(b \ge 2\) there are infinitely many \(t \ge 1\) such that the \(b\)-repinteger \(k_b^{(t)} = k\,(b^{\ell t} - 1)/(b^{\ell} - 1)\), where \(\ell\) is the number of base-\(b\) digits of \(k\) (so \(k_b^{(t)}\) is \(k\) written \(t\) times in base \(b\)), is a Sierpiński number.

public theorem_wanted infinite_sierpinski_repintegers {k : ℕ} (hk : IsSierpinskiNumber k)
    {S : Finset (ℕ × ℕ × ℕ)} (hS : IsSierpinskiCoveringData k S) {b : ℕ} (hb : 2 ≤ b) :
    {t : ℕ | 1 ≤ t ∧ IsSierpinskiNumber (repstring b k 0 t)}.Infinite

Sources

• Bispels, Cohen, Harrington, Lowrance, Pontes, Schaumann and Wong, On Sierpiński and Riesel Repdigits and Repintegers, INTEGERS 26 (2026), Article A12, Theorem 3.3.

Uses 3

Statement usesSierpiński covering data · Sierpiński number · Repstrings

TheoremCovering-produced Riesel numbers give infinitely many Riesel repintegers#●ready to prove

Let \(k\) be a Riesel number that has Riesel covering data, so that \(k\) is produced as a Riesel number by the covering system method. Then for every natural number \(b \ge 2\) there are infinitely many \(t \ge 1\) such that the \(b\)-repinteger \(k_b^{(t)} = k\,(b^{\ell t} - 1)/(b^{\ell} - 1)\), where \(\ell\) is the number of base-\(b\) digits of \(k\) (so \(k_b^{(t)}\) is \(k\) written \(t\) times in base \(b\)), is a Riesel number.

public theorem_wanted infinite_riesel_repintegers {k : ℕ} (hk : IsRieselNumber k)
    {S : Finset (ℕ × ℕ × ℕ)} (hS : IsRieselCoveringData k S) {b : ℕ} (hb : 2 ≤ b) :
    {t : ℕ | 1 ≤ t ∧ IsRieselNumber (repstring b k 0 t)}.Infinite

Sources

• Bispels, Cohen, Harrington, Lowrance, Pontes, Schaumann and Wong, On Sierpiński and Riesel Repdigits and Repintegers, INTEGERS 26 (2026), Article A12, Theorem 3.4.

Uses 3

Statement usesRiesel covering data · Riesel number · Repstrings

TheoremBinary repintegers of 18107 that are Sierpiński numbers#●ready to prove

For every natural number \(t \equiv 25 \pmod{56}\), the binary repinteger \(18107_2^{(t)} = 18107\,(2^{15t} - 1)/(2^{15} - 1)\), whose binary expansion is that of \(18107\) (fifteen binary digits) written \(t\) times, is a Sierpiński number.

public theorem_wanted isSierpinskiNumber_repstring_18107 {t : ℕ} (ht : t ≡ 25 [MOD 56]) :
    IsSierpinskiNumber (repstring 2 18107 0 t)

Sources

• Bispels, Cohen, Harrington, Lowrance, Pontes, Schaumann and Wong, On Sierpiński and Riesel Repdigits and Repintegers, INTEGERS 26 (2026), Article A12, Theorem 3.5.

Uses 2

Statement usesSierpiński number · Repstrings

TheoremBinary repintegers of 18107 that are Riesel numbers#●ready to prove

For every natural number \(t \equiv 31 \pmod{56}\), the binary repinteger \(18107_2^{(t)} = 18107\,(2^{15t} - 1)/(2^{15} - 1)\), whose binary expansion is that of \(18107\) (fifteen binary digits) written \(t\) times, is a Riesel number.

public theorem_wanted isRieselNumber_repstring_18107 {t : ℕ} (ht : t ≡ 31 [MOD 56]) :
    IsRieselNumber (repstring 2 18107 0 t)

Sources

• Bispels, Cohen, Harrington, Lowrance, Pontes, Schaumann and Wong, On Sierpiński and Riesel Repdigits and Repintegers, INTEGERS 26 (2026), Article A12, Theorem 3.6.

