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Write a short program on your favorite RPN calculator to display the Fibonacci sequence:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ...
No manual initialization allowed. On the HP15C LE you might want to use R/S instead of PSE. Currently I have a couple of 7step HP12C programs and one 6step WP 34S program, but this can always be improved (except by creating a new firmware just for this task  FIB can of course be used if needed :).
Happy programming!
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Two completely untest 34S solutions both six steps.
01 iC 0 Clx
02 iC 1 FIB
03 x[<>] Y PSE 10
04 RCL+ Y RCL L
05 PSE 10 INC X
06 BACK 003 BACK 004
The left version is more suitable to other RPN calculators of course. Both are untested and likely broken but the ideas are probably okay.
 Pauli
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It works. It's similar to the one I came up with when using FIB. I forgot to mention the step count should include LBL when applicable, sorry!
001 LBL 99
002 0
003 FIB
004 PSE 10
005 INC L
006 RCL L
007 BACK 04
Gerson.
Edited: 7 Jan 2012, 9:20 p.m.
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For 42S:
00 { 10Byte Prgm }
01>LBL 01
02 CLST
03 SIGN
04>LBL 02
05 STO+ ST Y
06 X<>Y
07 PSE
08 GTO 02
09 .END.
Edited: 7 Jan 2012, 9:46 p.m.
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10 bytes! One step shorter here... but 1 byte longer.
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This is supposed to be just an exercise, not a challenge. Feel free to submit your solutions. Different approaches are always interesting, regardless of code size or register usage.
Here are my solutions:
WP34S
001 LBL 88
002 1
003 LN
004 STOP
005 RCL+ L
006 BACK 02
HP12C
01 0
02 n!
03 LASTx
04 PSE
05 x<>y
06 +
07 GTO 03
01 1
02 LN
03 PSE
04 LASTx
05 x<>y
06 +
07 GTO 03
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Here are a few for the HP41C:
01 LBL 00
02 CLST
03 E
04 ENTER
05 LBL 01
06 X<>Z
07 ST+ Z
08 VIEW Z
09 PSE
10 GTO 01
11 END
16 Bytes
01 LBl 00
02 .
03 E
04 LBL 01
05 X<>Y
06 ST+ Y
07 VIEW Y
08 PSE
09 GTO 01
10 END
16 Bytes
01 LBL 00
02 E
03 .
04 LBL 01
05 +
06 PSE
07 LASTX
08 X<>Y
09 GTO 01
10 END
14 Bytes
I used local labels and synthetic instructions to make the programs shorter.
. = 0
E = 1
Cheers,
Fouad
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Thanks for your participation!
I'd completely forgotten about this old thread. Not exactly the same problem, but some interesting ideas there, like Kioshi Ashima's as implemented on the WP 34S:
001 LBL 11
002 FIX 00
003 # Phi
004 FILL
005 2
006 +
007 /
008 PSE 10
009 *
010 BACK 02
This is based on the formula F(n) = round(phi^n/sqrt(5)) (See https://oeis.org/A000045). The first two Fibonacci numbers are not
properly rounded by FIX 0 though.
Regards,
Gerson.
Edited: 8 Jan 2012, 3:21 p.m.
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On an RPL machine, these two instructions
DUP2 +
will compute the *next* Fibonacci number, given the two previous ones on the stack.
So, a full program becomes
\<< 0 1
2 100 START
DUP2 +
NEXT
\>>
(Though, on an RPL machine you'd want to collect the resulting numbers in a list, of course, and not just dump them on the stack.)
With this being too easy for an RPL machine, here's the advanced RPL programming exercise:
if F(10) is 55, what are the last 4 digits of F(10^14)?
(You're welcome to try this on your RPN machine, too, and I will be multo impressed if you can solve this...)
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Quote:
if F(10) is 55, what are the last 4 digits of F(10^14)?
Just a guess: 6875.
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lol. For a guess, that's pretty good! Did a calculator help you with your guess?
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Yes. I used your program to check it up to F(10^5) on my HP 50g:
%%HP: T(3)A(R)F(,);
\<< 0 1 2 4 ROLL
START DUP2 + ROT DROP
NEXT
\>>
F(10^2) 354224848179261915075
F(10^3) final digits: 228875
F(10^4) final digits: 366875
F(10^5) final digits: 746875
These appear to follow a pattern.
P.S.: The latter has 20,899 digits (~10^5*log_{10}(phi)). It took about one hour or so on the real 50g (I haven't timed it).
Edited: 8 Jan 2012, 7:06 p.m.
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Brilliant, Gerson!
