Near-perfect and deficient-perfect numbers

Definition\((k,\ell)\)-deficient-perfect with a given divisor#✓fully proved

For natural numbers \(k\), \(\ell\), \(n\) and \(d\), the number \(n\) is \([k,\ell]\)-deficient-perfect with divisor \(d\) when \(d\) is a proper divisor of \(n\) (\(d \mid n\) and \(d < n\)) and \(\sigma_k(n) + d^k = n^k + \ell n\), where \(\sigma_k(n) = \sum_{e \mid n} e^k\) is the sum of the \(k\)-th powers of the divisors of \(n\).

def KLDeficientPerfectWith (k l n d : ℕ) : Prop :=
  d ∣ n ∧ d < n ∧
    (∑ e ∈ n.divisors, e ^ k) + d ^ k = n ^ k + l * n
Used by 3
Used by\((k,\ell)\)-deficient-perfect numbers · Prime powers that are \((1,\ell)\)-deficient-perfect with a given divisor · \(2^a p\) is not \((k,\ell)\)-deficient-perfect with divisor \(1\) or \(p\)

Definition\((k,\ell)\)-deficient-perfect numbers#✓fully proved

For natural numbers \(k\), \(\ell\) and \(n\), write \(\sigma_k(n) = \sum_{e \mid n} e^k\). The number \(n\) is \([k,\ell]\)-deficient-perfect when there is a divisor \(d\) of \(n\) with \(d < n\) such that \(\sigma_k(n) + d^k = n^k + \ell n\).

def IsKLDeficientPerfect (k l n : ℕ) : Prop :=
  ∃ d, KLDeficientPerfectWith k l n d
Uses 1 · Used by 6
Statement uses\((k,\ell)\)-deficient-perfect with a given divisor
Used byDeficient \(F_k\)-perfect numbers · Deficient \(F\)-perfect numbers · \((k,\ell)\)-deficient-perfect prime powers for \(k \ge 2\) · \((2,\ell)\)-deficient-perfect prime powers (corrected) · \(2\) is \((k,1)\)-deficient-perfect · Odd primes are not \((k,\ell)\)-deficient-perfect

Definition\((k,\ell)\)-near-perfect with a given divisor#✓fully proved

For natural numbers \(k\), \(\ell\), \(n\) and \(d\), the number \(n\) is \([k,\ell]\)-near-perfect with divisor \(d\) when \(d\) is a proper divisor of \(n\) (\(d \mid n\) and \(d < n\)) and \(\sigma_k(n) = n^k + d^k + \ell n\), where \(\sigma_k(n) = \sum_{e \mid n} e^k\) is the sum of the \(k\)-th powers of the divisors of \(n\).

def KLNearPerfectWith (k l n d : ℕ) : Prop :=
  d ∣ n ∧ d < n ∧
    (∑ e ∈ n.divisors, e ^ k) = n ^ k + d ^ k + l * n
Used by 5
Used by\((k,\ell)\)-near-perfect numbers · Prime powers that are \((k,\ell)\)-near-perfect · Prime powers that are \((2,\ell)\)-near-perfect · A divisibility condition for near-perfect numbers \(2p^a\) · For \(a \ge 2\), \(2^a p\) is not \((k,\ell)\)-near-perfect with divisor \(1\) or \(p\)

Definition\((k,\ell)\)-near-perfect numbers#✓fully proved

For natural numbers \(k\), \(\ell\) and \(n\), write \(\sigma_k(n) = \sum_{e \mid n} e^k\). The number \(n\) is \([k,\ell]\)-near-perfect when there is a divisor \(d\) of \(n\) with \(d < n\) such that \(\sigma_k(n) = n^k + d^k + \ell n\).

def IsKLNearPerfect (k l n : ℕ) : Prop :=
  ∃ d, KLNearPerfectWith k l n d
Uses 1 · Used by 6
Statement uses\((k,\ell)\)-near-perfect with a given divisor
Used byThe only \((1,\ell)\)-near-perfect number of the form \(2p^a\) is \(18\) · Near \(F_k\)-perfect numbers · Near \(F\)-perfect numbers · Primes are not \((k,\ell)\)-near-perfect · Twice an odd prime is not \((k,\ell)\)-near-perfect · No \((2,\ell)\)-near-perfect number of the form \(2^a p\)

DefinitionDeficient \(F_k\)-perfect numbers#✓fully proved

For natural numbers \(k\) and \(n\), write \(\sigma_k(n) = \sum_{e \mid n} e^k\). The number \(n\) is deficient \(F_k\)-perfect when it is \([k,3]\)-deficient-perfect, that is, when there is a divisor \(d\) of \(n\) with \(d < n\) such that \(\sigma_k(n) + d^k = n^k + 3n\).

def IsDeficientFkPerfect (k n : ℕ) : Prop :=
  IsKLDeficientPerfect k 3 n
Uses 1 · Used by 4
Statement uses\((k,\ell)\)-deficient-perfect numbers
Used byNo deficient \(F_k\)-perfect number \(p_1^a p_2^b\) with \(a + b\) odd or \(a, b\) both odd · No deficient \(F_k\)-perfect number \(p_1^a p_2\) with distinct odd primes · No product of distinct odd primes is deficient \(F_k\)-perfect · No deficient \(F_k\)-perfect number of the form \(2p^a\) for \(k > 2\)

DefinitionNear \(F_k\)-perfect numbers#✓fully proved

For natural numbers \(k\) and \(n\), the number \(n\) is near \(F_k\)-perfect when it is \([k,3]\)-near-perfect, that is, when some proper divisor \(d\) of \(n\) (\(d \mid n\) and \(d < n\)) satisfies \(\sigma_k(n) = n^k + d^k + 3n\), where \(\sigma_k(n) = \sum_{e \mid n} e^k\) is the sum of the \(k\)-th powers of the divisors of \(n\).

No condition \(k > 1\) is imposed here; the results that need it assume it.

def IsNearFkPerfect (k n : ℕ) : Prop :=
  IsKLNearPerfect k 3 n
Uses 1 · Used by 3
Statement uses\((k,\ell)\)-near-perfect numbers
Used byNo near \(F_k\)-perfect number \(p_1^a p_2^b\) with \(a + b\) odd or \(a, b\) both odd · No near \(F_k\)-perfect number of the form \(p_1^a p_2\) · No product of distinct odd primes is near \(F_k\)-perfect

DefinitionDeficient \(F\)-perfect numbers#✓fully proved

For a natural number \(n\), write \(\sigma_2(n) = \sum_{e \mid n} e^2\). The number \(n\) is deficient \(F\)-perfect when it is \([2,3]\)-deficient-perfect, that is, when there is a divisor \(d\) of \(n\) with \(d < n\) such that \(\sigma_2(n) + d^2 = n^2 + 3n\).

def IsDeficientFPerfect (n : ℕ) : Prop :=
  IsKLDeficientPerfect 2 3 n
Uses 1 · Used by 4
Statement uses\((k,\ell)\)-deficient-perfect numbers
Used byThe only deficient \(F\)-perfect number of the form \(2p^a\) is \(6\) · Deficient \(F\)-perfect numbers of the form \(2^a p\) · No prime is deficient \(F\)-perfect · No prime power is deficient \(F\)-perfect or near \(F\)-perfect

DefinitionNear \(F\)-perfect numbers#✓fully proved

A natural number \(n\) is near \(F\)-perfect when it is \([2,3]\)-near-perfect, that is, when some proper divisor \(d\) of \(n\) (\(d \mid n\) and \(d < n\)) satisfies \(\sigma_2(n) = n^2 + d^2 + 3n\), where \(\sigma_2(n) = \sum_{e \mid n} e^2\) is the sum of the squares of the divisors of \(n\).

