[VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5tier"

04012019, 12:49 AM
Post: #21




RE: [VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5tier"
You are right Albert and I can't find a way around.


04012019, 03:25 PM
(This post was last modified: 04032019 06:01 PM by Albert Chan.)
Post: #22




RE: [VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5tier"
Just figured out how to improve cin(x) accuracy for large x
cin(x) = arcsin(cin(sin(x))) = nest(arcsin, cin(nest(sin, x, n)), n) Pick enough nested sin's so cin argument is small, say below 0.1 radian cin[x0_] := Block[ {n=0, x=x0+0.0}, While[Abs[x] ≥ 0.1, x = Sin[x]; n++]; Nest[ArcSin, x  (1/18) x^3  (7/1080) x^5  (51/32285) x^7, n] ] Above cin(x) setup give about 12 digits accuracy: x cin(x) cin(cin(cin(x)))  sin(x) 0.0 0.0 +0.0 0.2 0.199553461081 1.9e16 0.4 0.396375366278 +1.8e14 0.6 0.587446695546 1.1e16 0.8 0.769025184826 9.1e14 1.0 0.935745970819 +1.4e13 Pi/2. 1.210368344457 +2.6e13 0.71 0.688778525307 1.6e13 2.019 1.026923318694 +6.4e13 Edit: changed x^7 coefficient from 0.00158 to 51/32285 to get better accuracy 

04012019, 04:42 PM
(This post was last modified: 04012019 06:29 PM by JF Garnier.)
Post: #23




RE: [VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5tier"
(04012019 03:25 PM)Albert Chan Wrote: Just figured out how to improve cin(x) accuracy for large x Excellent ! Here is the HP71 version and results, after decipher of your code (not familiar with that language...): 10 ! SSMC24 20 A=1/18 @ B=7/1080 @ C=.00158 30 DEF FNC(X) 40 N=0 50 X=SIN(X) @ N=N+1 @ IF ABS(X)>=.1 THEN 50 60 ! X=X+A*X^3+B*X^5+C*X^7 61 X=C*X^7+B*X^5+A*X^3+X ! better 70 FOR I=1 TO N @ X=ASIN(X) @ NEXT I 80 FNC=X 90 END DEF 100 ! 110 FOR X=.2 TO 1 STEP .2 120 Y=FNC(FNC(FNC(X))) 130 PRINT X;Y;SIN(X);YSIN(X) 140 NEXT X >RUN .2 .198669330795 .198669330795 0 .4 .389418342314 .389418342309 .000000000005 .6 .564642473542 .564642473395 .000000000147 .8 .717356091570 .717356090900 .000000000670 1. .841470984040 .841470984808 .000000000768 >FNC(PI/2);FNC(FNC(FNC(PI/2))) 1.2103683495 .999999998579 >FNC(0.71) .688778525229 >FNC(2.019) 1.02692332142 JF [Edited: reversed the order of the polynom term evaluation, for slightly better accuracy] 

04022019, 10:43 PM
Post: #24




RE: [VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5tier"
Hi, all: Continuing with my original solutions, today it's time for: Tier 4  The Challenge: Consider the npoint dataset (1, 2), (2, 3), (3, 5), (4, 7), (5, 11), (6, 13), ..., (n, p_{n}) (the prime numbers), and the (n1)^{st} degree polinomial fit to this dataset of the form: P(x) = a_{0} + a_{1} (x1) + a_{2} (x1) (x2) + ... + a_{n1} (x1) (x2) (x3) ... (x(n1)) Write a program that takes no inputs but computes and outputs the limit of the sum of the coefficients a_{0}, a_{1}, ... , a_{n1} when n tends to infinity. My original solution: My original solution for the HP71B is this 4liner (168 bytes): 1 DESTROY ALL @ OPTION BASE 0 @ REPEAT @ N=N+1 @ DIM C(N) @ T=S 2 FOR I=1 TO N @ C(I)=FPRIM(C(I1)+1) @ NEXT I @ S=0 3 FOR I=1 TO N1 @ FOR J=N TO I+1 STEP 1 @ C(J)=C(J)C(J1) @ NEXT J @ NEXT I 4 FOR I=1 TO N @ S=S+C(I)/FACT(I1) @ NEXT I @ UNTIL S=T @ DISP N;S >RUN 20 3.40706916561 { it converged to the limit after fitting the first 20 primes: 2, 3, 5, ..., 71) } Notes:
That's all for Tier 4, thanks a lot to Albert Chan for his interest in this particular tier and congratulations for providing a correct solution and some explanation but please, Albert, next time *do* provide actual code for an HP calculator of your choice, so that people can try your solution for themselves. In the next days I'll post my solutions for the remaining tiers. V. Find All My HPrelated Materials here: Valentin Albillo's HP Collection 