Ok.
So, for the tenacious ones: how about the last 14 decimal digits of F(10^14)?
EDIT: you could have used the MOD instruction to keep the number of digits down. That would compute F(10^5) *a lot* faster. But, still, 10^14 is out of reach with this approach.
Edited: 8 Jan 2012, 7:21 p.m.
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The last four digits of F(10^4) coincide with the last four digits of F(10000) because 10^14 mod 15000 equals 10000.So it suffices to compute F(10000), easily done on the real 50g. See Pisano periods here. There is a calculator for the cycle lengths (scroll down halfway through the page). For modulus 10000 the result is
Fibs mod 10000: cycle length is 15000
It will not compute the cycle length for modulus 10^14, but it can be extrapolated from lower power of 10. The length appears to be 1.5*10^14, which means this approach won't be of help in this case.
Cheers,
Gerson.
P.S.: The fourth and higher powers of 10 modulus 15000 equals 10000  so my guess was indeed right :)
Edited: 8 Jan 2012, 10:07 p.m.
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My approach was slightly different: It's clear that the last digit of F_{n} has to repeat since there are only 10x10 = 100 possibilities for the two previous values. I used the following program with initial values ST X: 1 and ST Y: 0:
X<>Y
RCL+ ST Y
10
MOD
It turns out the length of the period is 60. So I added a loop to find the period for the last two digits:
00 { 27Byte Prgm }
01 LBL "Fib"
02 X<>Y
03 60
04 X<> ST Z
05 LBL 00
06 X<>Y
07 RCL+ ST Y
08 100
09 MOD
10 DSE ST Z
11 GTO 00
12 END
This gave me the following sequence:
y: 20 40 60 80 0
x: 61 21 81 41 1
Thus the period is: 60 x 5 = 300
Next step was to change line 03 to 300 and line 08 to 1E3 to find another sequence of 5:
y: 600 200 800 400 0
x: 801 601 401 201 1
After changing line 03 to 1500 and line 08 to 1E4 I got a sequence of 10:
y: 8000 6000 4000 2000 0 8000 6000 4000 2000 0
x: 9001 8001 7001 6001 5001 4001 3001 2001 1001 1
Thus I ended up with a period of 60 x 5 x 5 x 10 = 15,000.
From here I did the same as Gerson: 10^{14} MOD 15,000 = 10,000.
Changed line 03 to 10000 and line 08 to 1E5 to find:
y: 30626
x: 66875
Cheers
Thomas
Edited: 9 Jan 2012, 3:07 a.m.
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Wow, Thomas, very nice! I *am* impressed.
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Yes, your guess was right. I volunteer your result is even right to 5 digits.
(Which still doesn't help much toward 14 digits...)
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If you look up the Fibonacci page on wikipedia, you'll see various methods for direct computation of F(n).
One is matrix exponentiation: F(n) = [[0 1][1 1]] ^ n
This program for the 50g will tell you what F(100) mod 10^14 is, for example:
\<< 1.E14 R\>I MODSTO
[[0 1][1 1]] 1.E2 R\>I POWMOD
{1 2} GET
\>>
It takes about 1s on Emu48, so I figure about 10s on a 50g.
Now, if you change 1.E2 to 1.E3 it will produce a negative result. No idea why.
If you try 1.E14 it will tell you: "POWMOD Error: Integer too large"
I conclude that on the 50g POWMOD first does the POW, then the MOD, which, uhm, renders this instruction pretty much useless for matrices. (You want to MOD while you're POW'ing, or else you don't avoid the giant intermediate numbers.)
Well, it works on ND1, and the result is: 88299560546875
Modular matrix exponentiation can be made very fast with the binary method, where you're, essentially, moving in powers of 2 with every iteration. There're only ~50 (~log2(10^14)) matrix multiplications necessary to arrive at 10^14.
This could also be made to work on the 50g with ~10 lines of code to write a proper version of POWMOD. I'm pretty sure this could be made to run in under a minute, for F(10^14).
There's a Project Euler problem (#304) where finding this number (almost this number, there's a different modulus) is the first part of the simple but brutal task of summing up the Fibonacci numbers for each of 100,000 primes following 10^14.
On the 50g, 10 of those primes
\<< 1.E14 R\>I 1 10 START NEXTPRIME NEXT \>>
take about a minute to find (5s in Emulator; not sure how long exactly on the 50g), so 100,000 is a bit... daunting.
(I'm not sure what's done on the 50g, but this speed suggests to me that they're not using MillerRabin for the probabilistic primality check that kicks in for large numbers, used by NEXTPRIME.)
