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Partitions — the series

7 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. The partition 5 + 4 + 2 + 1 and its conjugate. A row of dots for each part, and the same dots read down the columns instead.

    A diagram turned on its side

    Write a partition as rows of dots, then read the columns instead. Every theorem in this essay is that one move, and the move proves things that no formula suggests.

    part 1 · number
  2. The partition product's coefficients to q¹². A row of series coefficients computed by expanding a product, beside the same numbers obtained another way.

    Every partition, hidden in a product

    Multiply out one factor for each part size and the coefficient of q to the n is the number of partitions of n. Nothing is being approximated: the product is a bookkeeping device that does the counting by multiplying.

    part 2 · number
  3. The product of (1 − qᵏ), and what survives at 12. The coefficients of the pentagonal product drawn as signed bars, with the partitions into distinct parts that Franklin's move leaves unpaired.

    The terms that cancel almost everything

    Multiply out the product of 1 − q, 1 − q², 1 − q³ and so on, and nearly every coefficient is zero. What survives is a single plus or minus one at 1, 2, 5, 7, 12, 15 — and the reason is a way of pairing partitions off so that each pair cancels.

    part 3 · number
  4. p(n) to 60, against the Hardy–Ramanujan estimate. The number of partitions of each number up to sixty on a logarithmic scale, with the asymptotic estimate drawn over it and the ratio of the two tabulated.

    The size of a number with no formula

    There is no closed expression for the number of partitions of n. There is an expression for how large it is — with a square root in the exponent and a π in front — and it is accurate enough that rounding a few terms of its refinement gives the exact count.

    part 4 · number
  5. Every fifth partition count divides, and the rank that says why. A row of partition counts with the ones in a congruence class marked, and a histogram of partitions sorted by rank.

    Every fifth one divides

    p(4) is 5, p(9) is 30, p(14) is 135, and every partition count at a number leaving four on division by five is divisible by five. Ramanujan read it off a table; the explanation is a way of splitting those partitions into five equal heaps.

    part 5 · number
  6. Random partitions of 1,000, scaled, against their limit shape. The outlines of 4 uniformly random partitions of 1000, scaled by the square root of 1000, lying close to the curve e^(−cx) + e^(−cy) = 1 with c = π/√6.

    The shape a random partition takes

    There are about twenty-four thousand billion billion billion ways to write 1,000 as a sum of whole numbers. Pick one at random, draw its Ferrers diagram, shrink it by the square root of a thousand, and it is almost exactly the curve e^(−cx) + e^(−cy) = 1 with c = π/√6. So is the next one, and the next. A random partition of a large number has a shape, and the shape is known exactly.

    part 6 · number
  7. Two ways of counting that agree at every number. For n up to 40, the counts of partitions with gaps of at least two against partitions into parts congruent to 1 or 4 mod 5, on a logarithmic scale, equal at every n, with the second identity's counts beside them.

    Two counts that agree for no visible reason

    Write 10 as a sum of whole numbers that differ from each other by at least two, and there are six ways. Write 10 as a sum of numbers that each leave 1 or 4 on division by 5, and there are six ways. The same happens for 20 (thirty-one each), for 40 (three hundred and seventy-four each), for every number anyone has checked and every number there is. The two lists look nothing alike, and no one has found a simple way to turn one into the other.

    part 7 · number

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