def IsNearFPerfect (n : ℕ) : Prop :=
  IsKLNearPerfect 2 3 n
Uses 1 · Used by 4
Statement uses\((k,\ell)\)-near-perfect numbers
Used byNo near \(F\)-perfect number of the form \(2p^a\) · No near \(F\)-perfect number of the form \(2^a p\) · No near \(F\)-perfect number of the form \(2^a p^b\) with \(b\) even · No prime power is deficient \(F\)-perfect or near \(F\)-perfect

TheoremOdd primes are not \((k,\ell)\)-deficient-perfect#●ready to prove

Let \(p\) be an odd prime. Then for all natural numbers \(k\) and \(\ell\), \(p\) is not \([k,\ell]\)-deficient-perfect.

theorem_wanted not_isKLDeficientPerfect_of_prime_odd
    (k l p : ℕ) (hp : p.Prime) (hpo : Odd p) :
    ¬ IsKLDeficientPerfect k l p
Sources
  • Flora Jeba S, Anirban Roy, Pankaj Jyoti Mahanta, and Manjil P. Saikia, "On Near F_k-Perfect and Deficient F_k-Perfect Numbers", Proposition 1.
Uses 1
Statement uses\((k,\ell)\)-deficient-perfect numbers

Theorem\(2\) is \((k,1)\)-deficient-perfect#●ready to prove

For every natural number \(k\), the number \(2\) is \([k,1]\)-deficient-perfect.

The divisor is \(d = 1\), since \(1 + 2^k + 1^k = 2^k + 1 \cdot 2\).

theorem_wanted isKLDeficientPerfect_two (k : ℕ) :
    IsKLDeficientPerfect k 1 2
Sources
  • Flora Jeba S, Anirban Roy, Pankaj Jyoti Mahanta, and Manjil P. Saikia, "On Near F_k-Perfect and Deficient F_k-Perfect Numbers", Corollary 1.
Uses 1
Statement uses\((k,\ell)\)-deficient-perfect numbers

TheoremNo prime is deficient \(F\)-perfect#●ready to prove

If \(p\) is a prime, then \(p\) is not deficient \(F\)-perfect.

theorem_wanted not_isDeficientFPerfect_of_prime (p : ℕ)
    (hp : p.Prime) :
    ¬ IsDeficientFPerfect p
Sources
  • Flora Jeba S, Anirban Roy, Pankaj Jyoti Mahanta, and Manjil P. Saikia, "On Near F_k-Perfect and Deficient F_k-Perfect Numbers", Corollary 2.
Uses 1
Statement usesDeficient \(F\)-perfect numbers

TheoremPrimes are not \((k,\ell)\)-near-perfect#●ready to prove

Let \(k\) and \(\ell\) be natural numbers with \(\ell \ge 1\) and let \(p\) be a prime. Then \(p\) is not \([k,\ell]\)-near-perfect.

theorem_wanted not_isKLNearPerfect_of_prime (k l p : ℕ)
    (hl : 1 ≤ l) (hp : p.Prime) :
    ¬ IsKLNearPerfect k l p
Sources
  • Flora Jeba S, Anirban Roy, Pankaj Jyoti Mahanta, and Manjil P. Saikia, "On Near F_k-Perfect and Deficient F_k-Perfect Numbers", Proposition 2.
Uses 1
Statement uses\((k,\ell)\)-near-perfect numbers

TheoremNo prime power is deficient \(F\)-perfect or near \(F\)-perfect#●ready to prove

For every prime \(p\) and every natural number \(a\), the number \(n = p^a\) is neither deficient \(F\)-perfect nor near \(F\)-perfect: there is no divisor \(d\) of \(n\) with \(d < n\) and \(\sigma_2(n) + d^2 = n^2 + 3n\), and no divisor \(d\) of \(n\) with \(d < n\) and \(\sigma_2(n) = n^2 + d^2 + 3n\), where \(\sigma_2(n)\) is the sum of the squares of the divisors of \(n\).

theorem_wanted prime_pow_not_fPerfect (p a : ℕ)
    (hp : p.Prime) :
    ¬ IsDeficientFPerfect (p ^ a) ∧ ¬ IsNearFPerfect (p ^ a)
Sources
  • Flora Jeba S, Anirban Roy, Pankaj Jyoti Mahanta, and Manjil P. Saikia, "On Near F_k-Perfect and Deficient F_k-Perfect Numbers", Theorem 1.
Uses 2
Statement usesDeficient \(F\)-perfect numbers · Near \(F\)-perfect numbers

TheoremPrime powers that are \((2,\ell)\)-near-perfect#●ready to prove

Let \(p\) be a prime, let \(a\), \(\ell\) and \(d\) be natural numbers with \(\ell \ge 1\), and write \(\sigma_2(m)\) for the sum of the squares of the divisors of \(m\). Then \(d\) is a proper divisor of \(p^a\) (\(d \mid p^a\) and \(d < p^a\)) with \(\sigma_2(p^a) = (p^a)^2 + d^2 + \ell p^a\) if and only if \(a = 2\), \(\ell = 1\) and \(d = 1\).

theorem_wanted klNearPerfectWith_prime_pow_two_iff
    (p a l d : ℕ) (hp : p.Prime) (hl : 1 ≤ l) :
    KLNearPerfectWith 2 l (p ^ a) d ↔
      a = 2 ∧ l = 1 ∧ d = 1
Sources
  • Flora Jeba S, Anirban Roy, Pankaj Jyoti Mahanta, and Manjil P. Saikia, "On Near F_k-Perfect and Deficient F_k-Perfect Numbers", Theorem 2.
Uses 1
Statement uses\((k,\ell)\)-near-perfect with a given divisor

Theorem\((2,\ell)\)-deficient-perfect prime powers (corrected)#●ready to prove

Let \(p\) be a prime and let \(a\) and \(\ell\) be natural numbers. If \(p^a\) is \([2,\ell]\)-deficient-perfect, then \(\ell = 1\), and either \(a = 2\), or \(p = 2\) and \(a = 1\).

The printed statement allows only the exception \(a = 2\), \(\ell = 1\), and so omits the solution \(p = 2\), \(a = 1\), \(\ell = 1\), \(d = 1\) (the number \(2\) of Corollary 1).

theorem_wanted isKLDeficientPerfect_prime_pow_two_imp
    (p a l : ℕ) (hp : p.Prime) :
    IsKLDeficientPerfect 2 l (p ^ a) →
      l = 1 ∧ (a = 2 ∨ (p = 2 ∧ a = 1))
Sources
  • Flora Jeba S, Anirban Roy, Pankaj Jyoti Mahanta, and Manjil P. Saikia, "On Near F_k-Perfect and Deficient F_k-Perfect Numbers", Theorem 3.
Uses 1
Statement uses\((k,\ell)\)-deficient-perfect numbers

TheoremPrime powers that are \((k,\ell)\)-near-perfect#●ready to prove

Let \(p\) be a prime, let \(a\), \(k\), \(\ell\) and \(d\) be natural numbers with \(k \ge 1\) and \(\ell \ge 1\), and write \(\sigma_k(m) = \sum_{e \mid m} e^k\). Then \(d\) is a proper divisor of \(p^a\) (\(d \mid p^a\) and \(d < p^a\)) with \(\sigma_k(p^a) = (p^a)^k + d^k + \ell p^a\) if and only if \(2 \le a \le k\), \(d = 1\) and \(\ell = p^{k-a} \sum_{i=0}^{a-2} p^{ik}\).

