FIGURATE NUMBER A figurate number is represention of number as a regular geometric pattern, say, of dots. If the pattern is polytopic, the figurate is labeled a polytopic number.
The ith polytopic number has the formula: P_i(n) = (n + i = 1)!/n!(i - 1)!, for n = 1, 2, 3, ..., where n! is the [[factorial]]. The first three polytopic numbers are:

P₂(n) = ½n(n + 1). triangular numbers

P₃(n) = 1/6 n(n + 1)(n + 2), tetrahedral numbers

P₄(n) = 1/24 n(n + 1)(n + 2)(n + 3), pentatopic numbers
.
Our present words "square number" and "cubic number" derive labeling from their representation as a square or cube.
Another label for these numbers is '''Pythgorean geometry''', since [[Pythagoras]] is credited with initiating them, with the notion that these numbers are generated from a [[gnomon]] or basic unit.
For example, the gnomon of the square number is the odd number, of the general form 2n + 1, n = 1, 2, 3, ... . This is easily demonstrated as follows.
8 8 8 8 8 8 8 8 8 7 7 7 7 7 7 7 8 7 6 6 6 6 6 6 8 7 6 5 5 5 5 5 8 7 6 5 4 4 4 4 8 7 6 5 4 3 3 3 8 7 6 5 4 3 2 2 8 7 6 5 4 3 2 1
To transform from n-square to (n + 1)-square (say, 7-square to 8-square):

to the n-square (say 7-square), adjoin (n + 1)-rows (8-rows) of elements;
also, (n + 1)-columns (8-columns);
so the adjunctions are (n + 1) + (n + 1) = 2n + 2 = 2(n+1) elements (say, 2x8=16);
the square pattern is completed by adjoining a corner element (underlined above, say 8);
thus, to n-square, one has adjoined 2(n+1) + 1 elements (say, 16+1=17), which is an odd number;
so the gnomon of any square is the odd number: 1, 3, 5, ... .
The above square is 8 x 8 = 64 elements. And note the sum of the first 8 odd numbers:
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 64.
Conversely, one can calculate the square root of any number by subtracting odd numbers. Thus, 64 - 1 = 63; 63 - 3 = 60; 60 - 5 = 55; 55 - 7 = 48; 48 - 9 = 39; 39 - 11 = 28; 28 - 13 = 15; 15 - 15 = 0. The subtraction of the first 8 odd numbers from 64 yields 0; hence, the square-root of 64 is 8.
The tedium of increasing number of subtractions as number grows is short-cut by a method similar to the standard way of square-rooting taught in school. For example: 1225 = 35 x 35, Note the sum of the digits of this square root: 3 + 5 = 8. This shortcut reduces 35 subtractions to only 8 subtractions. Etc.
The shortcut involves two "tricks": a markoff trick, and resumptive trick.
The markoff trick is already known from the familiar squareroot algorithm. One marks off the target number in pairs of digits, from the right, as in marking 1225 as 12'25; then, calculation begins with the first digit-pair to the left. The reason is that squaring a one-digit number results in a 1- or 2-digit square. Thus, 1, 2, 3 have, respectively, the 1-digit squares of 1, 4, 9. But 4 has the 2-digit square of 16; and numbers 5, 6, 7, 8, 9 have 2-digit squares. To allow for this, one begins with two digits to provide one digit at each process-stage.
The resumptive trick (unique to this present algorithm) shifts from one pair of target number digits to its next (rightward) two digits, explained in calculating the square root of 1225.