Uses 2

Statement usesRiesel number · Repstrings

TheoremInfinitely many Sierpiński repdigits in every base \(b > 78557\)#●ready to prove

Let \(b > 78557\) be a natural number. Then there are infinitely many Sierpiński numbers of the form \(k(b^t - 1)/(b - 1)\) with \(1 \le k < b\) and \(t \ge 1\), that is, \(b\)-repdigits: the digit \(k\) written \(t\) times in base \(b\).

public theorem_wanted infinite_sierpinski_repdigits {b : ℕ} (hb : 78557 < b) :
    {N : ℕ | ∃ k t : ℕ, 1 ≤ k ∧ k < b ∧ 1 ≤ t ∧ N = repstring b k 0 t ∧
      IsSierpinskiNumber N}.Infinite

Sources

• Bispels, Cohen, Harrington, Lowrance, Pontes, Schaumann and Wong, On Sierpiński and Riesel Repdigits and Repintegers, INTEGERS 26 (2026), Article A12, Corollary 3.7.

Uses 2

Statement usesSierpiński number · Repstrings

TheoremInfinitely many Riesel repdigits in every base \(b > 509203\)#●ready to prove

Let \(b > 509203\) be a natural number. Then there are infinitely many Riesel numbers of the form \(k(b^t - 1)/(b - 1)\) with \(1 \le k < b\) and \(t \ge 1\), that is, \(b\)-repdigits: the digit \(k\) written \(t\) times in base \(b\).

public theorem_wanted infinite_riesel_repdigits {b : ℕ} (hb : 509203 < b) :
    {N : ℕ | ∃ k t : ℕ, 1 ≤ k ∧ k < b ∧ 1 ≤ t ∧ N = repstring b k 0 t ∧
      IsRieselNumber N}.Infinite

Sources

• Bispels, Cohen, Harrington, Lowrance, Pontes, Schaumann and Wong, On Sierpiński and Riesel Repdigits and Repintegers, INTEGERS 26 (2026), Article A12, Corollary 3.8.

Uses 2

Statement usesRiesel number · Repstrings

TheoremRepdigits with digit 41 that are Sierpiński numbers#●ready to prove

Let \(b\) and \(t\) be natural numbers with \(b \equiv 87 \pmod{11184810}\) and \(t \equiv 59 \pmod{120}\). Then the base-\(b\) repdigit \(41_b^{(t)} = 41\,(b^t - 1)/(b - 1)\) is a Sierpiński number. Since \(b \ge 87 > 41\), its \(t\) base-\(b\) digits all equal \(41\).

public theorem_wanted isSierpinskiNumber_repstring_41 {b t : ℕ} (hb : b ≡ 87 [MOD 11184810])
    (ht : t ≡ 59 [MOD 120]) : IsSierpinskiNumber (repstring b 41 0 t)

Sources

• Bispels, Cohen, Harrington, Lowrance, Pontes, Schaumann and Wong, On Sierpiński and Riesel Repdigits and Repintegers, INTEGERS 26 (2026), Article A12, Theorem 3.9.

Uses 2

Statement usesSierpiński number · Repstrings

TheoremRiesel repdigits with digit \(101\) in bases \(b \equiv 180 \pmod{11184810}\)#●ready to prove

Let \(b\) and \(t\) be natural numbers with \(b \equiv 180 \pmod{11184810}\) and \(t \equiv 171 \pmod{240}\). Then the \(b\)-repinteger \(101_b^{(t)}\), which is the repdigit \(101\,(b^t - 1)/(b - 1)\) because \(101 < b\), is a Riesel number.

public theorem_wanted isRieselNumber_repstring_101 {b t : ℕ} (hb : b ≡ 180 [MOD 11184810])
    (ht : t ≡ 171 [MOD 240]) : IsRieselNumber (repstring b 101 0 t)

Sources

• Bispels, Cohen, Harrington, Lowrance, Pontes, Schaumann and Wong, On Sierpiński and Riesel Repdigits and Repintegers, INTEGERS 26 (2026), Article A12, Theorem 3.10.