The source prints the conclusion \(a = k\), which is false: for \(p = 2\), \(a = 2\), \(k = 3\), \(\ell = 2\), \(d = 1\) one has \(\sigma_3(4) = 73 = 4^3 + 1^3 + 2 \cdot 4\) with \(a \ne k\). The printed form is the case \(a = k\) of the identity \(\sum_{i=1}^{a-1} p^{ik} = \ell p^a\).

theorem_wanted klNearPerfectWith_prime_pow_iff
    (p a k l d : ℕ) (hp : p.Prime) (hk : 1 ≤ k) (hl : 1 ≤ l) :
    KLNearPerfectWith k l (p ^ a) d ↔
      2 ≤ a ∧ a ≤ k ∧ d = 1 ∧
        l = p ^ (k - a) *
          ∑ i ∈ Finset.range (a - 1), p ^ (i * k)
Sources
  • Flora Jeba S, Anirban Roy, Pankaj Jyoti Mahanta, and Manjil P. Saikia, "On Near F_k-Perfect and Deficient F_k-Perfect Numbers", Theorem 4.
Uses 1
Statement uses\((k,\ell)\)-near-perfect with a given divisor

Theorem\((k,\ell)\)-deficient-perfect prime powers for \(k \ge 2\)#●ready to prove

Let \(p\) be a prime and let \(a\), \(k\) and \(\ell\) be natural numbers with \(k \ge 2\). If \(p^a\) is \([k,\ell]\)-deficient-perfect, then \(p = 2\) and \(a = 1\).

theorem_wanted isKLDeficientPerfect_prime_pow_imp
    (p a k l : ℕ) (hp : p.Prime) (hk : 2 ≤ k) :
    IsKLDeficientPerfect k l (p ^ a) → p = 2 ∧ a = 1
Sources
  • Flora Jeba S, Anirban Roy, Pankaj Jyoti Mahanta, and Manjil P. Saikia, "On Near F_k-Perfect and Deficient F_k-Perfect Numbers", Theorem 5.
Uses 1
Statement uses\((k,\ell)\)-deficient-perfect numbers

TheoremPrime powers that are \((1,\ell)\)-deficient-perfect with a given divisor#●ready to prove

Let \(p\) be a prime, let \(a\), \(\ell\) and \(d\) be natural numbers, and write \(\sigma_1(m)\) for the sum of the divisors of \(m\). Then \(d\) is a proper divisor of \(p^a\) (\(d \mid p^a\) and \(d < p^a\)) with \(\sigma_1(p^a) + d = p^a + \ell p^a\) if and only if \(p = 2\), \(\ell = 1\), \(d = 1\) and \(a \ge 1\).

This is the \(k = 1\) part of Theorem 5 of the source.

theorem_wanted klDeficientPerfectWith_one_prime_pow_iff
    (p a l d : ℕ) (hp : p.Prime) :
    KLDeficientPerfectWith 1 l (p ^ a) d ↔
      p = 2 ∧ l = 1 ∧ d = 1 ∧ 1 ≤ a
Sources
  • Flora Jeba S, Anirban Roy, Pankaj Jyoti Mahanta, and Manjil P. Saikia, "On Near F_k-Perfect and Deficient F_k-Perfect Numbers", Theorem 5.
Uses 1
Statement uses\((k,\ell)\)-deficient-perfect with a given divisor

TheoremTwice an odd prime is not \((k,\ell)\)-near-perfect#●ready to prove

Let \(k\) and \(\ell\) be natural numbers with \(\ell \ge 1\) and let \(p\) be an odd prime. Then \(2p\) is not \([k,\ell]\)-near-perfect.

theorem_wanted not_isKLNearPerfect_two_mul_prime (k l p : ℕ)
    (hl : 1 ≤ l) (hp : p.Prime) (hpo : Odd p) :
    ¬ IsKLNearPerfect k l (2 * p)
Sources
  • Flora Jeba S, Anirban Roy, Pankaj Jyoti Mahanta, and Manjil P. Saikia, "On Near F_k-Perfect and Deficient F_k-Perfect Numbers", Theorem 6.
Uses 1
Statement uses\((k,\ell)\)-near-perfect numbers

TheoremDeficient \(F\)-perfect numbers of the form \(2^a p\)#●ready to prove

Let \(p\) be a prime and \(a\) a natural number. If \(2^a p\) is deficient \(F\)-perfect, then \(p = 3\) and \(a = 1\), so the number is \(6\).

The number \(6\) is indeed deficient \(F\)-perfect, with \(d = 2\).

theorem_wanted isDeficientFPerfect_two_pow_mul_prime_imp
    (p a : ℕ) (hp : p.Prime) :
    IsDeficientFPerfect (2 ^ a * p) → p = 3 ∧ a = 1
Sources
  • Flora Jeba S, Anirban Roy, Pankaj Jyoti Mahanta, and Manjil P. Saikia, "On Near F_k-Perfect and Deficient F_k-Perfect Numbers", Theorem 7.
Uses 1
Statement usesDeficient \(F\)-perfect numbers

Theorem\(2^a p\) is not \((k,\ell)\)-deficient-perfect with divisor \(1\) or \(p\)#●ready to prove

Let \(k\), \(\ell\), \(a\) be natural numbers with \(\ell \ge 1\) and let \(p\) be an odd prime. Call \(n\) \([k,\ell]\)-deficient-perfect with divisor \(d\) when \(d\) divides \(n\), \(d < n\) and \(\sigma_k(n) + d^k = n^k + \ell n\), where \(\sigma_k(n)\) is the sum of the \(k\)-th powers of the divisors of \(n\). Then \(2^a p\) is not \([k,\ell]\)-deficient-perfect with divisor \(1\), and not with divisor \(p\).

theorem_wanted not_klDeficientPerfectWith_two_pow_mul
    (k l p a : ℕ) (hl : 1 ≤ l) (hp : p.Prime) (hpo : Odd p) :
    ¬ KLDeficientPerfectWith k l (2 ^ a * p) 1 ∧
      ¬ KLDeficientPerfectWith k l (2 ^ a * p) p
Sources
  • Flora Jeba S, Anirban Roy, Pankaj Jyoti Mahanta, and Manjil P. Saikia, "On Near F_k-Perfect and Deficient F_k-Perfect Numbers", Theorem 8.
Uses 1
Statement uses\((k,\ell)\)-deficient-perfect with a given divisor

TheoremNo near \(F\)-perfect number of the form \(2^a p\)#●ready to prove

For every prime \(p\) and every natural number \(a\), the number \(n = 2^a p\) is not near \(F\)-perfect: there is no divisor \(d\) of \(n\) with \(d < n\) and \(\sigma_2(n) = n^2 + d^2 + 3n\), where \(\sigma_2(n)\) is the sum of the squares of the divisors of \(n\).

theorem_wanted not_isNearFPerfect_two_pow_mul_prime
    (p a : ℕ) (hp : p.Prime) :
    ¬ IsNearFPerfect (2 ^ a * p)
Sources
  • Flora Jeba S, Anirban Roy, Pankaj Jyoti Mahanta, and Manjil P. Saikia, "On Near F_k-Perfect and Deficient F_k-Perfect Numbers", Theorem 9.
Uses 1
Statement usesNear \(F\)-perfect numbers

TheoremNo \((2,\ell)\)-near-perfect number of the form \(2^a p\)#●ready to prove

Let \(\ell\) and \(a\) be natural numbers with \(\ell \ge 1\), and let \(p\) be an odd prime. Then \(2^a p\) is not \([2,\ell]\)-near-perfect.

