jump to chart of x^3 connecting x^2 and x.

Triangular or Tetraktys Number

Call the xth Triangular / Tetraktys number:  T_x . 

The formula for the xth triangular number is (x(x+1))/2.  The Pythagorean picture of tetraktys number looks like this:

l

ll

lll

llll

A triangular number is the total number of dots including the xth row.  Rows have dots equaling the positive integers.

Click here for the difference of x^2 and x^3 and Catalan's Conjecture.

How the power of 3 relates to triangular numbers squared:

Shells of squares add up to squares.  However, when shells of squares are treated in triangular-number-amounts of shells, this is the implied organization:

+1 . . .

+3+5 . . .

+7+9+11 . . .

+13+15+17+19 . . . 

+21+23+25+27+29 . . .

+31+33+35+37+39+41 . . .

etc.

Each row x of this newly conceived layout equals exactly x^3!

Row 1 through x also equals a set of power-of-two-shells that is T_x^th-squared's shells.

So treating shells-of-the-power-of-2

--each and every xth row will equal x^3 and

--will parallel xth integer because it is x number of shells-of-2 per xth row

--while the row by row cumulative total is the triangular-numbers-squared (as long as all rows from the top are included.)

To algebraically prove the relationship, see the Euler/Pascal weighting formulas that create shells, powers and sums of powers which define the following:

Euler/Pascal weighting for shells-of-x^3 series:

1...[plus 0+0(4)+1]...[0+0(4)+1] = 1 (sums to 1)

+1+7...[plus 0+1(4)+3]...[0+1(4)+4] = 8 (sums to 9)

+1+7+19...[plus 1+3(4)+6]...[1+4(4)+10] = 27 (sums to 36)

+1+7+19+37...[plus 3+6(4)+10]...[4+10(4)+20] = 64 (sums to 100)

+1+7+19+37+61...[plus 6+10(4)+15]...[10+20(4)+35] = 125 (sums to 225)

+1+7+19+37+61+91...[plus 10+15(4)+21]...[20+35(4)+56] = 216 (sums to 441)

etc.

Euler/Pascal weighting for SHELLS THAT ARE x^3

that are simultaneously shells of triangular numbers squared:

1...[plus 0+0(4)+1] (+1) ...sums to 1

+3+5...[plus 0+1(4)+4] (+8) ...sums to 9

+7+9+11...[plus 1+4(4)+10] (+27) ...sums to 36

+13+15+17+19...[plus 4+10(4)+20] (+64) ...sums to 100

+21+23+25+27+29...[plus 10+20(4)+35] (+125) ...sums to 225

+31+33+35+37+39+41...[plus 20+35(4)+56] (+216) ...sums to 441

etc.

Euler/Pascal weighting for the summations-of-x^3 series:

1...[0+0(4)+1]...[0+0(4)+1] = 1

1+8...[plus 0+1(4)+4]...[0+1(4)+5] = 9

1+8+27...[plus 1+4(4)+10]...[1+5(4)+15] = 36

1+8+27+64...[plus 4+10(4)+20]...[5+15(4)+35] = 100

1+8+27+64+125...[plus 10+20(4)+35]...[15+35(4)+70] = 225

1+8+27+64+125+216...[plus 20+35(4)+56]...[35+70(4)+126] =441

etc.

The x^2 series' Euler/Pascal weighting is two neighboring triangular numbers weighted at Euler's row two, which means multiplied by 1 and 1 (so the "weight" is moot).

Every (x(x+1))/2 ^th of an y^2 will also have a "triangular-number-squared" definition which would be the 1st through xth summed of x³ because squared triangular numbers may be defined by EITHER shells for the power of 2 or by shells that are x³.

Shells and Euler/Pascal weighting for x^2 series per row:

1...[0+1]...[equals 0+0(4)+1] = 1

1+3...[1+3] . . . . . . . . . . . . . . . .

1+3+5...[3+6]...[equals  0+1(4)+5] = 9

1+3+5+7...[6+10] . . . . . . . . . . . . . . . .

1+3+5+7+9...[10+15] . . . . . . . . . . . . . . . . .

1+3+5+7+9+11...[15+21]...[equals 1+5(4)+15] = 36

1+3+5+7+9+11+13...[21+28] . . . . . . . . . . . . . . . . .

1+3+5+7+9+11+13+15...[28+36] . . . . . . . . . . . . . . . . .

1+3+5+7+9+11+13+15+17...[36+45] . . . . . . . . . . . . . . . . .

1+3+5+7+9+11+13+15+17+19...[45+55]...[equals 5+15(4)+35] = 100

etc.

Imagine taking some (x(x+1))/2 ^th row of one of an above x^2 value and subtracting the previous (x-1(x))/2 ^th row's value.

How apparently the procedure would automatically cull to precisely the certain correct set of odd numbers necessary to form a number from the x^3 series!  With y difference between one triangular number and the previous triangular number, the value of the difference of the two triangular numbers if they are both squared will be exactly y^3.

