Template talk:Sequence of the Day for September 2

Here is the original version of this page. It might be too long, so I posted a shortened version. A014222: $\scriptstyle a(0)\,=\,0\,$, $\scriptstyle a(n)\,=\,3^{a(n-1)},\,n\,\geq \,1.\,$

{ 0, 1, 3, 27, 7625597484987 }

where 7625597484987 is the largest writeable exponent in Graham's number.

Today is the 66^th anniversary of end of WWII. To break increasingly more complicated codes, perhaps World War II was the single most important stimulus to the search for powerful computers. A math topic that has become accessible to laymen through powerful computers is that of tetrations.

$\scriptstyle a(0)\,=\,3\,\uparrow \uparrow \,({-\infty })\,=\,0,\,a(1)\,=\,3\,\uparrow \uparrow \,0\,=\,1,\,a(2)\,=\,3\,\uparrow \uparrow \,1\,=\,3,\,a(3)\,=\,3\,\uparrow \uparrow \,2\,$ $\scriptstyle \,=\,3^{3},\,a(4)\,=\,3\,\uparrow \uparrow \,3\,$ $\scriptstyle \,=\,3^{3^{3}}\,$ (tetrations of height 0,1,2, and 3 respectively (nonnegative integers) following that of a height of negative infinity), where a height of 0 gives the empty product (the multiplicative identity, i.e. 1) and where $\scriptstyle a\,\uparrow \uparrow \,(-b)\,\equiv \,1/(a\,\uparrow \uparrow \,b),\,b\,\geq \,0.\,$ On the tetration page, the notion of rational numbered heights, like $\scriptstyle {3}\,\uparrow \uparrow \,(5/3),$ is brought up and the proposed solution with the introduction of the tetra-logarithmic function and the tetra-exponential function (if such functions could have a well-defined algorithm).

?????

Could this be solved with the introduction of the tetra-logarithmic function $\scriptstyle \log ^{*}(z)\,$ and the tetra-exponential function $\scriptstyle \exp ^{*}(z)\,$, with possibly $b\uparrow \uparrow {r}{\overset {?}{\,\equiv \,}}\exp ^{*}(r~\log ^{*}(b)),\quad r\in \mathbb {R} ,\,$ or may be $b\uparrow \uparrow {r}{\overset {?}{\,\equiv \,}}\exp ^{*}((\log ^{*}(b))^{r}),\quad r\in \mathbb {R} ?\,$ Or in the case of a rational number $\scriptstyle {\frac {n}{d}}\,$ $b\uparrow \uparrow {\tfrac {n}{d}}{\overset {?}{\,\equiv \,}}(b***n)///d=(b///d)***n,\quad {\tfrac {n}{d}}\in \mathbb {Q} ?\,$

This leads us to further inquire into the relationship between tetration with a rational numbered height and exponentiation with a rational exponent. What purpose is there in the tetration equivalent $\scriptstyle n\,$^^$\scriptstyle {(1/n)}\,$ of $\scriptstyle n^{(1/n)}\,$? It is almost trivial to realize that $\scriptstyle n^{1/n}\,$ can represent the volume of an $\scriptstyle n\,$-cube raised to the power of the reciprocal of its dimension, but what about $\scriptstyle n\,$^^$\scriptstyle {(1/n)}\,$?

There seems to be a parallel between $\scriptstyle n^{1/n}\,$ and $\scriptstyle n\,$^^$\scriptstyle {(1/n)}\,$, in that $\scriptstyle n^{1/n}\,$ has an inverse, mapping $\scriptstyle \mathbb {R} _{\geq 1}\,\to \,\mathbb {R} _{\geq 1}\,$, and so does $\scriptstyle n\,$^^$\scriptstyle {(1/n)}\,$, because all the higher order equivalents of $\scriptstyle n^{n}\,$ (e.g. $\scriptstyle n\,$^^$\scriptstyle {n}\,$, $\scriptstyle n\,$^^^$\scriptstyle {n}\,$, $\scriptstyle n\,$^^^^$\scriptstyle {n}\,$,) mapping $\scriptstyle \mathbb {R} _{\geq 1}\,\to \,\mathbb {R} _{\geq 1}\,$, are bijective (one-to-one and onto), and all the higher order equivalents of $\scriptstyle n^{(1/n)}\,$ (e.g. $\scriptstyle n\,$^^$\scriptstyle {(1/n)}\,$, $\scriptstyle n\,$^^^$\scriptstyle {(1/n)}\,$, $\scriptstyle n\,$^^^^$\scriptstyle {(1/n)}\,$,) mapping $\scriptstyle \mathbb {R} _{\geq 1}\,\to \,\mathbb {R} _{\geq 1}\,$, thus are unique inverses. How far does this parallel go? A start in comparing the graphs of $\scriptstyle n\,$^...^$\scriptstyle (1/n)\,$ would be to look for an iterated equivalent of Steiner's Problem. In other words, is there a critical point of the graphs of $\scriptstyle n\,$^...^$\scriptstyle {(1/n)}\,$? If we lived in a simple world, they also would have a maximum at $\scriptstyle n\,=\,e\,$.