theorem_wanted not_isKLNearPerfect_two_two_pow_mul
    (l p a : ℕ) (hl : 1 ≤ l) (hp : p.Prime) (hpo : Odd p) :
    ¬ IsKLNearPerfect 2 l (2 ^ a * p)
Sources
  • Flora Jeba S, Anirban Roy, Pankaj Jyoti Mahanta, and Manjil P. Saikia, "On Near F_k-Perfect and Deficient F_k-Perfect Numbers", Theorem 10.
Uses 1
Statement uses\((k,\ell)\)-near-perfect numbers

TheoremFor \(a \ge 2\), \(2^a p\) is not \((k,\ell)\)-near-perfect with divisor \(1\) or \(p\)#●ready to prove

Let \(k\), \(\ell\), \(a\) be natural numbers with \(\ell \ge 1\) and \(a \ge 2\), and let \(p\) be an odd prime. Call \(n\) \([k,\ell]\)-near-perfect with divisor \(d\) when \(d\) divides \(n\), \(d < n\) and \(\sigma_k(n) = n^k + d^k + \ell n\), where \(\sigma_k(n)\) is the sum of the \(k\)-th powers of the divisors of \(n\). Then \(2^a p\) is not \([k,\ell]\)-near-perfect with divisor \(1\), and not with divisor \(p\).

theorem_wanted not_klNearPerfectWith_two_pow_mul
    (k l p a : ℕ) (hl : 1 ≤ l) (hp : p.Prime) (hpo : Odd p)
    (ha : 1 < a) :
    ¬ KLNearPerfectWith k l (2 ^ a * p) 1 ∧
      ¬ KLNearPerfectWith k l (2 ^ a * p) p
Sources
  • Flora Jeba S, Anirban Roy, Pankaj Jyoti Mahanta, and Manjil P. Saikia, "On Near F_k-Perfect and Deficient F_k-Perfect Numbers", Theorem 11.
Uses 1
Statement uses\((k,\ell)\)-near-perfect with a given divisor

TheoremThe only deficient \(F\)-perfect number of the form \(2p^a\) is \(6\)#●ready to prove

The number \(6\) is deficient \(F\)-perfect, and for every prime \(p\) and every natural number \(a\), if \(2p^a\) is deficient \(F\)-perfect then \(2p^a = 6\).

The divisor for \(6\) is \(d = 2\).

theorem_wanted isDeficientFPerfect_two_mul_prime_pow
    (p a : ℕ) (hp : p.Prime) :
    IsDeficientFPerfect 6 ∧
      (IsDeficientFPerfect (2 * p ^ a) → 2 * p ^ a = 6)
Sources
  • Flora Jeba S, Anirban Roy, Pankaj Jyoti Mahanta, and Manjil P. Saikia, "On Near F_k-Perfect and Deficient F_k-Perfect Numbers", Theorem 12.
Uses 1
Statement usesDeficient \(F\)-perfect numbers

TheoremNo near \(F\)-perfect number of the form \(2p^a\)#●ready to prove

For every prime \(p\) and every natural number \(a\), the number \(n = 2p^a\) is not near \(F\)-perfect: there is no divisor \(d\) of \(n\) with \(d < n\) and \(\sigma_2(n) = n^2 + d^2 + 3n\), where \(\sigma_2(n)\) is the sum of the squares of the divisors of \(n\).

theorem_wanted not_isNearFPerfect_two_mul_prime_pow
    (p a : ℕ) (hp : p.Prime) :
    ¬ IsNearFPerfect (2 * p ^ a)
Sources
  • Flora Jeba S, Anirban Roy, Pankaj Jyoti Mahanta, and Manjil P. Saikia, "On Near F_k-Perfect and Deficient F_k-Perfect Numbers", Theorem 13.
Uses 1
Statement usesNear \(F\)-perfect numbers

TheoremThe only \((1,\ell)\)-near-perfect number of the form \(2p^a\) is \(18\)#●ready to prove

For natural numbers \(\ell\) and \(n\), say that \(n\) is \([1,\ell]\)-near-perfect when some divisor \(d\) of \(n\) with \(d < n\) satisfies \(\sigma(n) = n + d + \ell n\), where \(\sigma(n)\) is the sum of the divisors of \(n\). Then \(18\) is \([1,1]\)-near-perfect, and for every odd prime \(p\) and all natural numbers \(a\) and \(\ell \ge 1\), if \(2p^a\) is \([1,\ell]\)-near-perfect then \(2p^a = 18\).

The divisor for \(18\) is \(d = 3\). This is part 1 of Theorem 14 of the source.

theorem_wanted isKLNearPerfect_one_two_mul_prime_pow
    (p a l : ℕ) (hp : p.Prime) (hpo : Odd p) (hl : 1 ≤ l) :
    IsKLNearPerfect 1 1 18 ∧
      (IsKLNearPerfect 1 l (2 * p ^ a) → 2 * p ^ a = 18)
Sources
  • Flora Jeba S, Anirban Roy, Pankaj Jyoti Mahanta, and Manjil P. Saikia, "On Near F_k-Perfect and Deficient F_k-Perfect Numbers", Theorem 14.
Uses 1
Statement uses\((k,\ell)\)-near-perfect numbers

TheoremA divisibility condition for near-perfect numbers \(2p^a\)#●ready to prove

Let \(p\) be an odd prime, let \(a\), \(k\), \(\ell\) and \(d\) be natural numbers with \(k > 1\) and \(\ell \ge 1\), and write \(\sigma_k(m) = \sum_{e \mid m} e^k\). If \(d\) is a proper divisor of \(2p^a\) (\(d \mid 2p^a\) and \(d < 2p^a\)) with \(\sigma_k(2p^a) = (2p^a)^k + d^k + \ell \cdot 2p^a\), then \(p^2 \mid 1 + 2^k\), and \(a\) is even if and only if \(d\) is odd.

This is part 2 of Theorem 14 of the source.

theorem_wanted klNearPerfectWith_two_mul_prime_pow_imp
    (p a k l d : ℕ) (hp : p.Prime) (hpo : Odd p) (hk : 1 < k)
    (hl : 1 ≤ l) :
    KLNearPerfectWith k l (2 * p ^ a) d →
      p ^ 2 ∣ 1 + 2 ^ k ∧ (Even a ↔ Odd d)
Sources
  • Flora Jeba S, Anirban Roy, Pankaj Jyoti Mahanta, and Manjil P. Saikia, "On Near F_k-Perfect and Deficient F_k-Perfect Numbers", Theorem 14.
Uses 1
Statement uses\((k,\ell)\)-near-perfect with a given divisor

TheoremNo near \(F\)-perfect number of the form \(2^a p^b\) with \(b\) even#●ready to prove

For every prime \(p\) and all natural numbers \(a\) and \(b\) with \(b\) even, the number \(n = 2^a p^b\) is not near \(F\)-perfect: there is no divisor \(d\) of \(n\) with \(d < n\) and \(\sigma_2(n) = n^2 + d^2 + 3n\), where \(\sigma_2(n)\) is the sum of the squares of the divisors of \(n\).

theorem_wanted not_isNearFPerfect_two_pow_mul_prime_pow
    (p a b : ℕ) (hp : p.Prime) (hb : Even b) :
    ¬ IsNearFPerfect (2 ^ a * p ^ b)
Sources
  • Flora Jeba S, Anirban Roy, Pankaj Jyoti Mahanta, and Manjil P. Saikia, "On Near F_k-Perfect and Deficient F_k-Perfect Numbers", Theorem 15.
Uses 1
Statement usesNear \(F\)-perfect numbers

TheoremNo deficient \(F_k\)-perfect number \(p_1^a p_2\) with distinct odd primes#●ready to prove

Let \(k > 1\) be a natural number, let \(p_1 \ne p_2\) be odd primes and let \(a\) be a natural number. Then \(p_1^a p_2\) is not deficient \(F_k\)-perfect.

