Integer Partition Generator

Name: Integer Partition Generator
Availability: InStock
Rating: 4.9 (506 reviews)
Author: Generatr

n = λ₁ + λ₂ + ...

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4.9(506 reviews)

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p(n)

Integer partitions

n (1–30)

p(8)

Distinct: 6Odd parts: 6

Partitions

7 + 1

6 + 2

6 + 1 + 1

5 + 3

5 + 2 + 1

5 + 1 + 1 + 1

4 + 4

4 + 3 + 1

4 + 2 + 2

4 + 2 + 1 + 1

4 + 1 + 1 + 1 + 1

3 + 3 + 2

3 + 3 + 1 + 1

3 + 2 + 2 + 1

3 + 2 + 1 + 1 + 1

3 + 1 + 1 + 1 + 1 + 1

2 + 2 + 2 + 2

2 + 2 + 2 + 1 + 1

2 + 2 + 1 + 1 + 1 + 1

2 + 1 + 1 + 1 + 1 + 1 + 1

1 + 1 + 1 + 1 + 1 + 1 + 1 + 1

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About Integer Partition Generator

An integer partition generator listing all ways to write n as a sum of positive integers (order doesn't matter). Shows partition count p(n), Young diagrams, conjugate partitions, and distinct/odd parts. All calculations are client-side.

Integer Partition Generator Features

All partitions
Count p(n)
Young diagram
Conjugate
Distinct/odd

Integer partitions: ways to write n as sum of positive integers (order doesn't matter). p(4)=5: 4, 3+1, 2+2, 2+1+1, 1+1+1+1. The partition function p(n) grows exponentially: p(100)=190,569,292,356. Euler's pentagonal theorem gives a recurrence.

How to Use

Enter n:

Partitions: All representations
Count: p(n)
Types: Distinct, odd parts

Euler's Contributions

Euler proved: # partitions with distinct parts = # partitions with odd parts. Generating function: Σp(n)xⁿ = Π1/(1−xᵏ). Pentagonal theorem gives recurrence for p(n).

Young Diagrams

Each partition has a Young diagram: rows of boxes. Conjugate: transpose the diagram. Self-conjugate partitions ↔ partitions with distinct odd parts.

Step-by-Step Instructions

1Enter n.
2View all partitions.
3Count p(n).
4See distinct parts.
5Explore Young diagrams.

Integer Partition Generator — Frequently Asked Questions

How fast does p(n) grow?+

Hardy-Ramanujan: p(n) ~ exp(π√(2n/3))/(4n√3). p(10)=42, p(50)=204,226, p(100)≈1.9×10¹¹. The growth is sub-exponential but faster than polynomial.

What is Euler's identity for partitions?+

The number of partitions of n into distinct parts equals the number into odd parts. Proof via generating functions: Π(1+xᵏ) = Π1/(1−x^(2k−1)). Example: p(5, distinct)=3 (5, 4+1, 3+2) = p(5, odd)=3 (5, 3+1+1, 1+1+1+1+1).

What are restricted partitions?+

Partitions with constraints: distinct parts (5=4+1, 3+2), parts ≤ k, exactly m parts, etc. Each restriction has its own generating function and combinatorial identities. Ramanujan discovered remarkable congruences like p(5n+4)≡0 (mod 5).

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