Bacon Bits

G (TREEX!(TREEX!(TREEX!(TREEX!(TREEX!(TREEX!(TREEX!(TREE$(Gx$(X ) - Bacon Bits1

G = Graham’s function.

In this version of the function (Which shall be called **Bacon Bits** (Recursive Exponentiation) going forward), each level of each power tower gets reset and repeats after each number. For example with 3⁴, instead of just 3x3x3x3 = **81**, it would be (3x3=9)(9x3x3=81)(81x3x3x3) = **2,187**.

So to help picture this, if used in Graham’s number, it would mean that as each layer of the **Vertical** Sun Tower in the **1st** part of **G1** is worked out **Horizontally**, each of those numbers in the **Horizontal** display, would be worked out **Vertically** to feed into the next **Horizontal** number. If we take infinitesimal 3⁴ and make it 3⁵, that would change what would normally be 3x3x3x3x3 = **243,** into (3x3=9)(9x3x3=81)(81x3x3x3=2,187)(2,187x3x3x3x3) = **177,147**.

Take it one step further with 3⁶ and, what would be **729** into (3x3=9)(9x3x3=81)(81x3x3x3=2,187)(2,187x3x3x3x3)( 177,147x3x3x3x3x3) =**43,046,721**. With 3⁷, instead of **2,187**, we get **31,381,059,609**. Let’s skip to 3¹⁰, just to see where it takes us. With a regular exponent, we would be at **59,049**, which is a dramatic increase in and of itself. Using our **RecEx** function, it would be first **68,630,377,364,883,** then **45,028,390,600,000,000,** and finally **886,293,812,000,000,000,000.**

**59,049** VS **886,293,812,000,000,000,000** at just 10 small steps in.

This version of the 1st (infinitesimal) part of G1 is normally 3²⁷, but using the Bacon Bit function(BB), it will be 3⁸¹. But for the sake of comparison, let’s see what 3²⁷ is using BB vs normal exponentiation.

88,446,705,700,000,000,000,000,000,000,000, 000,000,000,000,000,000,000,000,000,000,000, 000,000,000,000,000,000,000,000,000,000,000, 000,000,000,000,000,000,000,000,000,000,000, 000,000,000,000,000,000,000,000,000,000,000, 000 +

( 88 Quattuorqiunquagintillion, 446 tresqiunquagintillion, 705 duoqiunquagintillion, 700 unqiunquagintillion + )

Vs

7,625,597,484,987

( 7 trillion, 625 billion, 597 million, 484 thousnad, 987 )

Meaning that the height of the 1st power tower, which we call the sun tower (150 million Km tall) in G1, will be a number that is 66 levels of magnitude beyond a Googol.

8¹⁶⁶

Vs

1¹⁰⁰

(1 Million Googols would only be 1¹⁰⁶)

A Googol itself already being a number so big that if we filled the observable universe (94 billion light years across, with light traveling at 300,000 Km per second) with sand and wrote a 0 on every single grain of sand, we would still need 10 billion universes to write out an amount of anything totaling 1¹⁰⁰. So to write out 1¹⁰³, we’d need 10 trillion. To write out 1¹⁰⁶, we’d need 10 quadrillion universes and so on.

And here with 8¹⁶⁶, we are needing roughly 80 duovigintillion universes full of sand just to count up to that number.

In G1, if each 3 is 2cm high, we get our 150 Million Km Sun Tower. To help put this in perspective, I’m going to give a comparison using the absolute largest size of any grain of sand (smallest is 1/16 of a mm) , which is 2mm. So if we change our 0s to 3s for G1, we’ll need to bump them up at least 10 times their current size. So now we are needing

We are now needing 800 duovigintillion universes just to write out our “Sun Tower”, so basically for G1 with our Bacon Bit function, it’s an 800 Duovigintillion Universe Tower. That is again only using 3²⁷  in BB, but we’ll be using 3⁸¹, and just going to 3²⁸ would take us from

88,446,705,700,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,

000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000

(88 Quattuorqiunquagintillion)

To

674,458,977,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,

000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000

(674 Octoqiunquagintillion)

Which itself, just 3²⁸ , would now require a 67 Quattuordecillion Universe Tower to write out our 3s. That’s almost 10 trillion times larger than our 800 Duovigintillion Universe Tower.