theorem_wanted not_isDeficientFkPerfect_prime_pow_mul
    (k p₁ p₂ a : ℕ) (hk : 1 < k) (hp₁ : p₁.Prime)
    (hp₂ : p₂.Prime) (ho₁ : Odd p₁) (ho₂ : Odd p₂)
    (hne : p₁ ≠ p₂) :
    ¬ IsDeficientFkPerfect k (p₁ ^ a * p₂)
Sources
  • Flora Jeba S, Anirban Roy, Pankaj Jyoti Mahanta, and Manjil P. Saikia, "On Near F_k-Perfect and Deficient F_k-Perfect Numbers", Theorem 16.
Uses 1
Statement usesDeficient \(F_k\)-perfect numbers

TheoremNo near \(F_k\)-perfect number of the form \(p_1^a p_2\)#●ready to prove

Let \(k > 1\), let \(p_1\) and \(p_2\) be distinct odd primes and let \(a\) be a natural number. Then \(p_1^a p_2\) is not near \(F_k\)-perfect.

theorem_wanted not_isNearFkPerfect_prime_pow_mul
    (k p₁ p₂ a : ℕ) (hk : 1 < k) (hp₁ : p₁.Prime)
    (hp₂ : p₂.Prime) (ho₁ : Odd p₁) (ho₂ : Odd p₂)
    (hne : p₁ ≠ p₂) :
    ¬ IsNearFkPerfect k (p₁ ^ a * p₂)
Sources
  • Flora Jeba S, Anirban Roy, Pankaj Jyoti Mahanta, and Manjil P. Saikia, "On Near F_k-Perfect and Deficient F_k-Perfect Numbers", Theorem 17.
Uses 1
Statement usesNear \(F_k\)-perfect numbers

TheoremNo deficient \(F_k\)-perfect number of the form \(2p^a\) for \(k > 2\)#●ready to prove

Let \(k > 2\), let \(p\) be an odd prime and let \(a\) be a natural number. Then \(2p^a\) is not deficient \(F_k\)-perfect.

theorem_wanted not_isDeficientFkPerfect_two_mul_pow
    (k p a : ℕ) (hk : 2 < k) (hp : p.Prime) (hpo : Odd p) :
    ¬ IsDeficientFkPerfect k (2 * p ^ a)
Sources
  • Flora Jeba S, Anirban Roy, Pankaj Jyoti Mahanta, and Manjil P. Saikia, "On Near F_k-Perfect and Deficient F_k-Perfect Numbers", Theorem 18.
Uses 1
Statement usesDeficient \(F_k\)-perfect numbers

TheoremNo deficient \(F_k\)-perfect number \(p_1^a p_2^b\) with \(a + b\) odd or \(a, b\) both odd#●ready to prove

Let \(k > 1\) be a natural number, let \(p_1 \ne p_2\) be odd primes and let \(a, b \ge 1\) be natural numbers such that \(a + b\) is odd or both \(a\) and \(b\) are odd. Then \(n = p_1^a p_2^b\) is not deficient \(F_k\)-perfect: there is no divisor \(d\) of \(n\) with \(d < n\) and \(\sigma_k(n) + d^k = n^k + 3n\), where \(\sigma_k(n) = \sum_{e \mid n} e^k\).

theorem_wanted not_isDeficientFkPerfect_pow_mul_pow
    (k p₁ p₂ a b : ℕ) (hk : 1 < k) (hp₁ : p₁.Prime)
    (hp₂ : p₂.Prime) (ho₁ : Odd p₁) (ho₂ : Odd p₂)
    (hne : p₁ ≠ p₂) (ha : 1 ≤ a) (hb : 1 ≤ b)
    (hpar : Odd (a + b) ∨ (Odd a ∧ Odd b)) :
    ¬ IsDeficientFkPerfect k (p₁ ^ a * p₂ ^ b)
Sources
  • Flora Jeba S, Anirban Roy, Pankaj Jyoti Mahanta, and Manjil P. Saikia, "On Near F_k-Perfect and Deficient F_k-Perfect Numbers", Theorem 19.
Uses 1
Statement usesDeficient \(F_k\)-perfect numbers

TheoremNo near \(F_k\)-perfect number \(p_1^a p_2^b\) with \(a + b\) odd or \(a, b\) both odd#●ready to prove

Let \(k > 1\), let \(p_1 \ne p_2\) be odd primes and let \(a, b \ge 1\) be natural numbers such that \(a + b\) is odd or both \(a\) and \(b\) are odd. Then \(p_1^a p_2^b\) is not near \(F_k\)-perfect.

theorem_wanted not_isNearFkPerfect_pow_mul_pow
    (k p₁ p₂ a b : ℕ) (hk : 1 < k) (hp₁ : p₁.Prime)
    (hp₂ : p₂.Prime) (ho₁ : Odd p₁) (ho₂ : Odd p₂)
    (hne : p₁ ≠ p₂) (ha : 1 ≤ a) (hb : 1 ≤ b)
    (hpar : Odd (a + b) ∨ (Odd a ∧ Odd b)) :
    ¬ IsNearFkPerfect k (p₁ ^ a * p₂ ^ b)
Sources
  • Flora Jeba S, Anirban Roy, Pankaj Jyoti Mahanta, and Manjil P. Saikia, "On Near F_k-Perfect and Deficient F_k-Perfect Numbers", Theorem 20.
Uses 1
Statement usesNear \(F_k\)-perfect numbers

TheoremNo product of distinct odd primes is deficient \(F_k\)-perfect#●ready to prove

Let \(k > 1\) and let \(s\) be a finite set of odd primes. Then \(\prod_{p \in s} p\) is not deficient \(F_k\)-perfect.

theorem_wanted not_isDeficientFkPerfect_prod_primes
    (k : ℕ) (hk : 1 < k) (s : Finset ℕ)
    (hs : ∀ p ∈ s, p.Prime ∧ Odd p) :
    ¬ IsDeficientFkPerfect k (∏ p ∈ s, p)
Sources
  • Flora Jeba S, Anirban Roy, Pankaj Jyoti Mahanta, and Manjil P. Saikia, "On Near F_k-Perfect and Deficient F_k-Perfect Numbers", Theorem 21.
Uses 1
Statement usesDeficient \(F_k\)-perfect numbers

TheoremNo product of distinct odd primes is near \(F_k\)-perfect#●ready to prove

Let \(k > 1\) and let \(s\) be a finite set of odd primes. Then \(\prod_{p \in s} p\) is not near \(F_k\)-perfect.

theorem_wanted not_isNearFkPerfect_prod_primes
    (k : ℕ) (hk : 1 < k) (s : Finset ℕ)
    (hs : ∀ p ∈ s, p.Prime ∧ Odd p) :
    ¬ IsNearFkPerfect k (∏ p ∈ s, p)
Sources
  • Flora Jeba S, Anirban Roy, Pankaj Jyoti Mahanta, and Manjil P. Saikia, "On Near F_k-Perfect and Deficient F_k-Perfect Numbers", Theorem 22.
Uses 1
Statement usesNear \(F_k\)-perfect numbers

Theorem\(2\)-near perfect numbers of the form \(2^k p\)#●ready to prove

Let \(k \ge 1\), let \(p\) be a prime and let \(n = 2^k p\). Suppose \(d_1 \ne d_2\) are divisors of \(n\) with \(\sigma(n) = 2n + d_1 + d_2\), where \(\sigma\) is the sum-of-divisors function (that is, \(n\) is \(2\)-near perfect with omitted divisors \(d_1, d_2\)). Then, after possibly swapping \(d_1\) and \(d_2\), one of the following holds, where all equations are between integers: (1) \(p = 2^k - 1\), \(d_1 = 1\) and \(d_2 = p\); (2) for some natural numbers \(a, b\), \(p = 2^{k+1} - 2^a - 2^b - 1\), \(d_1 = 2^a\) and \(d_2 = 2^b\); (3) for some natural numbers \(a, b\), \(p(1 + 2^b) = 2^{k+1} - 2^a - 1\), \(d_1 = 2^a\) and \(d_2 = 2^b p\); (4) for some natural numbers \(a, b\), \(p(1 + 2^a + 2^b) = 2^{k+1} - 1\), \(d_1 = 2^a p\) and \(d_2 = 2^b p\).