Mark off 1225 as 12'25; begin calculation with left pair of digits, namely, 12.
Begin subtracting odd numbers: 12 - 1 = 11; 11 - 3 = 8; 8 - 5 = 3; but the next odd number, 7, cannot be subtracted, leaving the nonzero [[Difference]] of 3, invoking the resumptive trick.
The left-most digit of the square root, 3, actually representing 30, because the second digit from the right in [[Decimal numeration]] is the "tens digit".
To difference 3 (= 8 - 5), adjoin next two marked off digits (25), obtaining 325, and resume odd number subtraction.
The last "successful" subtrahend was 5; but the next odd number, 7, cannot be subtracted, so '''interpolate''' between 5 and 7 for number 6. (This is "first part" of the resumptive trick.)
Since (noted above) the successful 3 subtractions actually represent the 2-digit 30, treat the interppolated 6 as 60; resume odd number subtraction with the first odd number in the sixities, namely, 61. (This is "seond and final part" of the resumptive trick or subalgorithm.)
We find 325 - 61 = 264; 264 - 63 = 201; 201 - 65 = 136; 136 - 67 = 69; 60 - 69 = 0.
Having passed from 325 to 0 by five subtractions, the second digit is 5: and 30 + 5 - 35, that is, the square root of 1225 is 35, obtained in exactly 3 + 5 = 8 subtractions by applying the markoff and resumptive tricks or subalgorithms.
To see again, consider 144 = (12)^{2. The square-root is easily calculated by twelve
subtractions: 144 - (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25) = 144 - 144 = 0.
However, the mark-off and resumptive tricks reduce this to 1 + 2 = 3 odd number subtractions.
Markoff 144 as 1'44.
Starting with left-most pair, start subtraction 1 - 1 = 0; so leftmost digit of square-root
is 1, representing 10.
Bring down second pair of digits: 0 + 44 = 44 and beging subtracting odd numbers.
Interpolate between "successful" 1, "failed" 3, namely, 2, to represent the twenties,
whose first odd number is 21, so resuming the subtraction with 21.
44 - 21 = 23; 23 - 23 = 0, resulting in two subtractions, so second digit is 2: 10 + 2 =
12 as square-root of 144, obtained by 1 + 2 = 3 sbubtracting of odd numbers.
A '''special case''' involves a '''zero difference''', illustrated in 10² = 100.
Marking off 100 as 1'00, then 1 - 1 = 0 (difference), which combines with the second pair of
digits as 000, But this in decimal notation is simply 0, resulting in 10 as square root,
achieving in 1 + 0 = 1 subtraction. Another example is 20² = 400. Marking off 400 as
4'00. then 4 - (1 + 3) = 0 (difference), which combines with second pair of digits as 0000,
representing 0, yielding root 20, obtaine by 2 + 0 = 2 subtractions, i.e., 4 - (1 + 3).
Cubes of natural numbers or positive integers can be generated from S = 1, 3, 5, 6, 7, ...,
2n + 1, ...., n = 1,2, 3, ..., by "moving sums", similar to the "moving averages" of [[Statics]]:

First member of S: 1 = 1³.
Mext two members of S: 3 + 5 = 8 = 2³.
Next three members of S: 7 + 9 + 11 = 27 = 3³.
Next four members of S: 13 + 15 + 17 + 19 = 64 = 4³.
Next five of S: 21 + 23 + 25 + 27 + 29 = 125 = 5³.
Next six of S: 31 + 33 + 35 + 37 + 39 + 41 = 216 = 6³.
Next seven of S: 43 + 45 + 47 + 49 + 51 + 53 + 55 = 343 = 7³.
Etc.

Obviously, "moving differences" of S yield cube-roots. Is there a shortcut? Can S be manipulated
generate fourth powers?

This procedure (taking many words to explain, but quickly executed) is not restricted to
calculating suqare roots of natural numbers or positive inegers. It can even be applied toward
calculating the irrational square root of 2, to any number of decimal places. Etc.

Children construct figurate numbers from pebbles, bottle caps, etc.

As a bonus, children can use figurate numbers to discover the Commutative Law and
associative Law for addition and multiplication -- laws
usually dictated to them -- by buiding rows and tables of dots.

Using asterisks in place of dots or bottle caps or pebbles the additive commutativity of
2 + 3 = 3 + 2 = 5 becomes:
* * | * * * --> * * * | * * --> * * * * *
And the multiplicative commutativity of 2 X 3 = 3 X 2 = 6
becomes:
* * --> * * * --> * * * * * *
* * * * *
* *
The concepts of figurate numbers and gnomon implicitly anticipate the modern concept of}