04032019, 02:05 AM
Post: #25




RE: [VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5tier"
(04022019 10:43 PM)Valentin Albillo Wrote: I think you meant sum converged using 19 primes (20 primes to confirm 12digits convergence) sum using 19 primes = 414453 270752 384363 / 19! ≈ 3.40706 916563 sum using 20 primes = 414453 270752 580132 / 19! ≈ 3.40706 916563 Also, forward difference tables may be built incrementally. C(1) = p1 C(2) = p2  p1 C(3) = p3  2 p2 + p1, C(4) = p4  3 p3 + 3 p2  p1, C(5) = p5  4 p4 + 6 p3  4 p2 + p1, ... Above can be simplified without a prime table: C(1) = p1 C(2) = p2  C(1) C(3) = p3  C(1)  2 C(2) C(4) = p4  C(1)  3 C(2)  3 C(3) C(5) = p5  C(1)  4 C(2)  6 C(3)  4 C(4) ... 

04032019, 02:57 AM
(This post was last modified: 04272019 05:02 PM by Albert Chan.)
Post: #26




RE: [VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5tier"
I only have a HP12C, which is not powerful enough to make primes, build delta tables ...
XCas code: terms(n) := { local c, s, p, j, k; c := flatten(matrix(n,0)); s := 0; p := 0; for(j:=0; j<n; j++) { p := nextprime(p); c[j] := p; for(k:=0; k<j; k++) c[j] := c[j]  comb(j,k) * c[k]; s += c[j] / float(j!); print(p, s); } } terms(20) → 02 2.0 03 3.0 05 3.5 07 3.33333333333 11 3.45833333333 13 3.38333333333 17 3.41527777778 19 3.40476190476 23 3.4076140873 29 3.40696097884 31 3.40708691578 37 3.40706684905 41 3.4070693814 43 3.40706915834 47 3.40706916344 53 3.40706916625 59 3.40706916552 61 3.40706916564 67 3.40706916563 71 3.40706916563 Edit: replaced Python code to XCas, so HP prime user can try out. 

04052019, 01:58 AM
Post: #27




RE: [VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5tier"
Hi, all: At long last, today it's time for my final original solution, namely: Tier 5  The Challenge: Consider the function cin(x) which has the defining property that cin(cin(cin(x))) = sin(x). Write a program or function which accepts an argument x in the range [Pi, Pi] and outputs the corresponding value of cin(x) correct to at least 810 digits in the whole range. Use it to tabulate cin(x) for x = 0.0, 0.2, 0.4, ..., 1.0 and also to compute cin(Pi/2), cin(0.71), cin(2.019). My original solution: My original solution for the HP71B is the following userdefined function (plus initialization code): 1 DESTROY ALL @ OPTION BASE 1 @ DIM C(7) @ READ C 2 DATA 1,1/18,7/1080,643/408240,13583/29393280,29957/215550720,24277937/648499737600 3 DEF FNC(X) @ L=0 @ M=1/3 @ REPEAT @ X=SIN(X) @ L=L+1 @ UNTIL ABS(X)<M 4 S=0 @ FOR Z=1 TO 7 @ S=S+C(Z)*X^(2*Z1) @ NEXT Z 5 FOR Z=1 TO L @ S=ASIN(S) @ NEXT Z @ FNC=S @ END DEF Instead of tabulating it for 0.0, 0.2, ..., 1.0 as I originally asked, let's better tabulate it for x from 0 to Pi/2 in steps of Pi/10: 6 FOR X=0 TO PI/2 STEP PI/10 7 Y=FNC(FNC(FNC(X))) @ DISP X;FNC(X);Y;SIN(X);YSIN(X) @ NEXT X >FIX 10 >RUN x cin(x) cin(cin(cin(x))) sin(x) Error  0.0000000000 0.0000000000 0.0000000000 0.0000000000 0 0.3141592654 0.3124163699 0.3090169944 0.3090169944 1.0E12 0.6283185307 0.6138343796 0.5877852523 0.5877852523 2.2E11 0.9424777961 0.8897456012 0.8090169944 0.8090169944 4.1E11 1.2566370614 1.1122980783 0.9510565164 0.9510565163 1.0E10 1.5707963268 1.2103683445 1.0000000000 1.0000000000 1.0E11 So we've got 10 correct decimals or better, as the error in cin(x) is even smaller than the error in cin(cin(cin(x)))sin(x) which doesn't exceed 10^{10}. As for the discrete values asked in the challenge: >FIX 10 @ FNC(PI/2); FNC(0.71); FNC(2.019) 1.2103683445 0.6887785253 1.0269233188 Notes:
That's all for Tier 5, I could say a whole lot more about this topic and post additional code and results but this post is long enough as it is so I'll stop right now. Thank you very much to Albert Chan, JF Garnier, Oulan and Gerson W. Barbosa for your valuable contributions and to Werner for your interest, I hope you enjoyed it all ! V. Find All My HPrelated Materials here: Valentin Albillo's HP Collection 

