Partitions with distinct even parts

TheoremCongruences modulo 24 for partitions with distinct even parts#●ready to prove

Write \(\mathrm{ped}(m)\) for the number of partitions of \(m\) in which the even parts are distinct (odd parts may repeat). For every natural number \(n\), each of \(\mathrm{ped}(225n+43)\), \(\mathrm{ped}(225n+88)\), \(\mathrm{ped}(225n+133)\) and \(\mathrm{ped}(225n+223)\) is congruent to \(0\) modulo \(24\).

public theorem_wanted partitionsEvenPartsDistinct_mod24_progressions (n : ℕ) :
    Nat.ModEq 24 (distinctEvenPartsCount (225 * n + 43)) 0 ∧
      Nat.ModEq 24 (distinctEvenPartsCount (225 * n + 88)) 0 ∧
      Nat.ModEq 24 (distinctEvenPartsCount (225 * n + 133)) 0 ∧
      Nat.ModEq 24 (distinctEvenPartsCount (225 * n + 223)) 0

Sources

• Hemjyoti Nath, New Congruences for Partitions Where the Even Parts Are Distinct, INTEGERS 24 (2024), Article A79, Theorem 1.1.

Uses 1

Statement usesPartitions whose even parts are distinct

TheoremPower-of-five congruence modulo 24 for partitions with distinct even parts#●ready to prove

Write \(\mathrm{ped}(m)\) for the number of partitions of \(m\) in which the even parts are distinct (odd parts may repeat). For all natural numbers \(k \ge 1\) and \(n\),

\[ \mathrm{ped}(9n+7) \equiv \mathrm{ped}\left(9 \cdot 5^{2k} n + \frac{57 \cdot 5^{2k} - 1}{8}\right) \pmod{24}. \]

The quotient is exact, since \(57 \cdot 5^{2k} - 1\) is divisible by \(8\).

public theorem_wanted partitionsEvenPartsDistinct_mod24_five_power
    (k n : ℕ) (hk : 1 ≤ k) :
    Nat.ModEq 24 (distinctEvenPartsCount (9 * n + 7))
      (distinctEvenPartsCount
        (9 * 5 ^ (2 * k) * n + (57 * 5 ^ (2 * k) - 1) / 8))

Sources

• Hemjyoti Nath, New Congruences for Partitions Where the Even Parts Are Distinct, INTEGERS 24 (2024), Article A79, Theorem 1.1.

Uses 1

Statement usesPartitions whose even parts are distinct

2 ready to prove

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