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Panic in the Year Zero Zero
Solution to the
1, 2, 3, 4, 25 Sequence
This is the solution to the first puzzle I ran on my Puzzle Page
The puzzle here was to determine the next three terms in this sequence, or more broadly the gimmick behind the sequence. Scroll down slowly if you want to try to solve this yourself, or just skip down to the solution. You should look at just the first eleven terms, then scratch your head for a few days before coming back to see more.
Though it is theoretically possible, I doubt one person in a billion can solve this in fewer than 11 terms. But if you think you're that one, scroll down far enough to look at these ten numbers:
1, 2, 3, 4, 25, 67, 85, 34, 53, 92, ...
Can you figure that out? Maybe you should take a few days looking at this one before you look at more terms.
Hints:
- The sequence is infinitely long.
- No mundane (Earthly) knowledge is required. (In fact it might hurt)
- Once you know the secret it's easy, but until then it's hard.
- The solution is "nice", I can describe it clearly in less than 75 ascii characters. I can describe it cryptically in 28 ascii characters.
- I was asked whether I named the sequence "1,2,3,4,25" because I started with those five numbers and designed a sequence for them, or I developed a sequence and the first five numbers happened to be 1,2,3,4,25. Well it's a little of both, but more of the latter. The sequence is defined by a simple algorithm. I made some arbitrary choices in the algorithm to cause it to produce this sequence.
As Bob now knows, the solution is "nice". It is not a many termed polynomial nor is it a many stepped algorithm.
Here are the first eleven terms:
1, 2, 3, 4, 25, 67, 85, 34, 53, 92, 211, ...
Looks a lot easier doesn't it? Okay it's not that easy, but I believe I would have solved this with eleven terms, as should at least 1% of homo sapiens.
But no one solved it until the twelfth term:
1, 2, 3, 4, 25, 67, 85, 34, 53, 92, 211, 262, ...
Congratulations to Davis, the first person in the world to solve this puzzle. He accomplished this thoroughly laudible feat on 1997-11-15 using twelve terms (thru 262).
Not enough? Stop here and pnnder for a few days before looking further down.
Okay, here are the first 16 terms:
1, 2, 3, 4, 25, 67, 85, 34, 53, 92, 211, 262, 233, 254, 245, 667, ...
And congratulations to Al Russell who solved this on 1997-12-01 using sixteen terms (thru 667). The only other person ever to solve this sequence without the benefit of the 28th term.
This is where this puzzle went bizarre. No one more solved this until the 38th term. Why don't you try it with 20 terms:
1, 2, 3, 4, 25, 67, 85, 34, 53, 92, 211, 262, 233, 254, 245, 667, 675, 694, 863, 852, ...
Not enough? Look at 24 terms:
1, 2, 3, 4, 25, 67, 85, 34, 53, 92, 211, 262, 233, 254, 245, 667, 675, 694, 863, 852, 311, 332, 393, 764, ...
How about looking at 27 terms:
1, 2, 3, 4, 25, 67, 85, 34, 53, 92, 211, 262, 233, 254, 245, 667, 675, 694, 863, 852, 311, 332, 393, 764, 705, 567, 505, ...
Please, please, don't scroll down further without giving this a lot of thought. IMHO, the 28th term reveals too much. I just can't be impressed by someone solving this after seeing the 28th term. So don't look any farther until you have given a really good try at solving this in 27 terms.
Really.
Stop! You are about to reveal the 28th term.
Last warning. Here it comes.
Okay, here are 28 terms:
1, 2, 3, 4, 25, 67, 85, 34, 53, 92, 211, 262, 233, 254, 245, 667, 675, 694, 863, 852, 311, 332, 393, 764, 705, 567, 505, 064, ...
It's simple now, isn't it? No? Oh well, look at 33 terms:
1, 2, 3, 4, 25, 67, 85, 34, 53, 92,
211, 262, 233, 254, 245, 667, 675, 694, 863, 852,
311, 332, 393, 764, 705, 567, 505, 064, 093, 932,
411, 452, 2163, ...