04052019, 07:40 PM
(This post was last modified: 04072019 04:12 PM by Albert Chan.)
Post: #28




RE: [VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5tier"
Below Lua code scale cin argument to [sin(0.5), 0.5], do cin, then undo asin/sin's
local sin, asin = math.sin, math.asin function cin(x) local y, n = x*x, 0 while y > 0.25 do x=sin(x); y=x*x; n=n+1 end if y < 0.0324 then  x < 0.18 local z = y*(0.00013898 + y*0.00003744) + 13583/29393280 x = x  x*y*(1/18 + y*(7/1080 + y*(643/408240 + y*z))) return n==0 and x or asin(x) end while y < 0.229848847 do x=asin(x); y=x*x; n=n1 end y = y  0.2399  x = [sin(0.5), 0.5] y = 0.013724194890539722 + y*( 0.058965322546572385 + y*( 0.007795773378183463 + y*( 0.002109528417736682 + y*( 0.000663984666232017 + y*( 0.000199482968029459 ))))) x = x  x*y  x = cin(x) for i=1,n do x = asin(x) end for i=1,n do x = sin(x) end return x end Result *very* accurate. Example: x = 2.019 cin(x) = 1.02692 331869 35764 cin(cin(x)) = 0.956628 929996 1186 cin(cin(cin(x))) = 0.90122 698939 98129 math.sin(x) = 0.90122 698939 98126 