Uses 2

Statement usesRiesel number · Repstrings

TheoremInfinitely many Sierpiński repunits from primitive divisors of \(2^{2^j} - 1\)#●ready to prove

Let \(\tau\) be a natural number. For \(1 \le j \le \tau\) let \(p_j\) be a primitive prime divisor of \(2^{2^j} - 1\) in the sequence \((2^m - 1)_{m \ge 0}\), and let \(q \ne p_\tau\) be a primitive prime divisor of \(2^{2^\tau} - 1\) in the same sequence. Let \(b > 2\) be a natural number with \(q \nmid b\). For \(1 \le j \le \tau\) put \(\ell_j = 1\) if \(p_j \mid b\), \(\ell_j = p_j\) if \(p_j \nmid b\) and \(b \equiv 1 \pmod{p_j}\), and otherwise \(\ell_j = \operatorname{ord}_{p_j}(b)\), the multiplicative order of \(b\) modulo \(p_j\); let \(L = \operatorname{lcm}(2, \ell_1, \dots, \ell_\tau)\). Suppose that either (i) \(b \equiv 1 \pmod q\) and \(q \nmid L\), or (ii) \(b \not\equiv 1 \pmod q\) and there is a natural number \(s \equiv 1 \pmod L\) with \((b^s - 1)/(b - 1) \equiv -1 \pmod q\). Then there are infinitely many Sierpiński numbers among the \(b\)-repunits \((b^t - 1)/(b - 1)\) with \(t \ge 1\).

public theorem_wanted infinite_sierpinski_repunits_of_primitive_divisors (τ : ℕ) (p : ℕ → ℕ)
    (hp : ∀ j ∈ Finset.Icc 1 τ,
      IntegerSequence.IsPrimitivePrimeDivisor (fun m => (2 : ℤ) ^ m - 1) (2 ^ j) (p j))
    (q : ℕ) (hq : IntegerSequence.IsPrimitivePrimeDivisor (fun m => (2 : ℤ) ^ m - 1) (2 ^ τ) q)
    (hqp : q ≠ p τ) (b : ℕ) (hb : 2 < b) (hbq : ¬ q ∣ b)
    (hcase :
      let ℓ : ℕ → ℕ := fun j =>
        if p j ∣ b then 1 else if b ≡ 1 [MOD p j] then p j else orderOf (b : ZMod (p j))
      let L : ℕ := Nat.lcm 2 ((Finset.Icc 1 τ).lcm ℓ)
      (b ≡ 1 [MOD q] ∧ ¬ q ∣ L) ∨
        (¬ b ≡ 1 [MOD q] ∧ ∃ t : ℕ, (((b ^ t - 1) / (b - 1) : ℕ) : ℤ) ≡ -1 [ZMOD q] ∧
          t ≡ 1 [MOD L])) :
    {N : ℕ | ∃ t : ℕ, 1 ≤ t ∧ N = repstring b 1 0 t ∧ IsSierpinskiNumber N}.Infinite

Sources

• Bispels, Cohen, Harrington, Lowrance, Pontes, Schaumann and Wong, On Sierpiński and Riesel Repdigits and Repintegers, INTEGERS 26 (2026), Article A12, Theorem 3.11.

