Talk:Steady squares: Difference between revisions
Why not 1001 or 900625? |
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Then g<sub>n</sub> is always 2n%10 which never equals n.--[[User:Nigel Galloway|Nigel Galloway]] ([[User talk:Nigel Galloway|talk]]) 14:15, 23 December 2021 (UTC)
:Excellent stuff. Likewise on the latter point we cannot "zero-fill" the single digit g<sub>0</sub>=5 or g<sub>0</sub>=6 because neither g<sub>1</sub> nor g<sub>n+1</sub> can ever be 0 (with either being sufficient grounds for permanent elimination). Going back to the first paragraph, another way to say that (as a guide rather than a proof) might be, using the triple-zero-fill example above:
55742..90625^2 = ...6100055742..90625
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 I suspect the number of candidates at each iteration settles into a simple pattern, at least it does not seem to grow much. --[[User:Petelomax|Pete Lomax]] ([[User talk:Petelomax|talk]]) 01:38, 24 December 2021 (UTC)
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Revision as of 01:41, 24 December 2021
Reduce search range by observation of the results
By taking a look at the results,I tried to search for results by prepending a digit to n.
And there comes the solution:By prepending a digit to n, than n*n must end in n to be a steady square.
So only solutions of one digit before can be solutions.
<lang pascal>
function CalcSquare(n:LongInt;Pot10:byte);
//pot10 is 10^(count of digits of n -1)
//prepend one digit to n
var
dgt: LongInt;
begin
//ex.: n= 5,Pot10 = 1, dgt= 2;-> n2= (2*10*1+5)^2 = 625 For dgt := 1 to 9 do begin // n= 5 , dgt= 2*10 -> n2= 625 n2 = sqr(dgt*10*Pot10+n); n2 = dgt*dgt*100*Pot10*Pot10+2*dgt*10*pot10*n+n*n;
// term: dgt*dgt*100*Pot10*Pot10+2*dgt*10*pot10*n// always has ends in Pot10+1 "0" digits // so n*n must end in n, to be a steady square
end;
</lang>
- 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).--Nigel Galloway (talk) 12:46, 21 December 2021 (UTC)
- Ah, you beat me to it (I posted my Phix entry before reading this). I was hopeful, as you (Horst) seem to suggest, that we only have to look at the k-1 digit solutions on each iteration, and that way maybe even find all possible such numbers there could ever be, that is if any iteration found none we'd be done, but alas zero-fill totally spanners that illusion (eg 90625 off the back of 625). Oh, unless I've missed a logical reason why it is not worth checking (eg) 1001 or (say) 900625... Then again the series could be and probably is infinite anyway. --Pete Lomax (talk) 14:53, 21 December 2021 (UTC)
- Nevermind that last bit:
1000557423423230896109004106619977392256259918212890625^2 = 10011151575673345586...92256259918212890625 (109 digits)
- is a triple-zero-fill, and I've found examples all the way up to (a quite random) 548 digit n with a 1096 digit square. --Pete Lomax (talk) 15:49, 21 December 2021 (UTC)
- Pete you were right.My first thought was, there where only a finite set of steady squares.
Thanks for showing, that prepending 1..x zeros ( and increasing Pot10 the same way ) to n, which is generating a stead square, lets n*n still be a steady square.
I am happy for not writing "proof" ;-) --Horst.h (talk) 10:32, 22 December 2021 (UTC)
- Pete you were right.My first thought was, there where only a finite set of steady squares.
- is a triple-zero-fill, and I've found examples all the way up to (a quite random) 548 digit n with a 1096 digit square. --Pete Lomax (talk) 15:49, 21 December 2021 (UTC)
Why not 1001 or 900625
Consideration of this question leads to an improvement of the algorithm. Consider a number of the form g2g1g0. Uncontentiously I assert that g0 is 1, 5 or 6. Prepending n=0..9 to each and squaring these to obtain xg1g0 and filtering for g1=n, I obtain 01, 25, 76 which are candidates for the next iteration. Note that using n=0 for g0=5 excludes 05 because 0 does not equal 2 and that it is only necessary to check g1 against n. Extending for g3 makes the candidate list 001, 625, and 376. At g4 interestingly 0625 is included because g4 in 390625 passes g4=n for n=0. Note 00625 will fail at the next iteration, thus eliminating 900625.
Considering g0=1 and gn-1..g1=0
101* 201* 1001* 101 201 1001 ----- --- ---- 101+ 201+ 1001+ 10100 20200 1001000 ----- ----- ------- 10201= 20401= 1002001=
Then gn is always 2n%10 which never equals n.--Nigel Galloway (talk) 14:15, 23 December 2021 (UTC)
- Excellent stuff. Likewise on the latter point we cannot "zero-fill" the single digit g0=5 or g0=6 because neither g1 nor gn+1 can ever be 0 (with either being sufficient grounds for permanent elimination). Going back to the first paragraph, another way to say that (as a guide rather than a proof) might be, using the triple-zero-fill example above:
55742..90625^2 = ...6100055742..90625 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 I suspect the number of candidates at each iteration settles into a simple pattern, at least it does not seem to grow much. --Pete Lomax (talk) 01:38, 24 December 2021 (UTC)