04072019, 04:58 PM
Post: #29




RE: [VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5tier"
Though I did not participate in this challenge, I have taken the liberty of adapting Valantin's Albert's programs into RPL with a twist unlimited precision.
This program does not run until convergence but a fixed number of iterations, which is the number n that is input into the program. The program requires the external libraries ListExt, GoferLists, and Long Float. %%HP: T(3)A(R)F(.); @ Generate list of primes: \<< \> n \<< 2 2 n START DUP NEXTPRIME NEXT n \>LIST @ Inverse binomial transform of above list: DUP HEAD SWAP 2 n START \GDLIST DUP HEAD SWAP NEXT DROP n \>LIST @ List of factorials 0 through n: 1 n 1  LSEQ \<< * \>> Scanl1 + @ Divide to form list of ratios: / @ Accumulate above list: \<< + EVAL \>> Scanl1 @ Convert to list of LongFloats: 1. \<< \>FNUM \>> DOLIST \>> \>> To begin, store a number into the variable DIGITS which sets the precision that LongFloat uses. In this example, I used 50. for DIGITS and 60 for the number of iterations. I then used the following simple program to turn the resulting list into a string suitable for display or printing: \<< \>STR 3. OVER SIZE 2.  SUB " " 13. CHR 10. CHR + SREPL DROP \>> The result: 2 3 35000000000000000000000000000000000000000000000000.E49 33333333333333333333333333333333333333333333333333.E49 34583333333333333333333333333333333333333333333333.E49 33833333333333333333333333333333333333333333333333.E49 34152777777777777777777777777777777777777777777778.E49 34047619047619047619047619047619047619047619047619.E49 34076140873015873015873015873015873015873015873016.E49 34069609788359788359788359788359788359788359788360.E49 34070869157848324514991181657848324514991181657848.E49 34070668490460157126823793490460157126823793490460.E49 34070693813966383410827855272299716744161188605633.E49 34070691583365194476305587416698527809638920750032.E49 34070691634410012386202862393338583814774290964767.E49 34070691662452161790786129410468034806659145283484.E49 34070691655244232025812713643401474089304777135465.E49 34070691656406347881896434650869758246042092914175.E49 34070691656257873262373750840742575708166882908384.E49 34070691656273966717789714824786163263074495403684.E49 34070691656272442599684037804120145048719207427525.E49 34070691656272570305882488238644909673728392032261.E49 34070691656272560845399289953795750049707556470279.E49 34070691656272561452781304231311072994035926413957.E49 34070691656272561421162961843454416475655340721354.E49 34070691656272561422168859510781044267796095465102.E49 34070691656272561422203623227227792237954574766870.E49 34070691656272561422193499936605021028268165932935.E49 34070691656272561422194680710735493212204028544306.E49 34070691656272561422194575066153490058695686270606.E49 34070691656272561422194583144322191113049301542936.E49 34070691656272561422194582596611005303547258196512.E49 34070691656272561422194582630026111767164206858021.E49 34070691656272561422194582628184366702426127844481.E49 34070691656272561422194582628275666348866780110705.E49 34070691656272561422194582628271661692858309901367.E49 34070691656272561422194582628271810600936280034971.E49 34070691656272561422194582628271806504336081216950.E49 34070691656272561422194582628271806529151697629234.E49 34070691656272561422194582628271806536170465629760.E49 34070691656272561422194582628271806535499300745912.E49 34070691656272561422194582628271806535542548688840.E49 34070691656272561422194582628271806535540241528738.E49 34070691656272561422194582628271806535540348557090.E49 34070691656272561422194582628271806535540344239572.E49 34070691656272561422194582628271806535540344383289.E49 34070691656272561422194582628271806535540344380187.E49 34070691656272561422194582628271806535540344380140.E49 34070691656272561422194582628271806535540344380151.E49 34070691656272561422194582628271806535540344380150.E49 34070691656272561422194582628271806535540344380149.E49 34070691656272561422194582628271806535540344380150.E49 34070691656272561422194582628271806535540344380150.E49 34070691656272561422194582628271806535540344380150.E49 34070691656272561422194582628271806535540344380149.E49 34070691656272561422194582628271806535540344380150.E49 34070691656272561422194582628271806535540344380150.E49 34070691656272561422194582628271806535540344380151.E49 34070691656272561422194582628271806535540344380150.E49 34070691656272561422194582628271806535540344380151.E49 It can be observed that:  LongFloat numbers are not very userfriendly.  There is noise in the last digit, so really 49digit accuracy in this case.  Rate of converge increases, only about 56 iterations required to confirm 49 digits. 

04082019, 03:36 PM
(This post was last modified: 04102019 02:41 AM by Albert Chan.)
Post: #30




RE: [VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5tier"
I posted cin(x) puzzle to the Lua mailing list, and got an elegant solution from Egor Skriptunoff.
Taylor coefficients built on the fly, without any need for CAS. http://luausers.org/lists/lual/201904/msg00063.html Below code modified a bit for speed, accuracy, and extended cin(x) for tin(x): Quote:local sin, asin = math.sin, math.asin lua> maclaurin_coefs = maclaurin_of_tin() lua> for i=50,125,25 do  match post #28 Coefs : print(2*i+1, maclaurin_coefs(i)) : end 101 0.08337562280550574 151 388536047335.2163 201 6.555423874650777e+027 251 3.536522049267692e+046 lua> function nest(f,x,n) for i=1,n do x=f(x);print(i, x) end end lua> nest(egor, 2.019, 2)  egor = tin 1 0.9894569770589354 2 0.9012269893998129 lua> maclaurin_coefs = maclaurin_of_cin() lua> nest(egor, 2.019, 3)  egor = cin 1 1.0269233186935764 2 0.9566289299961186 3 0.9012269893998129 lua> math.sin(2.019) 0.9012269893998126 