Uses 3

Statement usesPrimitive prime divisor of an integer sequence · Sierpiński number · Repstrings

TheoremInfinitely many Sierpiński repunits in bases \(b \equiv 1, 147, 265, 378 \pmod{641}\) with \(b \not\equiv 1 \pmod 5\)#●ready to prove

Infinitely many Sierpiński repunits in bases \(b \equiv 1, 147, 265, 378 \pmod{641}\) with \(b \not\equiv 1 \pmod 5\). Let \(b > 2\) be a natural number with \(b \equiv c \pmod{641}\) for some \(c \in \{1, 147, 265, 378\}\) and \(b \not\equiv 1 \pmod 5\). Then there are infinitely many Sierpiński numbers among the \(b\)-repunits \((b^t - 1)/(b - 1)\) with \(t \ge 1\).

public theorem_wanted infinite_sierpinski_repunits_mod_641 {b : ℕ} (hb : 2 < b)
    (h641 : ∃ c ∈ ({1, 147, 265, 378} : Finset ℕ), b ≡ c [MOD 641]) (h5 : ¬ b ≡ 1 [MOD 5]) :
    {N : ℕ | ∃ t : ℕ, 1 ≤ t ∧ N = repstring b 1 0 t ∧ IsSierpinskiNumber N}.Infinite

Sources

• Bispels, Cohen, Harrington, Lowrance, Pontes, Schaumann and Wong, On Sierpiński and Riesel Repdigits and Repintegers, INTEGERS 26 (2026), Article A12, Corollary 3.12.

Uses 2

Statement usesSierpiński number · Repstrings

TheoremInfinitely many Riesel repunits from primitive divisors of \(2^{2^j} - 1\)#●ready to prove

Let \(\tau\) be a natural number. For \(1 \le j \le \tau\) let \(p_j\) be a primitive prime divisor of \(2^{2^j} - 1\) in the sequence \((2^m - 1)_{m \ge 0}\), let \(q \ne p_\tau\) be a primitive prime divisor of \(2^{2^\tau} - 1\) in the same sequence, and put \(P = p_1 \cdots p_\tau\). Let \(b > 2\) be a natural number with \(b \equiv 1 \pmod P\), and suppose that (i) \(q \mid b\), or (ii) \(b \equiv 1 \pmod q\), or (iii) \(q \nmid b\), \(b \not\equiv 1 \pmod q\) and \(\gcd(P, \operatorname{ord}_q(b)) = 1\), where \(\operatorname{ord}_q(b)\) is the multiplicative order of \(b\) modulo \(q\). Then there are infinitely many Riesel numbers among the \(b\)-repunits \((b^t - 1)/(b - 1)\) with \(t \ge 1\).

public theorem_wanted infinite_riesel_repunits_of_primitive_divisors (τ : ℕ) (p : ℕ → ℕ)
    (hp : ∀ j ∈ Finset.Icc 1 τ,
      IntegerSequence.IsPrimitivePrimeDivisor (fun m => (2 : ℤ) ^ m - 1) (2 ^ j) (p j))
    (q : ℕ) (hq : IntegerSequence.IsPrimitivePrimeDivisor (fun m => (2 : ℤ) ^ m - 1) (2 ^ τ) q)
    (hqp : q ≠ p τ) (b : ℕ) (hb : 2 < b) (hbP : b ≡ 1 [MOD ∏ j ∈ Finset.Icc 1 τ, p j])
    (hcase : q ∣ b ∨ b ≡ 1 [MOD q] ∨
      (¬ q ∣ b ∧ ¬ b ≡ 1 [MOD q] ∧
        Nat.Coprime (∏ j ∈ Finset.Icc 1 τ, p j) (orderOf (b : ZMod q)))) :
    {N : ℕ | ∃ t : ℕ, 1 ≤ t ∧ N = repstring b 1 0 t ∧ IsRieselNumber N}.Infinite

Sources

• Bispels, Cohen, Harrington, Lowrance, Pontes, Schaumann and Wong, On Sierpiński and Riesel Repdigits and Repintegers, INTEGERS 26 (2026), Article A12, Theorem 3.13.

