View source for Talk:Super-d numbers

== Big Doubt (Solved) ==

According to the definition in https://oeis.org/A032745, I don't understand why 32767 is considered a super-5 number, since: 
5*32767^5 = 188865838036335554720440
and there is no "55555" substring in it.
Same thing for super-6 number 27257.  --[[User:Dejan94|Dejan94]] ([[User talk:CDejan94|talk]]) 8:04, 27 March 2022 (UTC)

: Seems to me you need a better calculator.

    raku -e'say 5 * 32767 ** 5'
    18886583803633<span style="color:red;">55555</span>43035

    raku -e'say 6 * 27257 ** 6'
    2460478505381<span style="color:red;">666666</span>506497894

And please, sign your edits.
--[[User:Thundergnat|Thundergnat]] ([[User talk:Thundergnat|talk]]) 13:46, 27 March 2022 (UTC)

That's hilarious because R (my language) is giving me those wrong results ad I'm getting mad:
<lang R>
> 5 * 32767 ** 5
[1] 188865838036335554720440
> 5 * 32767 ^ 5
[1] 188865838036335554720440

> 6 * 27257 ** 6
[1] 2460478505381666522282046264
</lang>
What should I do?
Sorry for the sign, I'm newbie here and I didn't know how to do some things or special characters.
--[[User:Dejan94|Dejan94]] ([[User talk:Dejan94|talk]]) 14:02, 27 March 2022 (UTC)

: I don't have a good answer for you as I only have a passing familiarity with R. A Google search for '''R language integer precision''' turns up some possibly useful information, though again, I am not the best one to judge. --[[User:Thundergnat|Thundergnat]] ([[User talk:Thundergnat|talk]]) 14:14, 27 March 2022 (UTC)

:: Thanks mate, I will investigate! It's funny because R is used for complex statistical and mathematical computing and I'm really shocked about this
--[[User:Dejan94|Dejan94]] ([[User talk:Dejan94|talk]]) 14:20, 27 March 2022 (UTC)

==== Solution ====
I did it! I installed "Rmpfr" package for arbitrary precision floating point numbers and I used it to augment precision! Now I can add the solution of the task in R. Thank you [[User:Thundergnat|Thundergnat]] for suggestion, I also learned a new thing and a flaw of R!
--[[User:Dejan94|Dejan94]] ([[User talk:Dejan94|talk]]) 19:14, 27 March 2022 (UTC)

== super-d numbers==
It is noted elsewhere that '''super-d''' numbers are always expressed in decimal &nbsp; (base ten). &nbsp; &nbsp; -- [[User:Gerard Schildberger|Gerard Schildberger]] ([[User talk:Gerard Schildberger|talk]]) 07:34, 12 October 2019 (UTC)
: I updated the description of super-d numbers to reflect this. Thanks. --[[User:Chunes|Chunes]] ([[User talk:Chunes|talk]]) 20:50, 12 October 2019 (UTC)

== Mostly addition method ==
One can get a list of squares by summing up the list of odd numbers, like this:
<pre>
   0 + 1 =  1
   1 + 3 =  4
   4 + 5 =  9
   9 + 7 = 16
  16 + 9 = 25
</pre>
Another way of tabulating it:
<pre>
square, 1st difference, 2nd difference
     0,              1,              2
     1,              3,              2
     4,              5,              2
     9,              7,              2
    16,              9,              2
    25,             11,              2
</pre>
The 2nd difference, 2, is also 2!<br/>
On to cubes:
<pre>
  cube, 1st difference, 2nd difference, 3rd difference
     0,              1,              6,              6
     1,              7,             12,              6
     8,             19,             18,              6
    27,             37,             24,              6
    64,             61,             30,              6
   125,             91,             36,              6
   216,            127,             42,              6
</pre>
The 3rd difference, 6, is also 3!<br/>

This pattern continues on, with the number of differences required (to calculate by adding) increasing by one and the last difference being n!  To calculate a list of any powers, one can start at zero with the proper array of initial values, or start at "n" with a proper array of initial values.  When starting at zero, some of the initial values will be negative.  When starting at "n", all initial values will be positive.

Here is the initial array for each "n", starting at zero:
<pre>
2:         0        -1         2 
3:         0         1        -6        6 
4:         0        -1        14      -36       24 
5:         0         1       -30      150     -240      120 
6:         0        -1        62     -540     1560    -1800      720 
7:         0         1      -126     1806    -8400    16800   -15120    5040 
8:         0        -1       254    -5796    40824  -126000   191520 -141120    40320 
9:         0         1      -510    18150  -186480   834120 -1905120 2328480 -1451520 362880 
</pre>
Here is the initial array for each "n", starting at "n":
<pre>
2:         4         3         2 
3:        27        19        12        6 
4:       256       175       110       60       24 
5:      3125      2101      1320      750      360      120 
6:     46656     31031     19502    11340     5880     2520      720 
7:    823543    543607    341796   201726   109200    52080    20160    5040 
8:  16777216  11012415   6927230  4131036  2298744  1164240   514080  181440    40320 
9: 387420489 253202761 159338640 95750430 54313560 28594440 13608000 5594400  1814400 362880 
</pre>

When working backwards by summing instead of subtracting, the tables look like this because we work right to left on the array:<br/>
Squares:
<pre>
0:         0        -1         2
1:         1         1         2
2:         4         3         2
3:         9         5         2
4:        16         7         2
5:        25         9         2
</pre>
Cubes:
<pre>
0:         0         1        -6        6 
1:         1         1         0        6 
2:         8         7         6        6 
3:        27        19        12        6 
4:        64        37        18        6 
5:       125        61        24        6 
</pre>

Regarding the term "mostly addition", for this task there is some multiplication by a small integer (2 thru 9 in ''scale()''), but the powers of "n" are calculated by only addition.  The initial tables are calculated with multiplication and a power function (''ipow()'').

A side note, one can also calculate cubes by summing groups of odd numbers like this:
<pre>
  1 =  1
  8 =  3 +  5
 27 =  7 +  9 + 11
 64 = 13 + 15 + 17 + 19
125 = 21 + 23 + 25 + 27 + 29
</pre>
But I haven't worked out how (or if) a similar method works for higher powers.
--[[User:Enter your username|Enter your username]] ([[User talk:Enter your username|talk]]) 16:08, 21 August 2020 (UTC)