The paper states only this necessary direction; the four situations are not sufficient as printed (\(k = 3\), \(a = b = 1\) gives the prime \(p = 11\) in situation 2, but \(d_1 = d_2 = 2\)).

public theorem_wanted two_near_perfect_two_pow_mul_prime_cases {k p d₁ d₂ : ℕ} (hk : 0 < k)
    (hp : p.Prime) (hd₁ : d₁ ∣ 2 ^ k * p) (hd₂ : d₂ ∣ 2 ^ k * p) (hne : d₁ ≠ d₂)
    (hσ : ArithmeticFunction.sigma 1 (2 ^ k * p) = 2 * (2 ^ k * p) + d₁ + d₂) :
    ∃ e₁ e₂ : ℕ, ((e₁ = d₁ ∧ e₂ = d₂) ∨ (e₁ = d₂ ∧ e₂ = d₁)) ∧
      (((p : ℤ) = 2 ^ k - 1 ∧ e₁ = 1 ∧ e₂ = p) ∨
        (∃ a b : ℕ, (p : ℤ) = 2 ^ (k + 1) - 2 ^ a - 2 ^ b - 1 ∧ e₁ = 2 ^ a ∧ e₂ = 2 ^ b) ∨
        (∃ a b : ℕ, (p : ℤ) * (1 + 2 ^ b) = 2 ^ (k + 1) - 2 ^ a - 1 ∧
          e₁ = 2 ^ a ∧ e₂ = 2 ^ b * p) ∨
        (∃ a b : ℕ, (p : ℤ) * (1 + 2 ^ a + 2 ^ b) = 2 ^ (k + 1) - 1 ∧
          e₁ = 2 ^ a * p ∧ e₂ = 2 ^ b * p))
Sources
  • Vedant Aryan, Dev Madhavani, Savan Parikh, Ingrid Slattery and Joshua Zelinsky, On 2-Near Perfect Numbers, INTEGERS 24 (2024), Article A61, Theorem 1.

Theorem\(2\)-near perfect numbers of the form \(2^k p^2\)#●ready to prove

Let \(p\) be a prime, let \(k\) be a natural number and let \(n = 2^k p^2\). If \(n\) has divisors \(d_1 \ne d_2\) with \(\sigma(n) = 2n + d_1 + d_2\), where \(\sigma\) is the sum-of-divisors function (that is, \(n\) is \(2\)-near perfect with omitted divisors \(d_1, d_2\)), then \(n\) is \(18\), \(36\) or \(200\).

public theorem_wanted two_near_perfect_two_pow_mul_prime_sq {k p d₁ d₂ : ℕ} (hp : p.Prime)
    (hd₁ : d₁ ∣ 2 ^ k * p ^ 2) (hd₂ : d₂ ∣ 2 ^ k * p ^ 2) (hne : d₁ ≠ d₂)
    (hσ : ArithmeticFunction.sigma 1 (2 ^ k * p ^ 2) = 2 * (2 ^ k * p ^ 2) + d₁ + d₂) :
    2 ^ k * p ^ 2 = 18 ∨ 2 ^ k * p ^ 2 = 36 ∨ 2 ^ k * p ^ 2 = 200
Sources
  • Vedant Aryan, Dev Madhavani, Savan Parikh, Ingrid Slattery and Joshua Zelinsky, On 2-Near Perfect Numbers, INTEGERS 24 (2024), Article A61, Theorem 2.

TheoremSquare values of the discriminant \(2^{2k+2} + 2^{k+2} - 2^{a+2} - 7\)#●ready to prove

Let \(a\) and \(k\) be natural numbers with \(1 \le a \le k\). If \(2^{2k+2} + 2^{k+2} - 2^{a+2} - 7\) is the square of an integer, then \(k = a = 1\).

public theorem_wanted eq_one_of_isSquare_discriminant {a k : ℕ} (ha : 0 < a) (hak : a ≤ k)
    (hD : IsSquare ((2 : ℤ) ^ (2 * k + 2) + 2 ^ (k + 2) - 2 ^ (a + 2) - 7)) :
    k = 1 ∧ a = 1
Sources
  • Vedant Aryan, Dev Madhavani, Savan Parikh, Ingrid Slattery and Joshua Zelinsky, On 2-Near Perfect Numbers, INTEGERS 24 (2024), Article A61, Lemma 2.

TheoremDivisibility constraints on odd solutions of \((2^{k+1} - 1)(p^2 + p + 1) = 2^{k+1} p^2 + 2^b p + 1\)#●ready to prove

Divisibility constraints on odd solutions of \((2^{k+1} - 1)(p^2 + p + 1) = 2^{k+1} p^2 + 2^b p + 1\). Let \(b\) and \(k\) be natural numbers and let \(p\) be an odd natural number such that \((2^{k+1} - 1)(p^2 + p + 1) = 2^{k+1} p^2 + 2^b p + 1\). Then, as integers, \(p\) divides \(2^k - 1\) and \(p + 1\) divides \(2^b - 2\).

public theorem_wanted dvd_of_two_near_perfect_eq_six {b k p : ℕ} (hp : Odd p)
    (h : ((2 : ℤ) ^ (k + 1) - 1) * (p ^ 2 + p + 1) = 2 ^ (k + 1) * p ^ 2 + 2 ^ b * p + 1) :
    (p : ℤ) ∣ 2 ^ k - 1 ∧ (p + 1 : ℤ) ∣ 2 ^ b - 2
Sources
  • Vedant Aryan, Dev Madhavani, Savan Parikh, Ingrid Slattery and Joshua Zelinsky, On 2-Near Perfect Numbers, INTEGERS 24 (2024), Article A61, Lemma 3.

TheoremNo odd solutions of \((2^{k+1} - 1)(p^2 + p + 1) = 2^{k+1} p^2 + 2^b p + 1\) for \(2 \le b \le k - 1\)#●ready to prove

No odd solutions of \((2^{k+1} - 1)(p^2 + p + 1) = 2^{k+1} p^2 + 2^b p + 1\) for \(2 \le b \le k - 1\). Let \(b\) and \(k\) be natural numbers with \(2 \le b\) and \(b + 1 \le k\), and let \(p\) be an odd natural number. Then \((2^{k+1} - 1)(p^2 + p + 1) \ne 2^{k+1} p^2 + 2^b p + 1\).

public theorem_wanted not_two_near_perfect_eq_seven {b k p : ℕ} (hp : Odd p) (hb : 2 ≤ b)
    (hbk : b + 1 ≤ k) :
    ((2 : ℤ) ^ (k + 1) - 1) * (p ^ 2 + p + 1) ≠ 2 ^ (k + 1) * p ^ 2 + 2 ^ b * p + 1
Sources
  • Vedant Aryan, Dev Madhavani, Savan Parikh, Ingrid Slattery and Joshua Zelinsky, On 2-Near Perfect Numbers, INTEGERS 24 (2024), Article A61, Lemma 4.

