# 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. --Dejan94 (talk) 8:04, 27 March 2022 (UTC)

- Seems to me you need a better calculator.

```
raku -e'say 5 * 32767 ** 5'
188865838036335555543035
```

```
raku -e'say 6 * 27257 ** 6'
2460478505381666666506497894
```

And please, sign your edits. --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. --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. --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

--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 Thundergnat for suggestion, I also learned a new thing and a flaw of R! --Dejan94 (talk) 19:14, 27 March 2022 (UTC)

## super-d numbers

It is noted elsewhere that **super-d** numbers are always expressed in decimal (base ten). -- Gerard Schildberger (talk) 07:34, 12 October 2019 (UTC)

## Mostly addition method

One can get a list of squares by summing up the list of odd numbers, like this:

0 + 1 = 1 1 + 3 = 4 4 + 5 = 9 9 + 7 = 16 16 + 9 = 25

Another way of tabulating it:

square, 1st difference, 2nd difference 0, 1, 2 1, 3, 2 4, 5, 2 9, 7, 2 16, 9, 2 25, 11, 2

The 2nd difference, 2, is also 2!

On to cubes:

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

The 3rd difference, 6, is also 3!

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:

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

Here is the initial array for each "n", starting at "n":

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

When working backwards by summing instead of subtracting, the tables look like this because we work right to left on the array:

Squares:

0: 0 -1 2 1: 1 1 2 2: 4 3 2 3: 9 5 2 4: 16 7 2 5: 25 9 2

Cubes:

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

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:

1 = 1 8 = 3 + 5 27 = 7 + 9 + 11 64 = 13 + 15 + 17 + 19 125 = 21 + 23 + 25 + 27 + 29

But I haven't worked out how (or if) a similar method works for higher powers. --Enter your username (talk) 16:08, 21 August 2020 (UTC)