Uses 3

Statement usesPrimitive prime divisor of an integer sequence · Riesel number · Repstrings

TheoremInfinitely many Riesel repunits in bases \(b \equiv 16518444216571 \pmod{2^{64} - 1}\)#●ready to prove

Let \(b\) be a natural number with \(b \equiv 16518444216571 \pmod{18446744073709551615}\), where \(18446744073709551615 = 2^{64} - 1\). Then there are infinitely many Riesel numbers among the \(b\)-repunits \((b^t - 1)/(b - 1)\) with \(t \ge 1\).

public theorem_wanted infinite_riesel_repunits_mod_two_pow_64_sub_one {b : ℕ}
    (hb : b ≡ 16518444216571 [MOD 18446744073709551615]) :
    {N : ℕ | ∃ t : ℕ, 1 ≤ t ∧ N = repstring b 1 0 t ∧ IsRieselNumber N}.Infinite

Sources

• Bispels, Cohen, Harrington, Lowrance, Pontes, Schaumann and Wong, On Sierpiński and Riesel Repdigits and Repintegers, INTEGERS 26 (2026), Article A12, Corollary 3.14.

Uses 2

Statement usesRiesel number · Repstrings

TheoremBinary repstrings of 659 that are Sierpiński numbers#●ready to prove

For every natural number \(t\) with \(t \equiv 131 \pmod{2730}\) or \(t \equiv 1361 \pmod{2730}\), the binary repstring \(659_2^{(2;t)} = 659\,(2^{12t} - 1)/(2^{12} - 1)\) is a Sierpiński number. Its binary expansion is that of \(659\) (ten binary digits) written \(t\) times with two zeros between consecutive copies.

public theorem_wanted isSierpinskiNumber_repstring_659 {t : ℕ}
    (ht : t ≡ 131 [MOD 2730] ∨ t ≡ 1361 [MOD 2730]) : IsSierpinskiNumber (repstring 2 659 2 t)

Sources

• Bispels, Cohen, Harrington, Lowrance, Pontes, Schaumann and Wong, On Sierpiński and Riesel Repdigits and Repintegers, INTEGERS 26 (2026), Article A12, Theorem 4.3.

Uses 2

Statement usesSierpiński number · Repstrings

TheoremBinary repstrings of 659 that are Riesel numbers#●ready to prove

For every natural number \(t\) with \(t \equiv 1369 \pmod{2730}\) or \(t \equiv 2599 \pmod{2730}\), the binary repstring \(659_2^{(2;t)} = 659\,(2^{12t} - 1)/(2^{12} - 1)\) is a Riesel number. Its binary expansion is that of \(659\) (ten binary digits) written \(t\) times with two zeros between consecutive copies.

public theorem_wanted isRieselNumber_repstring_659 {t : ℕ}
    (ht : t ≡ 1369 [MOD 2730] ∨ t ≡ 2599 [MOD 2730]) : IsRieselNumber (repstring 2 659 2 t)

Sources

• Bispels, Cohen, Harrington, Lowrance, Pontes, Schaumann and Wong, On Sierpiński and Riesel Repdigits and Repintegers, INTEGERS 26 (2026), Article A12, Theorem 4.4.

Uses 2

Statement usesRiesel number · Repstrings

TheoremTheorem 4.6 (Sierpiński and Riesel repstring coverings correspond)#●ready to prove

Let \(k \ge 1\), \(\mathfrak{t} \ge 1\), \(w \ge 1\) and \(z \ge 0\) be natural numbers. There is a single finite set of triples that is Sierpiński covering data for the \(2\)-repstring \(k_2^{(z;t)}\) for every \(t \ge 1\) with \(t \equiv \mathfrak{t} \pmod{w}\) if and only if there is a single finite set of triples that is Riesel covering data for \(k_2^{(z;t')}\) for every \(t' \ge 1\) with \(t' \equiv -\mathfrak{t} \pmod{w}\).