TheoremSum and parity of the omitted divisors of a \(2\)-near perfect \(2^k p^2\)#●ready to prove

Let \(p\) be an odd prime, let \(k\) be a natural number, and let \(d_1 \ne d_2\) be divisors of \(n = 2^k p^2\) with \(\sigma(n) = 2n + d_1 + d_2\), where \(\sigma\) denotes the sum of divisors. Then \(d_1 + d_2 = -p^2 + (2^{k+1} - 1) p + (2^{k+1} - 1)\), and \(d_1\) and \(d_2\) have opposite parity.

public theorem_wanted omitted_divisors_sum_two_pow_mul_prime_sq {k p d₁ d₂ : ℕ}
    (hp : p.Prime) (hp2 : p ≠ 2) (hd₁ : d₁ ∣ 2 ^ k * p ^ 2) (hd₂ : d₂ ∣ 2 ^ k * p ^ 2)
    (hne : d₁ ≠ d₂)
    (hσ : ArithmeticFunction.sigma 1 (2 ^ k * p ^ 2) = 2 * (2 ^ k * p ^ 2) + d₁ + d₂) :
    ((d₁ + d₂ : ℕ) : ℤ) = -(p : ℤ) ^ 2 + (2 ^ (k + 1) - 1) * p + (2 ^ (k + 1) - 1) ∧
      d₁ % 2 ≠ d₂ % 2
Sources
  • Vedant Aryan, Dev Madhavani, Savan Parikh, Ingrid Slattery and Joshua Zelinsky, On 2-Near Perfect Numbers, INTEGERS 24 (2024), Article A61, Lemma 5.

TheoremOdd solutions \(p\) of \((2^{k+1} - 1)(p^2 + p + 1) = 2^{k+1} p^2 + 2^b p^2 + 1\) divide \(2^k - 1\)#●ready to prove

Odd solutions \(p\) of \((2^{k+1} - 1)(p^2 + p + 1) = 2^{k+1} p^2 + 2^b p^2 + 1\) divide \(2^k - 1\). Let \(k\) and \(b\) be positive natural numbers and let \(p\) be an odd natural number such that \((2^{k+1} - 1)(p^2 + p + 1) = 2^{k+1} p^2 + 2^b p^2 + 1\). Then \(p\) divides \(2^k - 1\).

public theorem_wanted dvd_of_two_near_perfect_eq_twelve {b k p : ℕ} (hk : 0 < k) (hb : 0 < b)
    (hp : Odd p)
    (h : ((2 : ℤ) ^ (k + 1) - 1) * (p ^ 2 + p + 1) = 2 ^ (k + 1) * p ^ 2 + 2 ^ b * p ^ 2 + 1) :
    (p : ℤ) ∣ 2 ^ k - 1
Sources
  • Vedant Aryan, Dev Madhavani, Savan Parikh, Ingrid Slattery and Joshua Zelinsky, On 2-Near Perfect Numbers, INTEGERS 24 (2024), Article A61, Lemma 6.

TheoremOdd solutions \(p\) of \((2^{k+1} - 1)(p^2 + p + 1) = 2^{k+1} p^2 + 2^b p^2 + 1\) satisfy \(p + 1 \mid 2^b + 2\)#●ready to prove

Odd solutions \(p\) of \((2^{k+1} - 1)(p^2 + p + 1) = 2^{k+1} p^2 + 2^b p^2 + 1\) satisfy \(p + 1 \mid 2^b + 2\). Let \(k\) and \(b\) be positive natural numbers and let \(p\) be an odd natural number such that \((2^{k+1} - 1)(p^2 + p + 1) = 2^{k+1} p^2 + 2^b p^2 + 1\). Then, as integers, \(p + 1\) divides \(2^b + 2\).

public theorem_wanted succ_dvd_of_two_near_perfect_eq_twelve {b k p : ℕ} (hk : 0 < k)
    (hb : 0 < b) (hp : Odd p)
    (h : ((2 : ℤ) ^ (k + 1) - 1) * (p ^ 2 + p + 1) = 2 ^ (k + 1) * p ^ 2 + 2 ^ b * p ^ 2 + 1) :
    (p + 1 : ℤ) ∣ 2 ^ b + 2
Sources
  • Vedant Aryan, Dev Madhavani, Savan Parikh, Ingrid Slattery and Joshua Zelinsky, On 2-Near Perfect Numbers, INTEGERS 24 (2024), Article A61, Lemma 7.

TheoremStrongly \(2\)-near perfect numbers of the form \(2^k p\)#●ready to prove

Let \(p\) be an odd prime, let \(k\) be a natural number and let \(n = 2^k p\). Suppose that \(n\) is strongly \(2\)-near perfect, that is, \(\sigma(n) = 2n + d + n/d\) for some divisor \(d\) of \(n\), where \(\sigma\) is the sum-of-divisors function. Then there is a natural number \(a \le k\) with \(p(1 + 2^{k-a}) = 2^{k+1} - 2^a - 1\), that is, \(p = (2^{k+1} - 2^a - 1)/(1 + 2^{k-a})\).

public theorem_wanted strong_two_near_perfect_two_pow_mul_prime_form {k p : ℕ}
    (hp : p.Prime) (hp2 : p ≠ 2)
    (hstrong : ∃ d, d ∣ 2 ^ k * p ∧
      ArithmeticFunction.sigma 1 (2 ^ k * p) = 2 * (2 ^ k * p) + d + 2 ^ k * p / d) :
    ∃ a ≤ k, (p : ℤ) * (1 + 2 ^ (k - a)) = 2 ^ (k + 1) - 2 ^ a - 1
Sources
  • Vedant Aryan, Dev Madhavani, Savan Parikh, Ingrid Slattery and Joshua Zelinsky, On 2-Near Perfect Numbers, INTEGERS 24 (2024), Article A61, Lemma 8.

TheoremPrime solutions of \(p(1 + 2^{k-a}) = 2^{k+1} - 2^a - 1\) satisfy \(k < 2a\)#●ready to prove

Let \(a \le k\) be natural numbers and let \(p\) be a prime with \(p(1 + 2^{k-a}) = 2^{k+1} - 2^a - 1\), that is, \(p = (2^{k+1} - 2^a - 1)/(1 + 2^{k-a})\). Then \(k < 2a\).

The source also assumes that \(n = 2^k p\) is strongly \(2\)-near perfect, that is, that \(\sigma(n) = 2n + d + n/d\) for some divisor \(d\) of \(n\), where \(\sigma\) is the sum of divisors. This follows from the equation for \(p\) by taking \(d = 2^a\).

public theorem_wanted lt_two_mul_of_strong_two_near_perfect_form {a k p : ℕ} (hp : p.Prime)
    (hak : a ≤ k) (hform : (p : ℤ) * (1 + 2 ^ (k - a)) = 2 ^ (k + 1) - 2 ^ a - 1) :
    k < 2 * a
Sources
  • Vedant Aryan, Dev Madhavani, Savan Parikh, Ingrid Slattery and Joshua Zelinsky, On 2-Near Perfect Numbers, INTEGERS 24 (2024), Article A61, Lemma 9.

TheoremIf \(b \ge 1\) and \(2^b + 1\) divides \(3 \cdot 2^a + 1\), then \(b = 2\)#●ready to prove

Let \(a\) and \(b\) be natural numbers with \(b \ge 1\). If \(2^b + 1\) divides \(3 \cdot 2^a + 1\), then \(b = 2\).