Once you have the F(NEXTPRIME(10^14)) and F(PREVPRIME(10^14)), you can use normal Fibonacci iteration (with MOD) to compute the spans between primes and eliminate the need to use matrix exponentiation to find each F(n). With an average of 33 numbers from prime to prime in this range, you're looking at having to compute 3.3 million Fibonacci numbers, though.
If you're interested, here's RPL+ code to do all that:
\<< 1234567891011 =m
1e14 =bias
1e5 =count
\<< [[0 1][1 1]] \>big SWAP
m modpow
DUP {1 1} GET
SWAP {2 1} GET \>> =F2
0 =sum
bias =:prev =a
a F2 EVAL
1 count START
1
a NEXTPRIME =:a prev 
START
DUP ROT +
m MOD
NEXT
a =prev
DUP sum + m MOD =sum
NEXT
DROP DROP
sum
\>>
Note that the matrix exponentiation method actually computes F(n1), F(n), F(n+1) at once. So, to continue with Fib iteration you can pick two of them from the result matrix, and you're ready to go.
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Quote:
So, for the tenacious ones: how about the last 14 decimal digits of F(10^14)?
We can combine the Fibonacci and the LucasSequence to find a fast algorithm to calculate both sequences using the following identities:
The idea is similar to binary exponentiation (a.k.a squareandmultiply algorithm). Here's a Pythonprogram:
#!/usr/bin/python
K = 10**14
B = 2**47
N = 0
M = 10**14
F = 0
L = 2
while B > 1:
(F, L) = (F*L % M, (5*F*F+L*L)/2 % M)
B /= 2
N *=2
print "%20d %20d %20d %20d %20d" % (N, B, K, F, L)
if K >= B:
(F, L) = ((F+L)/2 % M,(5*F+L)/2 % M)
K = B
N += 1
print "%20d %20d %20d %20d %20d" % (N, B, K, F, L)
And that's the output:
N B K F L
0 70368744177664 100000000000000 0 2
1 70368744177664 29631255822336 1 1
2 35184372088832 29631255822336 1 3
4 17592186044416 29631255822336 3 7
5 17592186044416 12039069777920 5 11
10 8796093022208 12039069777920 55 123
11 8796093022208 3242976755712 89 199
22 4398046511104 3242976755712 17711 39603
44 2199023255552 3242976755712 701408733 1568397607
45 2199023255552 1043953500160 1134903170 2537720636
90 1099511627776 1043953500160 67194370816120 26026380244498
180 549755813888 1043953500160 38521399707760 95774259272002
181 549755813888 494197686272 67147829489881 44190628905401
362 274877906944 494197686272 73003235747281 62383406970803
363 274877906944 219319779328 67693321359042 13699792853604
726 137438953472 219319779328 22604627687368 59229375788818
727 137438953472 81880825856 40917001738093 86126257112829
1454 68719476736 81880825856 68797008295097 27914836383243
1455 68719476736 13161349120 48355922339170 85949938929364
2910 34359738368 13161349120 29880280387880 1222581444498
5820 17179869184 13161349120 19590131884240 30816254472002
11640 8589934592 13161349120 19866585048480 83263801888002
11641 8589934592 4571414528 51565193468241 91298363565201
23282 4294967296 4571414528 94702926281441 21451378170403
23283 4294967296 276447232 58077152225922 47483004788804
46566 2147483648 276447232 53214104177288 83596643750418
93132 1073741824 276447232 75272696106384 856675174722
186264 536870912 276447232 32522299625248 84969227777282
372528 268435456 276447232 10340608015936 31006195307522
372529 268435456 8011776 20673401661729 41354617693601
745058 134217728 8011776 58418769896129 96892716347203
1490116 67108864 8011776 48860609677187 39467245923207
2980232 33554432 8011776 67118061778709 99561741164847
5960464 16777216 8011776 35902403842523 89664428533407
11920928 8388608 8011776 20398272665861 51376915027647
23841856 4194304 8011776 15043208059067 60138774356607
23841857 4194304 3817472 37590991207837 67677407325971
47683714 2097152 3817472 79233668834727 93180651092843
47683715 2097152 1720320 86207159963785 44674497633239
95367430 1048576 1720320 68859452249615 77012557631123
95367431 1048576 671744 72936004940369 10654909439599
190734862 524288 671744 76957202272031 55876229280803
190734863 524288 147456 66416715776417 20331120320479
381469726 262144 147456 89852350343743 94115666789443
762939452 131072 147456 10390253505149 64051296250247
762939453 131072 16384 37220774877698 58001281887996
1525878906 65536 16384 1539434313208 72834288896018
3051757812 32768 16384 34530356005744 10933216256322
6103515624 16384 16384 86304808313568 3648804967682
6103515625 16384 0 44976806640625 17586423267761
12207031250 8192 0 74957275390625 45489501953123
24414062500 4096 0 14520263671875 81231689453127
48828125000 2048 0 71563720703125 64288330078127
97656250000 1024 0 90838623046875 8721923828127
195312500000 512 0 85528564453125 49542236328127
390625000000 256 0 60955810546875 86651611328127
781250000000 128 0 57012939453125 87432861328127
1562500000000 64 0 78924560546875 85870361328127
3125000000000 32 0 92950439453125 88995361328127
6250000000000 16 0 50799560546875 82745361328127
12500000000000 8 0 36700439453125 95245361328127
25000000000000 4 0 38299560546875 70245361328127
50000000000000 2 0 11700439453125 20245361328127
100000000000000 1 0 88299560546875 20245361328127
My guess is that this algorithm should be fast on a HP50g as well. Due to the lack of arbitrary precision the HP42S doesn't qualify for this.