public theorem_wanted sierpinskiCoveringData_repstring_iff_riesel {k 𝔱 w : ℕ} (hk : 1 ≤ k)
    (h𝔱 : 1 ≤ 𝔱) (hw : 1 ≤ w) (z : ℕ) :
    (∃ S : Finset (ℕ × ℕ × ℕ), ∀ t : ℕ, 1 ≤ t → t ≡ 𝔱 [MOD w] →
        IsSierpinskiCoveringData (repstring 2 k z t) S) ↔
      ∃ S' : Finset (ℕ × ℕ × ℕ), ∀ t' : ℕ, 1 ≤ t' → (t' : ℤ) ≡ -(𝔱 : ℤ) [ZMOD w] →
        IsRieselCoveringData (repstring 2 k z t') S'

Sources

• Bispels, Cohen, Harrington, Lowrance, Pontes, Schaumann and Wong, On Sierpiński and Riesel Repdigits and Repintegers, INTEGERS 26 (2026), Article A12, Theorem 4.6.

Uses 3

Statement usesRiesel covering data · Sierpiński covering data · Repstrings

TheoremNarayana numbers that are Riesel numbers#●ready to prove

Let \((N_t)\) be Narayana's cows sequence, with \(N_0 = 0\) and \(N_1 = N_2 = 1\). If \(t\) is congruent modulo \(718391520\) to one of the \(27\) residues \(3629412\), \(13386852\), \(73151172\), \(96325092\), \(101203812\), \(119499012\), \(175604292\), \(198778212\), \(212194692\), \(240247332\), \(268299972\), \(328064292\), \(337821732\), \(360995652\), \(384169572\), \(402464772\), \(425638692\), \(467107812\), \(476865252\), \(495160452\), \(564682212\), \(587856132\), \(592734852\), \(611030052\), \(629325252\), \(652499172\), \(703725732\), then \(N_t\) is a Riesel number: \(N_t\) is odd and \(N_t \cdot 2^n - 1\) is not prime for every \(n \ge 1\).

With this indexing, \(N_t\) is odd exactly when \(t \bmod 7 \in \{1, 2, 3, 5\}\).

public theorem_wanted isRieselNumber_narayanaCowsSequence {t : ℕ}
    (ht : t % 718391520 ∈ ({3629412, 13386852, 73151172, 96325092, 101203812, 119499012,
      175604292, 198778212, 212194692, 240247332, 268299972, 328064292, 337821732, 360995652,
      384169572, 402464772, 425638692, 467107812, 476865252, 495160452, 564682212, 587856132,
      592734852, 611030052, 629325252, 652499172, 703725732} : Finset ℕ)) :
    IsRieselNumber (narayanaCowsSequence t)

Sources

• Katherine Franzone and Joshua Harrington, Sierpiński and Riesel Numbers in Narayana's Cow Sequence, INTEGERS 24 (2024), Article A10, Theorem 2.

Uses 2

Statement usesRiesel number · Narayana's cows sequence

TheoremInfinitely many Narayana numbers are Sierpiński numbers#●ready to prove

There are infinitely many natural numbers \(k\) that are Sierpiński numbers and terms of Narayana's cows sequence \((N_t)\), that is, \(k = N_t\) for some \(t\); here a Sierpiński number is an odd \(k\) such that \(k \cdot 2^n + 1\) is not prime for every \(n \ge 1\).

public theorem_wanted infinite_setOf_isSierpinskiNumber_narayanaCowsSequence :
    {k : ℕ | IsSierpinskiNumber k ∧ ∃ t, narayanaCowsSequence t = k}.Infinite

Sources

• Katherine Franzone and Joshua Harrington, Sierpiński and Riesel Numbers in Narayana's Cow Sequence, INTEGERS 24 (2024), Article A10, Theorem 3.

Uses 2

Statement usesSierpiński number · Narayana's cows sequence

11 fully proved19 ready to prove

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