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Pascals triangle — the series

6 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. Pascal's triangle mod 2, 32 rows. Only the odd entries are drawn; the pattern that appears is the Sierpiński triangle.

    Pascal's triangle, in two colours

    Shade the odd numbers in Pascal's triangle and a fractal appears. Nothing was designed to produce it, and the same shape arrives independently from a completely different construction.

    part 1 · discrete
  2. A run down a diagonal, and the entry it adds to. 9 rows of Pascal's triangle with 5 entries shaded and the entry they add to marked. The claim is checked by adding the shaded entries: 1 + 3 + 6 + 10 + 15 = 35.

    The run that lands one place along

    Add up a run of entries down one of Pascal's diagonals and the total is another entry of the triangle — one row further down and one place along. The same triangle holds four more sums of that kind, and each is a different question answered by the same additive rule.

    part 2 · discrete
  3. 35 routes across a 4 by 3 grid. A grid with each cell holding the number of monotone routes reaching it. The far corner holds 35, which is the binomial coefficient of 7 choose 3.

    Every entry counts the routes to it

    Turn Pascal's triangle forty-five degrees and it becomes a grid of street corners, with each entry counting the ways of walking there. Identities between the entries then become statements about routes, and the statements are proved by cutting the routes in one place.

    part 3 · discrete
  4. 3 carries in base 2, and 2 divides it 3 times. The addition of 5 and 7 written in base 2, column by column, with the carries marked. There are 3, and 2 divides the binomial coefficient 792 exactly 3 times.

    The carries decide the divisibility

    How many times a prime divides a binomial coefficient is not a fact about the coefficient at all. It is a count of the carries that happen when two numbers are added in that prime's base, which is a question about column addition and has nothing to do with choosing anything.

    part 4 · discrete
  5. Pascal's triangle modulo 4, where one digit at a time is not enough. 32 rows of Pascal's triangle coloured by remainder modulo 4 — hue for the last base-2 digit, depth for the second. The digit-by-digit product that gives every remainder modulo 2 gets the remainder modulo 4 wrong at 100 of the 243 entries 2 does not divide.

    A remainder read two digits at a time

    Lucas' theorem reads a binomial coefficient's remainder on division by a prime off its digits one at a time. On division by the prime's square the same reading is wrong at four odd entries in ten. What replaces it still reads digits — in overlapping pairs, with the prime taken out first and a sign that the carries decide.

    part 5 · discrete
  6. A polygon of 4 points averaged down to its curve at t = 0.40. A control polygon of 4 points, the 3 rounds of weighted averaging at t = 0.40 drawn as nested polylines, the single point they end at, and the whole curve those points trace. The weights on the control points are 0.216, 0.432, 0.288, 0.064.

    Averaging down the triangle

    Change one word in the rule that builds Pascal's triangle — take a share of each entry above instead of adding them — and the triangle stops counting and starts averaging. The same rule then draws smooth curves from polygons and approximates every continuous function by polynomials, at a rate that no amount of smoothness can improve.

    part 6 · discrete

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