Okay. look at 38 terms:
1, 2, 3, 4, 25, 67, 85, 34, 53, 92,
211, 262, 233, 254, 245, 667, 675, 694, 863, 852,
311, 332, 393, 764, 705, 567, 505, 064, 093, 932,
411, 452, 2163, 2194, 2275, 2667, 2645, 2854, ...
Third solver was Stephen Burns
on 1998-04-07 using thirty-eight terms (thru 2854).
1, 2, 3, 4, 25, 67, 85, 34, 53, 92,
211, 262, 233, 254, 245, 667, 675, 694, 863, 852,
311, 332, 393, 764, 705, 567, 505, 064, 093, 932,
411, 452, 2163, 2194, 2275, 2667, 2645, 2854, 2333, 2762, ...
Forth solver was Bob Frapples
on 1998-04-16 using forty terms (thru 2762).
Fifth solver was Yossi Klein
on 1998-04-28 using forty terms (thru 2762).
1, 2, 3, 4, 25, 67, 85, 34, 53, 92,
211, 262, 233, 254, 245, 667, 675, 694, 863, 852,
311, 332, 393, 764, 705, 567, 505, 064, 093, 932,
411, 452, 2163, 2194, 2275, 2667, 2645, 2854, 2333, 2762,
2511, 2592, ...
Sixth solver Hepo
on 1998-05-04 using forty-two terms (thru 2592).
1, 2, 3, 4, 25, 67, 85, 34, 53, 92,
211, 262, 233, 254, 245, 667, 675, 694, 863, 852,
311, 332, 393, 764, 705, 567, 505, 064, 093, 932,
411, 452, 2163, 2194, 2275, 2667, 2645, 2854, 2333, 2762,
2511, 2592, 2053, ...
Seventh solver Sean Dolyniuk
on 1998-05-11 using forty-three terms (thru 2053).
1, 2, 3, 4, 25, 67, 85, 34, 53, 92,
211, 262, 233, 254, 245, 667, 675, 694, 863, 852,
311, 332, 393, 764, 705, 567, 505, 064, 093, 932,
411, 452, 2163, 2194, 2275, 2667, 2645, 2854, 2333, 2762,
2511, 2592, 2053, 2934, ...
Eighth solver Jan May on 1998-05-17 using
forty-four terms (thru 2934).
1, 2, 3, 4, 25, 67, 85, 34, 53, 92,
211, 262, 233, 254, 245, 667, 675, 694, 863, 852,
311, 332, 393, 764, 705, 567, 505, 064, 093, 932,
411, 452, 2163, 2194, 2275, 2667, 2645, 2854, 2333, 2762,
2511, 2592, 2053, 2934, 2485, 6167, 6225, 6614, ...
Ninth solver Doug Miller on 1998-05-31 using
forty-eight terms (thru 6614).
1, 2, 3, 4, 25, 67, 85, 34, 53, 92, 211, 262, 233, 254, 245, 667, 675, 694, 863, 852, 311, 332, 393, 764, 705, 567, 505, 064, 093, 932, 411, 452, 2163, 2194, 2275, 2667, 2645, 2854, 2333, 2762, 2511, 2592, 2053, 2934, 2485, 6167, 6225, 6614, 6813, ...
And finally, our tenth solver Cormac Hand put this to rest on 1998-06-04 using forty-nine terms (thru 6813).
