Talk:Steady squares: Difference between revisions
→Reduce search range by observation of the results: Added syntaxhighlight tag on pascal code snippet.
(→Reduce search range by observation of the results: Added syntaxhighlight tag on pascal code snippet.) |
|||
| (14 intermediate revisions by 5 users not shown) | |||
Line 4:
And there comes the solution:By prepending a digit to n, than n*n must end in n to be a steady square.<BR>
So only solutions of one digit before can be solutions.
<
function CalcSquare(n:LongInt;Pot10:byte);
//pot10 is 10^(count of digits of n -1)
Line 20:
// so n*n must end in n, to be a steady square
end;
</syntaxhighlight>
:Note also that it must end in 1,5 or 6 (not 0 as any number ending in 0 squared ends with twice as many zeros).--[[User:Nigel Galloway|Nigel Galloway]] ([[User talk:Nigel Galloway|talk]]) 12:46, 21 December 2021 (UTC)
Line 49:
100055742..90625^2 = ...1100055742..90625
:The three zeros in the first line make it a zero-fill candidate for three rounds, after which the second line will either "take over the mantle" in the list of potential candidates, or there won't be anything that does. Or it might be more accurate to say it ''only'' becomes a candidate in three rounds time?
:From my experiments so far ... Oh, ffs, off course, how did I
:: It is also necessary to prove that there is only one digit that satisfied the requirement. That is the chains do not branch.--[[User:Nigel Galloway|Nigel Galloway]] ([[User talk:Nigel Galloway|talk]]) 19:51, 26 December 2021 (UTC)
==No Search Required==
I introduce a table:
for g<sub>0</sub>=5 for g<sub>0</sub>=6
(n+g)%10 g<sub>n</sub> g<sub>n</sub>
0 0 0
1 1 9
2 2 8
3 3 7
4 4 6
5 5 5
6 6 4
7 7 3
8 8 2
9 9 1
Starting with g<sub>0</sub>=6 I let g=0 and n=g<sub>0</sub>*g<sub>0</sub>/10=3.
Looking up 3 in the table g<sub>1</sub>=7. I let n=(n+g+2*g<sub>1</sub>*g<sub>0</sub>)/10=8, g=g<sub>1</sub>*g<sub>1</sub>=49.
Looking up 7 in the table g<sub>2</sub>=3. I let n=(n+g+2*g<sub>2</sub>*g<sub>0</sub>)/10=9, g=g<sub>2</sub>*g<sub>1</sub>+g<sub>1</sub>*g<sub>2</sub>=42.
Looking up 1 in the table g<sub>3</sub>=9. I let n=(n+g+2*g<sub>3</sub>*g<sub>0</sub>)/10=15, g=g<sub>3</sub>*g<sub>1</sub>+g<sub>2</sub>*g<sub>2</sub>+g<sub>1</sub>*g<sub>3</sub>=135.
Looking up 0 in the table g<sub>4</sub>=0. I let n=(n+g+2*g<sub>4</sub>*g<sub>0</sub>)/10=15, g=g<sub>4</sub>*g<sub>1</sub>+g<sub>3</sub>*g<sub>2</sub>+g<sub>2</sub>*g<sub>3</sub>+g<sub>1</sub>*g<sub>4</sub>=54.
Looking up 9 in the table g<sub>5</sub>=1. .... and so to infinity and beyond.
Starting with g<sub>0</sub>=5 I let g=0 and n=g<sub>0</sub>*g<sub>0</sub>/10=2.
Looking up 2 in the table g<sub>1</sub>=2. I let n=(n+g+2*g<sub>1</sub>*g<sub>0</sub>)/10=2, g=g<sub>1</sub>*g<sub>1</sub>=4.
Looking up 6 in the table g<sub>2</sub>=6. I let n=(n+g+2*g<sub>2</sub>*g<sub>0</sub>)/10=6, g=g<sub>2</sub>*g<sub>1</sub>+g<sub>1</sub>*g<sub>2</sub>=24.
Looking up 0 in the table g<sub>3</sub>=0. I let n=(n+g+2*g<sub>3</sub>*g<sub>0</sub>)/10=3, g=g<sub>3</sub>*g<sub>1</sub>+g<sub>2</sub>*g<sub>2</sub>+g<sub>1</sub>*g<sub>3</sub>=36.
Looking up 9 in the table g<sub>4</sub>=9. I let n=(n+g+2*g<sub>4</sub>*g<sub>0</sub>)/10=12, g=g<sub>4</sub>*g<sub>1</sub>+g<sub>3</sub>*g<sub>2</sub>+g<sub>2</sub>*g<sub>3</sub>+g<sub>1</sub>*g<sub>4</sub>=36.
Looking up 8 in the table g<sub>5</sub>=8. .... and so to infinity and beyond.
The existence of this table and method proves that these two sequences are infinite and unique.--[[User:Nigel Galloway|Nigel Galloway]] ([[User talk:Nigel Galloway|talk]]) 19:30, 26 December 2021 (UTC)
:Agreed and well done''!'' Phix entry added. --[[User:Petelomax|Pete Lomax]] ([[User talk:Petelomax|talk]]) 15:17, 31 December 2021 (UTC)
==RC's First Interesting Fact for 2022==
With the exception of g0 where 5 -> 6 if you tell me the nth digit in the 5 sequence, I can tell you the nth digit in the 6 sequence using the following mapping:
0 -> 9
1 -> 8
2 -> 7
3 -> 6
4 -> 5
5 -> 4
6 -> 3
7 -> 2
8 -> 1
9 -> 0
--[[User:Nigel Galloway|Nigel Galloway]] ([[User talk:Nigel Galloway|talk]]) 16:11, 3 January 2022 (UTC)
: Yes, the n.th digits sum up to 9 in [[Steady_Squares#Some_Examples]] for 80 digits without the last digit --[[User:Horst.h|Horst.h]] ([[User talk:Horst.h|talk]]) 16:41, 3 January 2022 (UTC)
<br>
: For each digit of the 5-series and 6-series (for the same number of digits), the first through the next-to-last digits (when present) sum up to 9, the final (ones) digit sums to 11.
: Also,
: for single digit, (10^1): 5 + 6 = 11
: for double digits, (10^2): 25 + 76 = 101
: for triple digits, (10^3): 625 + 376 = 1001
: ...pattern continues...
: So the 6-series (for any number of digits) can be obtained by subtracting the 5 series result (for that number of digits) from 10^(number of digits), then adding 1.
: For four digits (and a few others) the leading digit of the 5-series is a zero, so the 6-series result is 10001 - 0625 = 9376.
: When the first digit of the 5-series result is 9 (such as 90625), the corresponding 6-series result is a repeat of the previous number of digits (such as 9376), since the leading digit of the 6-series is zero.
--[[User:Enter your username|Enter your username]] ([[User talk:Enter your username|talk]]) 05:39, 5 January 2022 (UTC)
| |||