This is how integers, squares and cubes are entwined:

There is an amazing truism that x^3 is a certain set of x, consecutive, odd numbers:  the ((x(x-1))/2 +1) ^th to the (x(x+1)/2 ^th of the odd numbers equal exactly x^3.

The best way to grasp how the power of 3 manifests in parallel across definitions is to translate the definitions into their relation to sets of sixes.  See the blue topics:

VERY INTERESTINGLY, WHY CUBES ARE SETS OF SHELLS OF THE POWER OF TWO:

All cubes can be parsed out as a set of consecutively odd numbers.  The reason is the definition of x^3 has to do with accumulating 6's so that x^3 = 6*(x-1)-tetrahedral-number + x.  The cumulative value of x^3 transposes, through the character of odd-number-shells-of-the-power-of-2, into accumulated triangular sets of triangular numbers so that the x^2 values achieved by summed x^3 will be each of the triangular numbers, squared.

1^2 + 2^3 = 3^2

3^2 + 3^3 = 6^2 

6^2 + 4^3 = 10^2 

10^2 + 5^3 = 15^2

et cetera. 

Cubes being parse-able as shells-of-the-power-of-2 are the source of triangular-numbers-squared having shells that are cubes. 

Because cubes parse out to consecutive sets of odd numbers they can always be placed with some 1st through ath shells of the power of 2 and accumulate to c^2. 

Between two cases of neighboring (x-1)th and xth triangular-numbers there are x non-triangular integer values that could each be squared--each of which would increase by one shell-of-the-power-of-2, building up x shells from the ((x(x-1))/2 +1)th to the ((x(x+1)/2)th which is x^3.

Comparing

Integers at each row (A) and their summation (B), row to row creating triangular numbers

To

(C) are integers--jacked up as shells-of-the-power-of-2, accumulated in each row--that equal x^3 per row.  (D) is the summation of x^3, row added to all previous rows, and will be triangular numbers squared.

The Reason:

(C) is due to the fact that the powers of 3 are 6*(x-1)th tetrahedral + the xth integer.  See Euler/Pascal definitions.

A single (x-1)th tetrahedral number is the 1st through (x-1)th triangular numbers.  Six of these will equal 6*each triangular number--1st to the (x-1)th--plus the 1st through (x-1)th of x (the rest of the definition for x^3).  Neighboring triangular numbers define squares too.

Just as the tetraktys triangle gains one "1" per row for increasing integer amounts, so the parallel definition of x^3-that-is-a-shell gains "one per row x".  The amount gained is ((xth triangular)*6 plus 1).

Integer rows of (A) sum to triangular numbers in (B).  That creates the idea that (C) could be analyzed as sets of shells of squares that parallel (A).   The 6 sets of the 1 through (x-1)th triangular numbers plus 1 through (x-1)th of integers in (C) could be analyzed as sets of shells of squares via the definition:  2*(y-1)th integer plus yth one. 

By the shells-of-squares definition, each integer of each row would have 2 sets of (y-1) integers and one yth one.  To transfer x^3 values into shells-of-squares-values, think:

(C) shells of power of 2 . . . considered per row of a tetraktys-triangle layout (different colors indicate row extents)

(Squares are (C) summed )

But (D) is every x^3 summed (which was every row of (C))

For (D), the 2 * ((xth triangular) -1)th of the triangular numbers) plus one more xth triangular [which are the integers from 1 to xth summed] will be the total of the sums of 1 through x of x^3.

These double sets of 1st to (x-1)th triangular numbers, along with an extra xth triangular number, will form the 1st through xth sets of x^2, which also defines by neighboring triangular numbers adding together, squares by the Euler/Pascal/Worpitzky rules.

Since (D) only accumulated triangular sets of triangular numbers, the x^2 values achieved will be each of the triangular numbers, squared.

That is why (D) which is a series of the sums of 1 through x of x^3 is also triangular numbers squared.  And that is how cubes have a role as shells-TO-squares and how the shells OF squares form cubes.

One other B) to D) observation:

Obviously, the xth triangular number within (D) will, by pattern, be multiplied by integers that, in total, will also be equivalent to the very xth triangular number value that they are multiplying.  Of course these are triangular numbers squared--in the end the integers all combine to be an xth value triangular number * the same xth value that has been being multiplied.

For assessing portions or sets of triangular numbers, including the entire sets, or:

When x > a > 0 and x is the xth triangular number,

T_x - T_{x - a}  = (x - (a - (a - a/2 + 1))) a - a/2

(which looks like . . .)

Chart of the difference in values of the xth triangular number subtracting all previous triangular numbers except the "a" previous triangular numbers:

ax - (a(a-1))/2 = the sum from the (x-(a-1))th through the xth triangular numbers when a = the number of triangular numbers to compute and x is the xth triangular number from which to compute.

Send mail to Cecilia@noticingnumbers.net with questions or comments about this web site.
Copyright © 2004 noticingnumbers.net
Last modified: 12/16/05