The source prints the lemma with no range for \(b\); its proof divides \(a\) by \(b\), so \(b \ge 1\) is assumed here. The hypothesis is needed: for \(b = 0\), \(2^0 + 1 = 2\) divides \(3 \cdot 2^0 + 1 = 4\).

public theorem_wanted eq_two_of_two_pow_add_one_dvd {a b : ℕ} (hb : 1 ≤ b)
    (h : 2 ^ b + 1 ∣ 3 * 2 ^ a + 1) : b = 2
Sources
  • Vedant Aryan, Dev Madhavani, Savan Parikh, Ingrid Slattery and Joshua Zelinsky, On 2-Near Perfect Numbers, INTEGERS 24 (2024), Article A61, Lemma 10.

Theorem\(2\)-near perfect \(2^k p^2\) whose omitted divisors are powers of \(2\)#●ready to prove

Let \(p\) be an odd prime, let \(k\) be a natural number and let \(i \ne j\) be natural numbers with \(i \le k\) and \(j \le k\). If \(n = 2^k p^2\) satisfies \(\sigma(n) = 2n + 2^i + 2^j\), where \(\sigma\) is the sum-of-divisors function (that is, \(n\) is \(2\)-near perfect with omitted divisors \(2^i\) and \(2^j\)), then \(n = 18\) and either \(i = 0, j = 1\) or \(i = 1, j = 0\), so the omitted divisors are \(1\) and \(2\).

public theorem_wanted two_near_perfect_two_pow_mul_prime_sq_case_one {k p i j : ℕ}
    (hp : p.Prime) (hp2 : p ≠ 2) (hi : i ≤ k) (hj : j ≤ k) (hij : i ≠ j)
    (hσ : ArithmeticFunction.sigma 1 (2 ^ k * p ^ 2) = 2 * (2 ^ k * p ^ 2) + 2 ^ i + 2 ^ j) :
    2 ^ k * p ^ 2 = 18 ∧ ((i = 0 ∧ j = 1) ∨ (i = 1 ∧ j = 0))
Sources
  • Vedant Aryan, Dev Madhavani, Savan Parikh, Ingrid Slattery and Joshua Zelinsky, On 2-Near Perfect Numbers, INTEGERS 24 (2024), Article A61, Proposition 1.

TheoremNo \(2\)-near perfect \(2^k p^2\) with omitted divisors \(2^i\) and \(p\) or \(p^2\)#●ready to prove

Let \(p\) be an odd prime, let \(k\) and \(i\) be natural numbers with \(i \le k\), let \(m\) be \(1\) or \(2\), and let \(n = 2^k p^2\). Then \(\sigma(n) \ne 2n + 2^i + p^m\), where \(\sigma\) is the sum-of-divisors function; that is, \(n\) is not \(2\)-near perfect with omitted divisors \(2^i\) and \(p^m\).

public theorem_wanted not_two_near_perfect_two_pow_mul_prime_sq_case_two {k p i m : ℕ}
    (hp : p.Prime) (hp2 : p ≠ 2) (hi : i ≤ k) (hm : m = 1 ∨ m = 2) :
    ArithmeticFunction.sigma 1 (2 ^ k * p ^ 2) ≠ 2 * (2 ^ k * p ^ 2) + 2 ^ i + p ^ m
Sources
  • Vedant Aryan, Dev Madhavani, Savan Parikh, Ingrid Slattery and Joshua Zelinsky, On 2-Near Perfect Numbers, INTEGERS 24 (2024), Article A61, Proposition 2.

Theorem\(2\)-near perfect \(2^k p^2\) with omitted divisors \(1\) and \(2^b p^2\)#●ready to prove

Let \(p\) be an odd prime and let \(b, k\) be integers with \(1 \le b \le k\). If \(n = 2^k p^2\) satisfies \(\sigma(n) = 2n + 1 + 2^b p^2\), where \(\sigma\) is the sum-of-divisors function (that is, \(n\) is \(2\)-near perfect with omitted divisors \(1\) and \(2^b p^2\)), then \(n = 36\) and \(2^b p^2 = 18\).

public theorem_wanted two_near_perfect_two_pow_mul_prime_sq_case_three {k p b : ℕ}
    (hp : p.Prime) (hp2 : p ≠ 2) (hb : 1 ≤ b) (hbk : b ≤ k)
    (hσ : ArithmeticFunction.sigma 1 (2 ^ k * p ^ 2) =
      2 * (2 ^ k * p ^ 2) + 1 + 2 ^ b * p ^ 2) :
    2 ^ k * p ^ 2 = 36 ∧ 2 ^ b * p ^ 2 = 18
Sources
  • Vedant Aryan, Dev Madhavani, Savan Parikh, Ingrid Slattery and Joshua Zelinsky, On 2-Near Perfect Numbers, INTEGERS 24 (2024), Article A61, Proposition 3.

Theorem\(2\)-near perfect \(2^k p^2\) with omitted divisors \(p^m\) and \(2^b p^j\)#●ready to prove

Let \(p\) be an odd prime, let each of \(m\) and \(j\) be \(1\) or \(2\), and let \(b, k\) be integers with \(1 \le b \le k\). If \(n = 2^k p^2\) satisfies \(\sigma(n) = 2n + p^m + 2^b p^j\), where \(\sigma\) is the sum-of-divisors function (that is, \(n\) is \(2\)-near perfect with omitted divisors \(p^m\) and \(2^b p^j\)), then \(n = 200\).

public theorem_wanted two_near_perfect_two_pow_mul_prime_sq_case_five {k p m b j : ℕ}
    (hp : p.Prime) (hp2 : p ≠ 2) (hm : m = 1 ∨ m = 2) (hb : 1 ≤ b) (hbk : b ≤ k)
    (hj : j = 1 ∨ j = 2)
    (hσ : ArithmeticFunction.sigma 1 (2 ^ k * p ^ 2) =
      2 * (2 ^ k * p ^ 2) + p ^ m + 2 ^ b * p ^ j) :
    2 ^ k * p ^ 2 = 200
Sources
  • Vedant Aryan, Dev Madhavani, Savan Parikh, Ingrid Slattery and Joshua Zelinsky, On 2-Near Perfect Numbers, INTEGERS 24 (2024), Article A61, Proposition 4.

TheoremClassification of prime solutions of \(p(1 + 2^{k-a}) = 2^{k+1} - 2^a - 1\)#●ready to prove

Let \(a \le k\) be natural numbers and let \(p\) be a prime with \(p(1 + 2^{k-a}) = 2^{k+1} - 2^a - 1\), that is, \(p = (2^{k+1} - 2^a - 1)/(1 + 2^{k-a})\). Then \(k = a + 2\), \(2^k p / 2^a = 4p\), \(5p = 2^{a+3} - 2^a - 1\), and \(a \equiv 3 \pmod 4\).

The source states this for \(n = 2^k p\) strongly \(2\)-near perfect, that is, with \(\sigma(n) = 2n + d + n/d\) for some divisor \(d\) of \(n\), where \(\sigma\) is the sum of divisors. That hypothesis follows from the equation for \(p\) by taking \(d = 2^a\).

public theorem_wanted strong_two_near_perfect_two_pow_mul_prime_classification {a k p : ℕ}
    (hp : p.Prime) (hak : a ≤ k)
    (hform : (p : ℤ) * (1 + 2 ^ (k - a)) = 2 ^ (k + 1) - 2 ^ a - 1) :
    k = a + 2 ∧ 2 ^ k * p / 2 ^ a = 4 * p ∧ (5 * p : ℤ) = 2 ^ (a + 3) - 2 ^ a - 1 ∧
      a % 4 = 3
Sources
  • Vedant Aryan, Dev Madhavani, Savan Parikh, Ingrid Slattery and Joshua Zelinsky, On 2-Near Perfect Numbers, INTEGERS 24 (2024), Article A61, Proposition 5.
8 fully proved44 ready to prove

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