04092019, 06:53 PM
Post: #31




RE: [VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5tier"
(04072019 04:58 PM)John Keith Wrote: It can be observed that: They needn't be so. 34070691656272561422194582628271806535540344380151.E49 \<< ZZ\<\>F 51 FC? { "." } { "," } IFTE SWAP ROT \>STR DUP SIZE ROT + OVER 1 ROT SUB ROT + " " ROT + 1 ROT REPL \>> EVAL > 3.4070691656272561422194582628271806535540344380151  # EE7Dh 100 bytes, which can be optimized, of course. 

04092019, 11:37 PM
Post: #32




RE: [VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5tier"
.
Hi again, all (04052019 07:40 PM)Albert Chan Wrote: Below Lua code scale cin argument to [sin(0.5), 0.5], do cin, then undo asin/sin's [...] Result *very* accurate. Example: Indeed, impressive accuracy ! Thanks a lot for your Lua code, Albert Chan, I hope you'll adapt it to some HP calc's native programming language when you eventually get your hands on one (apart from the HP12C, that is). (04072019 04:58 PM)John Keith Wrote: Though I did not participate in this challenge, I have taken the liberty of adapting Valantin's Albert's programs into RPL with a twist unlimited precision.[...] The result: Yes, it does converge very fast and I love multiprecision computations and results. In fact, I don't understand why HP didn't ever include it in some of its advanced models right from the box (at least double precision as in some SHARP models which would do 20 digits without batting an eyelid.) Thanks a lot for your interest and your RPL highprecision results, much appreciated. (04082019 03:36 PM)Albert Chan Wrote: I posted cin(x) puzzle to the Lua mailing list, and got an elegant solution from Egor Skriptunoff. Taylor coefficients built on the fly, without any need for CAS. As I said before, truly excellent accuracy. Also thank you very much for posting my challenge to the Lua forums, for giving me credit for it, and for your outstandingly clear code which also includes an implementation and highprecision results fot the tin(x) function I mentioned in the challenge. Again, really appreciated. (04092019 06:53 PM)Gerson W. Barbosa Wrote:(04072019 04:58 PM)John Keith Wrote: It can be observed that: [...] LongFloat numbers are not very userfriendly. Very good effort to increase usability. As you know RPL is not my thing but I can appreciate your ingenuity. Thanks, Gerson. Best regards to all of you. V. . Find All My HPrelated Materials here: Valentin Albillo's HP Collection 

04102019, 04:11 PM
(This post was last modified: 04102019 04:11 PM by John Keith.)
Post: #33




RE: [VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5tier"
(04092019 11:37 PM)Valentin Albillo Wrote: Yes, it does converge very fast and I love multiprecision computations and results. In fact, I don't understand why HP didn't ever include it in some of its advanced models right from the box (at least double precision as in some SHARP models which would do 20 digits without batting an eyelid.) Thanks for your kind words, Valentin. The HP 49 and 50 do have exact integers whose size is limited only by memory. Though LongFloat is an external library and is a bit rough around the edges, its precision can be set up to 9999 digits. At that point, I think formatting becomes moot. 

04132019, 08:01 PM
Post: #34




RE: [VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5tier"
Hello Valentin,
I don't understand the term composite in the context of Tier 2. I first thought, that the result of SB must have at least 2 digits, but that can't be the point. Please explain what's meant by composite. Best regards Bernd 

04132019, 10:35 PM
(This post was last modified: 04142019 08:53 PM by Albert Chan.)
Post: #35




RE: [VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5tier"
I recently created nextprime.lua, which is needed for solving Tier 2 puzzle.
My Lua code available in https://github.com/achan001/PrimePi Quote:p = require 'nextprime' lua> function loop(n,f) for i=1,n do io.write(f(),' ') end print() end lua> seq=sb_find(7) lua> loop(10,seq) 7 4801 9547 9601 11311 11317 11941 11953 13033 13327 lua> seq=sb_find(31) lua> loop(10,seq) 31 619 709 739 769 829 859 919 1549 1579 

