Fascinating Figurate Numbers

In this article, we will cover the figurate numbers, which include the binomial coefficients (the triangular numbers, the tetrahedral numbers, the pentatopic numbers, and so on), and other polygonal numbers (the pentagonal numbers, the hexagonal numbers, the heptagonal numbers, and so on), as well as googolisms that I invented based on these numbers.

Triangular Numbers

The triangular numbers are the numbers that are sums of consecutive integers starting with 1. The first triangular numbers are

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210... These numbers get their name because a triangular number of objects can be arranged in a triangular shape. These are the numbers that can be seen in the third positions of the rows of Pascal's triangle, and the fraction 1/997002999 produces the triangular numbers in a repeating decimal (until they overlap):

0.000000001003006010015021028036045055066078091105120136153171190210231253276300325351378406435465...

The nth triangular number can be quickly calculated by n(n+1)/2. This was the shortcut Gauss used in the famous story where the class was asked to add all the whole numbers from 1 to 100.

Any number that is a sum of powers of 9 (a repunit in base 9) is triangular (in fact, the triangular root is a repunit in base 3):

1 + 9¹ = 10 (4th triangular number)

1 + 9¹ + 9² = 91 (13th triangular number)

1 + 9¹ + 9² + 9³ = 820 (40th triangular number)

1 + 9¹ + 9² + 9³ + 9⁴ = 7381 (121st triangular number)

1 + 9¹ + 9² + 9³ + 9⁴ + 9⁵ = 66430 (364th triangular number)

This is because (9ⁿ - 1)/8 = (3ⁿ - 1)(3ⁿ + 1)/8 = ((3ⁿ - 1)/2)(3ⁿ + 1)/4 = ((3ⁿ - 1)/2)((3ⁿ + 1)/2)/2 = M(M+1)/2 where M is (3ⁿ - 1)/2

Notice that the position of each of these numbers in the triangular number sequence is a repunit in base 3.

Googolisms I invented based on the triangular numbers

I named some googolisms based on triangular numbers. We will begin with the triangoogol, which is equal to 2 triangled 100 times. It is a number of exactly 163,489,217,887,235,173,465,886,368,894 digits. Below are the first and last 105 digits of the triangoogol. The first digits are calculated using logarithms (by starting with the first iterated triangle of 2 with more than the desired number of digits, taking its logarithm, multiplying by 2 to the power of the remaining number of triangles, then raising 10 to the fractional part), while the last digits are computed using modular arithmetic (with a difficulty I will explain later).

741579905318240666966034184074738593931118744313169686790005639068679627508588446895615847355004378100355...882120558152476660927697969785636092555

141276880773456396977294267827105086350274827345751276878885453031

If this number were written out (just for argument's sake) then the first comma would go after the first digit (i. e. 7,415,799,053,182,406,669,660,341,......,885,453,031).

The triangrol is equal to the number 2325779134965967427487810008002168938006567536111554301529048698796969115778520822121347163627529767530146944024732879347696758531031, or 2 triangled 10 times. It is the first iterated triangle of 2 to be greater than a googol.

The great triangrol is the first iterated triangle of 10 (and also 4) to be greater than googol, or the 8th iterated triangle of 10.

The triangrolplex is equal to 2 triangled 443 times. Its leading digits are 247591233247573347236447102420042348402140288166..., and it has about 2.929392929*10¹³² digits. It is the first iterated triangle of 2 to be greater than 10^triangrol. It ends with …11475081358932426146.

Below are more digits of the triangrolplex. These digits were computed using the same methods described earlier.

2475912332475733472364471024200423484021402881666258268129862250792053536430867493026944220713884699658946174384081131220868095248310331410389377591918268447638749483419874150167073211592197341053812089875487865877894448244695683944882025081933231263768499306075802756287747132059011093699426639701168882595057005656399650553062008721535538483646119048612665477929695641723270772116435288409269636789015648451905113001698027081220037936615804842572612157621842858471343364447018765386831879710550474084999938276630275370035989905074477238576243321719167496839378853051408339210162447452249180382458246442655953145939699982616899589368588086427317934236505449839781648025211366480849661655429815963161978723580945709711760263609075946029613837986154341545398074990577663588373592065681447137383732466

…

69081296564733072584956098237823010923240398681416473072596213471906899059854220978330768585392392092888901392944679111515126875947528097902828986547274874754472683597503038196501985037499011229997878043797453046217674840266791825682335629175964206849792727608733200826187943676031916303594497474895453551784425903080488960114409797282974498655838772764364155481947657247651631022586057756056638537140977861387597336323941957000867482138776433426381567493919918008131740973958550418763142125347281467178201835804314640480489070637746810598972954875156675179439190160099680241553686856992788873397380265841974793942230859338754761055776327138513103899211475081358932426146

I stopped at the triangrolplex because if we continued the pattern, after that it would not only be impossible to find the leading digits, but any digits. The last digits of iterated triangular numbers exhibit a nondeterministic tendency, with the parity of n triangled k times indeterminable without a complete computation of the moduli by powers of 2 of all iterated triangles of n up to k-1. For instance, it would be impossible to determine the parity of 2 triangled 1,000,000 times without knowing the outcome of 2 triangled 999,999 times mod 4, 999,998 times mod 8, 999,997 times mod 16, and so on. This is due to the coefficients of the polynomial expression denoting the n-th triangular number being 1/2.