Kind regards
Thomas
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How about another:
01 iC 0
02 iC 1
03 [cmplx]ENTER
04 +
05 PSE 10
06 BACK 003
Six steps, seven with an initial label.
 Pauli
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Nice use of ^{C}ENTER as DUP2 in RPL!
%%HP: T(3)A(R)F(,);
\<< \> n
\<< 0 1 1 n 2 
START DUP2 +
NEXT n \>LIST
\>>
\>>
Gerson.
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And another:
01 iC 0
02 iC 1
03 +
04 PSE 10
05 y[<>] L
06 BACK 003
Same length though.
 Pauli
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There's more than a way to skin a cat. On the HP15C:
001 f LBL A
002 g CLx
003 f MATRIX 1
004 f LBL 0
005 f PSE
006 RCL + 0
007 f x<> 0
008 GTO 0
Gerson.
Edited: 9 Jan 2012, 11:20 p.m.
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Nice work, again, Thomas.
Quote:
Due to the lack of arbitrary precision the HP42S doesn't qualify for this.
That's what I used to think too, until, almost exactly one year ago, an intrepid bunch (some active posters here) showed the way: Making Fib bits visible (on Silicium.org)
(I think this is my alltime favorite calculatorrelated discussion. It starts slow but quickly shows real ingenuity. I still can't get over those HP28C printouts.)
Now, I'd be willing to bet that F(10^14) mod 10^14 can even be solved on a 15C (LE).
Edited: 10 Jan 2012, 9:08 a.m.
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Thanks for link: it's an interesting post! I didn't mean to say it's impossible to do it using the HP42S. Or we could use Peter's MCODE MultiPrecision Library for HP41. I might give it a try but this is definitely not my first choice.
Thanks for the challenge as well!
Thomas
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I'm very tempted to add some more complex constants to the 34s "[cmplx] 1" e.g.
 Pauli
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That would be useful, not just for this task, so go ahead if there's room enough. By the way, is it possible to enter i (1 0) in one step only? Thanks!
Gerson.
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I can't think of a way to get i on the stack in one step. Can do it in two though.
We do have a complex version of the constants catalogue which puts zero in Y already. This doesn't include any small integers though. "CPX hshift CONST"
I've been thinking about possible functions to help with the commands in xrom stuff. If any more get added, some will be exposed to the user but don't expect them to necessarily be pleasant to use.
 Pauli
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Quote:
We do have a complex version of the constants catalogue which puts zero in Y already. This doesn't include any small integers though. "CPX hshift CONST"
Yes, I had checked that but then I realized it wouldn't make sense to include small integers there. Changing the behavior of the CPX key before numeric keys and [.] might be a option, for instance CPX 123 would put 0 in Y and 123 in X. [CPX] [R/S] might be a nonintuitive shortcut for entering i. I don't know the impact of these changes in the executable size though. Also, this may not be useful enough to be worth the trouble.
Gerson.
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A function that does pretty much nothing requires of the order of 100 bytes. This is function table overheads, catalogue overheads and the actual function prelude/epilogue that seems almost unavoidable.
Making small changes to the keyboard handler isn't so bad  much of this is table based now so adding a command to an unused location is usually cheap.
Adding a complex prefix to digit entry is quite a different change. No idea how large this would be.
The current work is moving as much functionality into internal keystroke programs as we can. Very slow but saving lots of space.
