Figurate number
From Academic Kids

A figurate number is a number that can be represented as a regular and discrete geometric pattern (e.g. dots). If the pattern is polytopic, the figurate is labeled a polytopic number, and may be a polygonal number or a polyhedral number.
The first few triangular numbers can be built from rows of 1, 2, 3, 4, 5, and 6 items:
*  *  *  *  *  *   * *  * *  * *  * *  * *    * * *  * * *  * * *  * * *     * * * *  * * * *  * * * *      * * * * *  * * * * *      * * * * * *
The nth regular rtopic number is given by the formula:
<math>P_r(n) = {{n + r  1} \choose r} = {n^{(r)} \over {r!}} \quad \mbox{for} \ n \ge 1<math>
<math>r!<math> is the factorial of <math>r<math>, <math>n \choose r<math> is a binomial coefficient, and <math>n^{(r)}<math> is the rising factorial.
Polytopic numbers for r = 2, 3, and 4 are:
 P_{2}(n) = 1/2 n(n + 1) (triangular numbers)
 P_{3}(n) = 1/6 n(n + 1)(n + 2) (tetrahedral numbers)
 P_{4}(n) = 1/24 n(n + 1)(n + 2)(n + 3) (pentatopic numbers)
Our present terms square number and cubic number derive from their geometric representation as a square or cube.
Contents 
Gnomon
Figurate numbers were a concern of Pythagorean geometry, since Pythagoras is credited with initiating them, and the notion that these numbers are generated from a gnomon or basic unit. The gnomon is the piece which needs to be added to a figurate number to transform it to the next biggest one.
For example, the gnomon of the square number is the odd number, of the general form 2n + 1, n = 1, 2, 3, ... . The square of size 8 composed of gnomons looks like this:
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 the nsquare (the square of size n) to the (n + 1)square, one adjoins 2n + 1 elements: one to the end of each row (n elements), one to the end of each column (n elements), and a single one to the corner. For example, when transforming the 7square to the 8square, we add 15 elements; these adjunctions are the 8s in the above figure.
Note that this gnomonic technique also provides a proof that the sum of the first n odd numbers is n^{2}; the figure illustrates 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 64 = 8^{2}.
Square roots
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 squareroot of 64 is 8.
The tedium of increasing number of subtractions as number grows is bypassed by a method similar to the standard way of squarerooting taught in school. For example: 1225 = 35 x 35, Note the sum of the digits of this square root: 3 + 5 = 8. This squareroot shortcut reduces 35 subtractions to only 8 subtractions. The shortcut involves two "tricks": a markoff trick, and resumptive trick.
The markoff trick is already known from the familiar square root 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 digitpair to the left. The reason is that squaring a onedigit number results in a 1 or 2digit square. Thus, 1, 2, 3 have, respectively, the 1digit squares of 1, 4, 9. But 4 has the 2digit square of 16; and numbers 5, 6, 7, 8, 9 have 2digit squares. To allow for this, one begins with two digits to provide one digit at each processstage.
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 from difference 3, so the resumptive trick is needed.
 The leftmost 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 2digit 30, treat the interpolated 6 as 60; resume odd number subtraction with the first odd number in the sixties, namely, 61. (This is "second and final part" of the resumptive trick or subalgorithm.)
 Result: 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 squareroot 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 markoff and resumptive tricks reduce this to 1 + 2 = 3 odd number subtractions.
 Markoff 144 as 1'44.
 Starting with leftmost pair, start subtraction 1  1 = 0; so leftmost digit of squareroot is 1, representing 10.
 Bring down second pair of digits: 0 + 44 = 44 and begin subtracting odd numbers.
 Interpolate between "successful" 1, "failed" 3, namely, 2, to represent the twenties, whose first odd number is 21, resuming the subtraction with 21.
 44  21 = 23; 23  23 = 0, resulting in two subtractions, so second digit is 2: 10 + 2 = 12 as squareroot of 144, obtained by 1 + 2 = 3 subtracting of odd numbers.
A special case involves a "zero difference", illustrated in 10^{2} = 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, achieved in 1 + 0 = 1 subtraction. Another example is 20^{2} = 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, achieved in 2 + 0 = 2 subtractions, i.e., 4  (1 + 3).
Cubes and cube roots
Cubes of natural numbers or positive integers can be generated from S = 1, 3, 5, 7, 9,..., 2n  1, ...; n = 1, 2, 3, ..., by "moving sums", similar to the "moving averages" of statistics:
 First member of S: 1 = 1^{3}.
 next two members of S: 3 + 5 = 8 = 2^{3}.
 Next three members of S: 7 + 9 + 11 = 27 = 3^{3}.
 Next four members of S: 13 + 15 + 17 + 19 = 64 = 4^{3}.
 Next five of S: 21 + 23 + 25 + 27 + 29 = 125 = 5^{3}.
 Next six of S: 31 + 33 + 35 + 37 + 39 + 41 = 216 = 6^{3}.
 Next seven of S: 43 + 45 + 47 + 49 + 51 + 53 + 55 = 343 = 7^{3}.
Thus, "moving differences" of S yield cuberoots.
This procedure (taking many words to explain, but quickly executed) is not restricted to calculating square roots of natural numbers or positive integers. It can even be applied toward calculating the irrational square root of 2, to any number of decimal places.
Demonstration of mathematical properties
School 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 building 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 * 3 = 3 * 2 = 6 becomes:
* * <> * * * <> * * * * * * * * * * * * *
Besides the subtractive method, the additive method can also approximate square roots of positive integers and solve quadratic equations.
The concepts of figurate numbers and gnomon implicitly anticipate the modern concept of recursion.
See also
References
 Gnomon, From Pharaohs to Fractals. Midhat J. Gazalé, Princeton University Press, Princeton, 1999.fr:Nombre figuré