I also defined box(n) as n triangled n times (or n boxed to 1 in the triangle, box, house, ... operator progression that I defined). Below are the values of some of the numbers that are equal to a number boxed.

1

6

231

1186570

347357071281165

2076895351339769460477611370186681

143892868802856286225154411591351342616163027795335641150249224655238508171

573249357811765072377666522361059195438648349138242383254432359289008321298201220014174056046893362736783713979710693432256284815101801226555769989930781799717441

19000479159436654182094538709078301648606160782962220459040910190503537900355586063528480304122816969675908529789983736849879883071931486211681488972464455547017098771687854065870488515885701949297209392059463594660134641914723617779945934233275476425426450532726956214657206353175342925158975647197432305812397497362391596445129949502257281332862850232369393970759981470200374118723859373135733302054874895905886944353296150415283667303327189958487311223414063173806827105293348288545895762349062809055653143550115728733817792159457120865530672397779581239572663154299380238092650157230645146677647225452141357523608492730207563870713722089967255484610633062891284574704274391143095428338586691237599416354264886557576085484922863609143965

Tetrahedral Numbers

Tetrahedral numbers are numbers that are sums of consecutive triangular numbers starting with 1. They get their name because a tetrahedral number of objects can be arranged into the shape of a tetrahedron. The first tetrahedral numbers are:

1, 4, 10, 20, 35, 56, 84, 120, 165, 220, 286, 364, 455, 560, 680, 816, 969, 1140, 1330, 1540, ...

The nth tetrahedral number can be quickly calculated by n(n+1)(n+2)/6. Unlike with the triangle, the last n digits of the kth tetrahedral number do not depend on whether the digit before the last n digits of k is even or odd.

There are only three tetrahedral numbers that are also square: 1, 4 (second tetrahedral number, 2^2), and 19600 (48th tetrahedral number, 140^2). There are also finitely many tetrahedral numbers that are also triangular.

The tetrahedral numbers are seen in the fraction 1/996005996001, or any fraction of the form 1/(10ⁿ - 1)⁴.

Googolisms I invented based on the tetrahedral numbers

I have also named googolisms based on the tetrahedral numbers. We will begin with the tetrahedroogol, which is equal to 3 tetrahedroned 100 times. The tetrahedroogol is a number of 111,969,417,824,932,076,575,821,557,634,511,960,861,863,938,060 digits, starting with 1710907228948061869110009387732..., and ending with ...809180.

Pentatopic Numbers

Pentatopic numbers are numbers that are sums of consecutive tetrahedral numbers starting with 1. They get their name because a pentatopic number of objects can (theoretically) be arranged into the shape of a pentatope. The first pentatopic numbers are:

1, 5, 15, 35, 70, 126, 210, 330, 495, 715, 1001, 1365, 1820, 2380, ...

The nth pentatopic number is quickly calculated by n(n+1)(n+2)(n+3)/24.

Simplex function

To go further, I defined the simplex function like so:

S(0, any number) = 1

S(1, n) = n

S(2, n) = 1 + 2 + 3 + 4 + ... + (n-2) + (n-1) + n

S(a, b) = S(a-1, 1) + S(a-1, 2) + S(a-1, 3) + ... + S(a-1, b-2) + S(a-1, b-1) + S(a-1, b)

S(a, b, c) = S(a, S(a, S(a, ... , S(a, S(a, b, c-1), c-1)...), c-1),c-1), c-1) w/ b nestings

The box, house, ... progression I mentioned earlier can be expressed in terms of this function:

box(n) = S(2, n, 2)

house(n) = S(2, n, 3)

rectangular-prism(n) = S(3, n, 2)

mansion(n) = S(3, n, 3)
I defined the box, house, mansion, etc. operators back in July 2016, while I devised the S function in November 2019.

S(3, 2, 3)

= S(3, S(3, 2, 2), 2)

= S(3, S(3, S(3, 2, 1), 1), 2)

= S(3, S(3, S(3, 2)), 2) (drop the 1s)

= S(3, S(3, 4), 2)

= S(3, 20, 2)

= tetr²⁰(3), where tetr(n) is the n-th tetrahedral number

~ 1.92901*10^{3,252,852,494}

Even just S(3, 2, 3) is a number with over three billion digits. Now try to imagine how big S(3, 3, 3) is:

S(3, 3, 3)

= S(3, S(3, S(3, 3, 2), 2), 2)

= S(3, S(3, S(3, S(3, S(3, 3, 1), 1), 1), 2), 2)

= S(3, S(3, S(3, S(3, S(3, 3))), 2), 2)

= S(3, S(3, S(3, S(3, 10)), 2), 2)

= S(3, S(3, S(3, 220), 2), 2)

= S(3, S(3, 1798940, 2), 2)

~ 10^{10^10^10^858312}