Notice that $\scriptstyle n\,$^^$\scriptstyle {(1/n)}\,$ is just a special case of $\scriptstyle n\,$^^$\scriptstyle r,\,r\,\in \,\mathbb {R} \,$ Could, then, the graph of the continuous version of our sequence ($\scriptstyle 3\,$^^$\scriptstyle r,\,r\,\in \,\mathbb {R} )\,$ have the same shape as the interpolated graph of the sequence?

I love the tidbit about World War II and the segue towards tetration. At some point after that, some of the stuff would be better off in the article on tetration. Alonso del Arte 04:51, 12 July 2011 (UTC)

As of 7/16/2011

Again the article has become too technical. Here is how it looked as of 7/16/2011:

A014222: $\scriptstyle a(0)\,=\,0\,$, $\scriptstyle a(n)\,=\,3^{a(n-1)},\,n\,\geq \,1.\,$

{ 0, 1, 3, 27, 7625597484987 }

Today is the 66^th anniversary of end of WWII. To break increasingly more complicated codes, WWII was an important stimulus to the search for powerful computers. Integer sequences of tetrations (like this one) are nearly impossible to extend very far without powerful computers!

For n>1, this sequence extended indefinitely, contains only the tetra-exponentials of Graham's number, where $\scriptstyle a(n)\,=\,(3$^^$\scriptstyle n,\,n\,\in \,\mathbb {Q} +)\,$, and 3^^^^3=3^^^(3^^^3)=3^^(3^^(3^^...(3^^3)...)), and the number of 3s in the expression on the right is 3^^^3=3^^(3^^3). Here n is a nonnegative integer; however, on the tetration page, the notion of rational numbered heights (like $\scriptstyle {3}\,$^^$\scriptstyle (5/3)$) is brought up.

This leads us to inquire into the relationship between tetration with a rational numbered height and exponentiation with a rational exponent. For instance, what purpose is there in the tetration equivalent $\scriptstyle n\,$^...^$\scriptstyle {(1/n)}\,$ (where the number of n's are arbitrary) of $\scriptstyle n^{(1/n)}\,$? We know for integer n, $\scriptstyle n^{1/n}\,$ can represent the volume of an $\scriptstyle n\,$-cube raised to the power of the reciprocal of its dimension, but what about $\scriptstyle n\,$^...^$\scriptstyle {(1/n)}\,$?

There seems to be a parallel between $\scriptstyle n^{1/n}\,$ and $\scriptstyle n\,$^...^$\scriptstyle {(1/n)}\,$, in that they have an inverse, mapping $\scriptstyle \mathbb {R} _{\geq 1}\,\to \,\mathbb {R} _{\geq 1}\,$. How far does this parallel go? A start in comparing the graphs of $\scriptstyle n\,$^...^$\scriptstyle (1/n)\,$ would be to look for an iterated equivalent of Steiner's Problem. In other words, is there a critical point of the graphs of $\scriptstyle n\,$^...^$\scriptstyle {(1/n)}\,$?

Now notice that $\scriptstyle n\,$^...^$\scriptstyle {(1/n)}\,$ is just a special cases of $\scriptstyle n\,$^...^$\scriptstyle r,\,r\,\in \,\mathbb {R} +\,$ Could, then, the graphs of the continuous version of our sequence ($\scriptstyle 3\,$^...^$\scriptstyle r,\,r\,\in \,\mathbb {R} )\,$ have the same shape as the interpolated graphs of ($\scriptstyle 3\,$^...^$\scriptstyle n,\,n\,\in \,\mathbb {Q} +)\,$?