User blog comment:Primussupremus/Is it possible to have pi up arrows in Knuth up arrow notation?/@comment-30754445-20170528205759

Knuth arrows aren't defined for non-integers. So strictly speaking, π [π arrows] π isn't defined.

However, we can generalize the definition of Knuth arrows to non-integers. In fact, there are several ways to do this, so π [π arrows] π has more than one possible value!

I know of at least two ways to define such a value:

1. Using Nayuta Ito's method:

π[π arrows]π = π[π-1 arrows]π[π-1 arrows]π[π-1 arrows](ππ-3)

~ π[π-1 arrows]π[π-1 arrows]π[π-1 arrows]1.17596059608

= π[π-1 arrows]π[π-1 arrows]π[π-2 arrows](π0.17596059608)

~ π[π-1 arrows]π[π-1 arrows]π[π-2 arrows]1.22314737593

= π[π-1 arrows]π[π-1 arrows]π[π-3 arrows](π0.22314737593)

~ π[π-1 arrows]π[π-1 arrows]π[π-3 arrows]1.29103402903

= π[π-1 arrows]π[π-1 arrows](π1+(π-3)×0.29103402903×1.291034029034-π

~ π[π-1 arrows]π[π-1 arrows]4.10077831

= π[π-1 arrows]π[π-2 arrows]π[π-2 arrows]π[π-2 arrows]π[π-2 arrows](π0.10077831)

~ π[π-1 arrows]π[π-2 arrows]π[π-2 arrows]π[π-2 arrows]π[π-2 arrows]1.12228181

= π[π-1 arrows]π[π-2 arrows]π[π-2 arrows]π[π-2 arrows]π[π-3 arrows](π0.12228181)

= π[π-1 arrows]π[π-2 arrows]π[π-2 arrows]π[π-2 arrows]π[π-3 arrows]1.15025038

= π[π-1 arrows]π[π-2 arrows]π[π-2 arrows]π[π-2 arrows](π1+(π-3)×0.15025038×1.150250384-π

~ π[π-1 arrows]π[π-2 arrows]π[π-2 arrows]π[π-2 arrows]3.6300369

= π[π-1 arrows]π[π-2 arrows]π[π-2 arrows]π[π-3 arrows]π[π-3 arrows]π[π-3 arrows](π0.6300369)

~ π[π-1 arrows]π[π-2 arrows]π[π-2 arrows]π[π-3 arrows]π[π-3 arrows]π[π-3 arrows]2.0569455

~ π[π-1 arrows]π[π-2 arrows]π[π-2 arrows]π[π-3 arrows]π[π-3 arrows]6.925066

~ π[π-1 arrows]π[π-2 arrows]π[π-2 arrows]π[π-3 arrows]43.217

~ π[π-1 arrows]π[π-2 arrows]π[π-2 arrows]π[π-3 arrows]43.217

~ π[π-1 arrows]π[π-2 arrows]π[π-2 arrows]74641

~ π[π-1 arrows]π[π-2 arrows](10↑↑74640)

~ π[π-1 arrows](10↑↑10↑↑74640)

~ 10↑↑↑10↑↑10↑↑74640 = GFF74640 ~ H2.5088

Alternatively, one could use a base-π variation of my letter notation:

π[π arrows] π = Jπ (in base π) = H(2×(π/2)π-3) base π

~ H2.1321 base π

= GG(π0.1321) base π

~ GG1.1632 base π

= GF(π0.1632) base π

~ GF1.2054 base π

= GE(π0.2054) base π

~ GE1.2651 base π

= G(π 0.2651 ) base π

~ G4.2553 base π

= FFFF(π 0.2553 ) base π

~ FFFF1.3395 base π

= FFFE(π 0.3395 ) base π

~ FFFE1.4749 base π

= FFF5.4108 base π

= FFEEEEE(π0.4108_ base π

~ FFEEEEE1.600 base π

~ FFEEEE6.246 base π

~ FFEEE1274.5 base π

~ FFEEE633 base 10

= 10↑↑10↑↑1010 10 633

(or if you wish: G3.648)

To summarize: there are at least two different reasonable values we could assign π[π arrows]π. Naturally, both of them are between 3[3 arrows]3 and 4[4 arrows]4:

3↑↑↑3 = Tritri ~ 10↑↑(7.6×1012) ~ FE12.88 ~ G2.31

π [π arrows] π (Psi Letters version) ~ 10↑↑10↑↑1010 10 633   = FFEEE633 ~ G3.65

π [π arrows] π (Nayuta Ito's version) ~ 10↑↑↑10↑↑10↑↑74640 = GFF74640 ~ H2.51

4↑↑↑↑4 = Tritet ~ 10↑↑↑10↑↑↑10↑↑10↑↑10↑(10154) ~ GGFFEE154 ~ H3.55