 Pauli
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XLBL"i"
iC 1
iC 0
RTN
This should do the trick with minimum overhead. We have to add a catalogue entry and a command table entry but that's not too expensive.
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Quote: This should do the trick with minimum overhead. We have to add a catalogue entry and a command table entry but that's not too expensive.
Not quite  the stack dynamics would be incorrect :(
XLBL"i"
xIN NILADIC_1_COMPLEX
iC 1
iC 0
xOUT xOUT_NORMAL
will work however. The code is 8 bytes, the command table is 10  12 bytes (I forget which) and the catalogue entry is 10 bits but can't live in the complex constants catalogue which is compiled automatically  the complex functions catalogue seems like an okay place to put it.
Pauli
Edited: 11 Jan 2012, 5:21 p.m.
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We've got a shorter solution on the 34s now:
01 [cmplx]i
02 RCL+ Y
03 PSE 10
04 x[<>] Y
05 BACK 003
 Pauli
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That's what [cmplx]i was for! If the steps 02 and 03 are swapped the sequence will begin with F(0). I'd searched for an i constant in the HP42S catalog but there was none. I think this first appeared in the HP28C. If now WP 34S has it, then it was worth posting this simple RPN exercise :)
BTW, how [cmplx]i is entered?
Gerson.
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Quote:
BTW, how [cmplx]i is entered?
It's in the complex catalogue (CPX 3) but I just mapped it to CPX dot as well. I'm sure Walter will find a better place on the keyboard.
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Remember that WP 34S instructions are generally 16 bits long. This allows to cramp so much functionality into a single instruction but costs more bytes than other designs. We are down to 10 bytes here, still pretty decent...
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We couldn't put it in the complex constants catalogue since that is automatically produced alongside the real version. Still, the complex catalogue is a decent place and Walter will likely find somewhere decent on the keyboard to locate it. "CPX 1" might be okay which would allow "CPX 0" as well :)
 Pauli
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We have way more than 256 different instructions kind of required this and by doing so the argument can be squeezed into the instruction as well.
We currently have 571 instructions and that is not counting integer, real & complex as different functions (there are 660 if you do that). We've almost got as many different functions as the 15C has total opcodes.
 Pauli
Edited: 12 Jan 2012, 5:38 p.m.
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HP 35s
F001 LBL F
F002 0
F003 1
F004 PSE
F005 eqn REGX+REGY
F006 GTO F004
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Great! The first 6step program for an unmodified HP calculator!
Since it does show F(0) and F(1) before looping (thanks to the 2line display) that's okay. Also, this allows us to match Allen's 10byte step solution for the HP42S above:
00 {10Byte Prgm }
01 LBL 00
02 CLST
03 N!
04 LBL 01
05 PSE
06 RCL+ ST L
07 GTO 01
08 .END.
Gerson.
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The 42S not, no, but Free42 decimal (as on the iPhone) can determine the last 12 digits this way.
Cheers, Werner
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I rather doubt the 35s solution is less than 10 bytes.
 Pauli
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Sorry! I think I didn't make myself clear. I meant showing F(0) and F(1) at the same time like Morten did would allow me to achieve the 10byte count in the HP42S. My post was really ambiguous, I realize now.
Gerson.
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Challenge accepted:
1E12
STO 00
1E14
XEQ "Fib"
gives:
y: 245361328127
x: 299560546875
Cheers
Thomas
00 { 90Byte Prgm }
01 LBL "Fib"
02 STO 01
03 LOG
04 2
05 LOG
06 /
07 IP
08 1
09 +
10 2
11 X<>Y
12 Y^X
13 STO 02
14 CLST
15 2
16 X<>Y
17 LBL 00
18 1
19 RCL 02
20 X<=Y?
21 GTO 02
22 RDN
23 RDN
24 RCL ST Y
25 X^2
26 X<>Y
27 STO* ST Z
28 X^2
29 STO+ ST Y
30 2
31 STO/ 02
32 STO/ ST Z
33 *
34 +
35 RCL 00
36 MOD
37 X<>Y
38 RCL 00
39 MOD
40 RCL 01
41 RCL 02
42 X<=Y?
43 GTO 01
44 RDN
45 RDN
46 GTO 00
47 LBL 01
48 
49 STO 01
50 RDN
51 STO+ ST Y
52 2
53 STO/ ST Z
54 *
55 RCL+ ST Y
56 RCL 00
57 MOD
58 X<>Y
59 RCL 00
60 MOD
61 GTO 00
62 LBL 02
63 RDN
64 RDN
65 END
Edited: 14 Jan 2012, 3:11 a.m.