04132019, 11:27 PM
Post: #36




RE: [VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5tier"
Hi, Bernd Grubert and Albert Chan: (04132019 08:01 PM)Bernd Grubert Wrote: I don't understand the term composite in the context of Tier 2. [...] Please explain what's meant by composite. With pleasure. In this context composite simply means not prime, i.e., if a number is not prime (thus it can be factored as the product of at least two not necessarily distinct prime factors) then it is considered composite. For instance: 25 is composite because it's not a prime, as it can be factored as 5 * 5 (two identical prime factors). 23 isn't composite because it's a prime, as its prime factoring is just itself, 23 (a single prime). Thanks for your interest. Should you have any further doubts, just tell me. (04132019 10:35 PM)Albert Chan Wrote: I recently created nextprime.lua, which is needed for solving Tier 2 puzzle. Nope, this computed sequence for base 7 and all others that follow are incorrect and thus not valid solutions for Tier 2. I think you misunderstood what's actually being asked, which I repeat here with some relevant highlighting for your convenience:
Best regards to all. V. Find All My HPrelated Materials here: Valentin Albillo's HP Collection 

04142019, 04:39 PM
Post: #37




RE: [VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5tier"
Hello Valentin,
Thanks for the explanation. Now everything is clear. Best regards Bernd 

04142019, 07:57 PM
Post: #38




RE: [VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5tier"
Somehow I had completely missed Tier 2 until I saw Bernd's post #35. Then I thought I had a good program until I saw Albert's reply and realized the uniqueness requirement, so back to the drawing board.
This problem turns out to be a good fit for the 50g and the Prime, both of which have NEXTPRIME and ISPRIME? as builtin functions. My program also uses the I>BL command plus a couple of other commands from ListExt. I have tried to keep stackrobatics to a minimum in the interest of readability. %%HP: T(3)A(R)F(.); \<< I\>R \> b n \<< { } 1 1. n START NEXTPRIME DUP b I\>BL LSUM DUP IF ISPRIME? THEN DROP ELSE ROT SWAP DUP2 IF POS THEN DROP SWAP ELSE + OVER + SWAP END END NEXT DROP DUP SIZE 2. / LDIST EVAL \>> \>> Inputs are the base on level 2 and the number of primes to check on level 1. Output are two separate lists, the composites and the primes. I would classify the size (163 bytes) and speed as reasonable if not exactly prizewinning, and it is sort of cheating as it uses so many preexisting commands. I shudder to think of writing such a program on a "classic" era machine. I have checked the first 100000 primes for 7 and 31, which take over 5 minutes each on the emulator, so my results are nowhere near as extensive as Albert's. Still a neat problem, I only wish I had noticed it earlier. 

04202019, 10:47 AM
(This post was last modified: 04202019 10:48 AM by Bernd Grubert.)
Post: #39




RE: [VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5tier"
Hello Valentin,
here is my solution to Tier 2. It is 192 bytes long, due to the lack of prime number checking and the remainder function on the HP15C. I have done the test runs on the HP15C emulator on a PC, since the processing time on my DM15L is far too long... Since the largest integer number the HP15C can exactly represent is 9,999,999,999. , this implementation of the MillerRabin algorithm can check only number up to 99,999. Due to memory limitations, on the real HP15C and the DM 15L the longest sequence is 26 values. For base 31 I got the sequence: 619, 18257, ...,(I stopped at 34139 after ~90 min., because I didn't want to wait any longer) For base 7 I got the sequence: 4801, ...,(I stopped at 23451 after ~60 min.) I have attached an HTMLdocumentation and a txtfile, that can be read into the emulator after changing the extension back to ".15c": Tier_2.htm (Size: 49.27 KB / Downloads: 1) and Tier_2.txt (Size: 6.5 KB / Downloads: 2) . Best regards Bernd 

04212019, 07:06 AM
(This post was last modified: 04212019 10:47 PM by Gilles.)
Post: #40




RE: [VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5tier"
Tier 1 :
Here is my solution without reading others responses. I image that there exists better way. This one is "bestial" ;D Always impressed how fast NewRPL is. Brutal force : 1/ HP50g NewRPL or RPL Code: « 2/ HP50g RPL with ListExt, shorter but slower Code: « 0 1000001111 1E10 FOR n n I>NL LDDUP SIZE 10 == { 1 + } IFT 11111 STEP » 

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