Talk:Steady squares: Difference between revisions
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: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 not spot ''that''? It is just a 5-chain and a 6-chain, with the 1-chain dying instantly. There will only ever be and always will be either one or two n-digit steady square numbers ending in 5 or 6 or both, maybe with the odd skip when both leap-frog. Apart from the first three, there will never be three steady squares of the same length, and if there are two, then one will end in 5 and the other in 6. ''DOH.'' I will however leave it up to someone else to come up with a proof that there will ''always'' be a ''single'' prefix digit that will work, for each chain. --[[User:Petelomax|Pete Lomax]] ([[User talk:Petelomax|talk]]) 02:05, 24 December 2021 (UTC)
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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=2, 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)
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