Going to 3²⁹ will get a number nearly 100 trillion times larger than 67 Quattuordecillion Universes.

A The highest that a normal calculator will let me go to is  3³⁶, which is 11,574,752,900,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000+.

That is a number that is around a Googol x Googol x Googol, which would mean 100 Billion x Googol x Googol Universes full of 2cm high 3s to write out our number at 3³⁶, which is still far far away from 3⁸¹.

Even just stopping there, can you imagine solving that number to get the number of power towers in part 2 of G1. And knowing that the 2nd power tower will be at least this height, and feed into the height of the next and so on.

Trying to go to 3³⁷ on a regular exponential calculator gave me a result of infinity, which is ofc not true but still hilarious.

The following are Step additions to Croutonillion-

Step 1(4,577). Take G (TREE!(TREE!(TREE!( (TREE!(TREE!(TREE!(TREE!(TREE$(Gx$(X ) - Bacon Bits 1

Step 2(4,578). Repeat Bacon Bits 1, Bacon Bits 1 times and call it Bacon Bits 2

Step 3(4,579). Repeat the above function until you have Bacon Bits X.

Step 4(4,580). Reapply Step 4466 with Bacon Bits X. Bacon Bits X is the new X being applied everywhere that X is mentioned in that step.

Step 5(4,581). Repeat this X times, replacing X with newly solved version of X each time. This shall be known as Bacon Bits X (X1).

Step 6(4,582). Repeat this until you reach Bacon Bits X (XX)

Step 7(4,583). Once we get to BB X(XX), keep it going until we get to X(X̅)which is X with 1,000 Xs. Repeat this until you get to X(X̅CIX̅), which is the X(X) function with all related Steps, repeated 99,000 times.

Step 8(4,584). Repeat all functions from the previous step, until reaching X(*M̅CIM̅) which is* the X(X) function with all related Steps, repeated 99,000,000 times. X(*M̅CIM̅) is the new* X.

Step 9(4,585). Repeat steps 1 – 4,584, X times.

Step 10(4,586). In each previous step, wherever there's X, insert Xx×. Now repeat steps 1 – 4,585, X(M̅CIM̅) times. The result is the new X

Step 11(4,587). In each previous step, wherever there's X, insert Xx×. Now repeat steps 1 – 4,586, X(M̅CIM̅) times. The result is the new X

Step 12(4,588). In each previous step, wherever there's X, insert Xx×. Now repeat steps 1 – 4,587, X(M̅CIM̅) times. The result is the new X

Step 13(4,589). In each previous step, wherever there's X, insert Xx×. Now repeat steps 1 – 4,588, X(M̅CIM̅) times. The result is the new X

Step 14(4,590). In each previous step, wherever there's X, insert Xx×. Now repeat steps 1 – 4,585, X(M̅CIM̅) times. The result is the new X

Step 15(4,591). Repeat the process for steps 10-14 (4,586-4,590) until you reach step X(M̅CIM̅). The result is then retroactively applied to steps 4,586 - step X(M̅CIM̅), as the new X. The result of that, is the newest X.

Step 16(4,592). Repeat steps 1 -- 4,591, G(TREE!(TREEX!......(TREEX!(GX......x.......×(M̅CIM̅)!^^^^^^X(M̅CIM̅) times. With 100 TREEX!s, 100 Xs, and 100 ^s. This is the new X

Step 17(4,593). G (Steps 1 - X(M̅CIM̅), with G being Graham’s Function. G1(Step 1)

Each power tower of steps feeds into the next. Every step created so far was just the 1st part of G1. Now X towers will be created, with the sum of X from the last tower, being the number of steps in the following tower. With step 1 from the following tower, always being more powerful than highest step of the previous tower, in the hierarchy of steps. That will be G1(Step 1). G2 (Step 2) will follow the same structure, but with X (G1(Step 1) ^s of layers of steps, and so on until GX(M̅CIM̅). The result will be the new X.