Solution
I am absolutely amazed at the volumes of logic people have put into solving this. Bob Frapples sent pages and pages of numbers, and I'm not going to invest hours of my time into decyphering his methodology. Stephen Burns sent this program:
#include <iostream.h>
int main()
{
int arr_seeds[10] = {1, 2, 6, 8, 3, 7, 5, 0, 9, 4};
int nJump = 1;
int nCumulative=0;
int nPos;
char szRes[6] = " ";
while (nJump < 634)
{
nPos=4;
szRes[nPos] = arr_seeds[(nCumulative) % 10] + 0x030;
int nTemp = (int) (nCumulative / 10);
while (nTemp > 0)
{
int nEl = (nTemp % 10);
nPos--;
szRes[nPos] = arr_seeds[nEl] + 0x030;
nTemp = (int) (nTemp / 10);
}
nCumulative += nJump;
nJump += 2;
cout << szRes << ", ";
}
return 0;
}
with output
1, 2, 3, 4, 25, 67, 85, 34, 53, 92,
211, 262, 233, 254, 245, 667, 675, 694, 863, 852,
311, 332, 393, 764, 705, 567, 505, 064, 093, 932,
411, 452, 2163, 2194, 2275, 2667, 2645, 2854, 2333, 2762,
2511, 2592, 2053, 2934, 2485, 6167, 6225, 6614, 6813, 6312,
6711, 6512, 6013, 6914, 6425, 8167, 8285, 8634, 8853, 8392,
8511, 8062, 8933, 8454, 3145, 3667, 3875, 3394, 3563, 3052,
3411, 7132, 7293, 7864, 7305, 7567, 7005, 7464, 5193, 5632,
5311, 5752, 5063, 5994, 0175, 0667, 0845, 0754, 0033, 0462,
9211, 9692, 9353, 9534, 9985, 4167, 4625, 4314, 4513, 4912,
21111, 21612, 21313, 21514, 21925, 22167, 22685, 22334, 22553, 22992,
(I trimmed Steve's output).
Yossi Klein figured out the gimmick, then verified it with a perl program:
#!/usr/local/bin/perl
my @a = qw(1 2 3 4 25 67 85 34 53 92 211
262 233 254 245 667 675 694 863 852 311
332 393 764 705 567 505 064 093 932 411
452 2163 2194 2275 2667 2645 2854 2333 2762 2511);
foreach $a (@a)
{
$a =~ tr /0-9/7014962538/;
print "$a\n";
}
I don't read perl, but with help I now understand this program. It translates the digits
so that the sequence is easily recognizable.
But as Davis said after seeing only 12 terms (thru 262):
I believe I have solved your sequence. First you created a sequence of integers starting with zero and increasing by one. Then you squared each number in the sequence to produce the following: 0,1,4,9,16,25,36,49,64,.... and then you used a cryptogram according to the following key: 0=1, 1=2, 4=3, 9=4, 6=5, 2=6, 5=7, 3=8, 8=9, 7=0 Then you simply substitued these numbers into the original sequence: 0 >> 1 1 >> 2 4 >> 3 9 >> 4 16 >> 25 25 >> 67 36 >> 85 49 >> 34 64 >> 53 81 >> 92 100 >> 211 121 >> 262 Therefore, the next numbers would be 233 ,250,205,667.Note that Davis made a mistake here. He subsequently corrected it to "233, 254, 245, 667".
So that's it folks. It's a sequence of squares with the digits encrypted according to the key Davis derived. Note he derived 7=0 long before I used it in a term.
Squares 0123456789->1268375094
I told you earthly knowledge was a hindrance, here. I'll bet if I had used letters or arbitrary symbols it would have been easier than using digits. Imagine this sequence was received from Mars. We didn't know what the symbols meant, so we called the first one 1, the second one 2, etc.
Stephen Burns produced a correct program without uncovering that the numbers were squares. I asked him
[D]o you know what sequence results from initial values of "int nJump = 1; int nCumulative = 0;" and an increment of "nCumulative += nJump; nJump += 2;"?and he replied
Unfortunately not, so I felt like a right berk when you revealed the gimmick, especially as I'd seen it before in a school IQ test (1, 4, 9, 16 ,..), and I used the same jump =3,5,7,9,.. method only to be told they were squares. Must have some sort of blind spot there.
Actually, that's the wonderful thing about math. You can approach a problem from a different direction and still reach the same answer. Steve was clever enough to solve it without using squares.
>> my counter has been incremented
