[VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5-tier" - Printable Version +- HP Forums (https://www.hpmuseum.org/forum) +-- Forum: HP Calculators (and very old HP Computers) (/forum-3.html) +--- Forum: General Forum (/forum-4.html) +--- Thread: [VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5-tier" (/thread-12656.html) |
RE: [VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5-tier" - Juan14 - 04-01-2019 12:49 AM You are right Albert and I can't find a way around. RE: [VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5-tier" - Albert Chan - 04-01-2019 03:25 PM 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 -1.9e-16 0.4 0.396375366278 +1.8e-14 0.6 0.587446695546 -1.1e-16 0.8 0.769025184826 -9.1e-14 1.0 0.935745970819 +1.4e-13 Pi/2. 1.210368344457 +2.6e-13 -0.71 -0.688778525307 -1.6e-13 2.019 1.026923318694 +6.4e-13 Edit: changed x^7 coefficient from -0.00158 to -51/32285 to get better accuracy RE: [VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5-tier" - J-F Garnier - 04-01-2019 04:42 PM (04-01-2019 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);Y-SIN(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 J-F [Edited: reversed the order of the polynom term evaluation, for slightly better accuracy] RE: [VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5-tier" - Valentin Albillo - 04-02-2019 10:43 PM Hi, all: Continuing with my original solutions, today it's time for: Tier 4 - The Challenge: Consider the n-point dataset (1, 2), (2, 3), (3, 5), (4, 7), (5, 11), (6, 13), ..., (n, pn) (the prime numbers), and the (n-1)st degree polinomial fit to this dataset of the form: P(x) = a0 + a1 (x-1) + a2 (x-1) (x-2) + ... + an-1 (x-1) (x-2) (x-3) ... (x-(n-1)) Write a program that takes no inputs but computes and outputs the limit of the sum of the coefficients a0, a1, ... , an-1 when n tends to infinity. My original solution: My original solution for the HP-71B is this 4-liner (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(I-1)+1) @ NEXT I @ S=0 3 FOR I=1 TO N-1 @ FOR J=N TO I+1 STEP -1 @ C(J)=C(J)-C(J-1) @ NEXT J @ NEXT I 4 FOR I=1 TO N @ S=S+C(I)/FACT(I-1) @ 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. RE: [VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5-tier" - Albert Chan - 04-03-2019 02:05 AM (04-02-2019 10:43 PM)Valentin Albillo Wrote: I think you meant sum converged using 19 primes (20 primes to confirm 12-digits 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) ... RE: [VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5-tier" - Albert Chan - 04-03-2019 02:57 AM I only have a HP-12C, 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. RE: [VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5-tier" - Valentin Albillo - 04-05-2019 01:58 AM 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 8-10 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 HP-71B is the following user-defined 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*Z-1) @ 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);Y-SIN(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 -1.0E-12 0.6283185307 0.6138343796 0.5877852523 0.5877852523 2.2E-11 0.9424777961 0.8897456012 0.8090169944 0.8090169944 4.1E-11 1.2566370614 1.1122980783 0.9510565164 0.9510565163 1.0E-10 1.5707963268 1.2103683445 1.0000000000 1.0000000000 1.0E-11 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, J-F Garnier, Oulan and Gerson W. Barbosa for your valuable contributions and to Werner for your interest, I hope you enjoyed it all ! V. RE: [VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5-tier" - Albert Chan - 04-05-2019 07:40 PM 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=n-1 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 RE: [VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5-tier" - John Keith - 04-07-2019 04:58 PM 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.E-49 33333333333333333333333333333333333333333333333333.E-49 34583333333333333333333333333333333333333333333333.E-49 33833333333333333333333333333333333333333333333333.E-49 34152777777777777777777777777777777777777777777778.E-49 34047619047619047619047619047619047619047619047619.E-49 34076140873015873015873015873015873015873015873016.E-49 34069609788359788359788359788359788359788359788360.E-49 34070869157848324514991181657848324514991181657848.E-49 34070668490460157126823793490460157126823793490460.E-49 34070693813966383410827855272299716744161188605633.E-49 34070691583365194476305587416698527809638920750032.E-49 34070691634410012386202862393338583814774290964767.E-49 34070691662452161790786129410468034806659145283484.E-49 34070691655244232025812713643401474089304777135465.E-49 34070691656406347881896434650869758246042092914175.E-49 34070691656257873262373750840742575708166882908384.E-49 34070691656273966717789714824786163263074495403684.E-49 34070691656272442599684037804120145048719207427525.E-49 34070691656272570305882488238644909673728392032261.E-49 34070691656272560845399289953795750049707556470279.E-49 34070691656272561452781304231311072994035926413957.E-49 34070691656272561421162961843454416475655340721354.E-49 34070691656272561422168859510781044267796095465102.E-49 34070691656272561422203623227227792237954574766870.E-49 34070691656272561422193499936605021028268165932935.E-49 34070691656272561422194680710735493212204028544306.E-49 34070691656272561422194575066153490058695686270606.E-49 34070691656272561422194583144322191113049301542936.E-49 34070691656272561422194582596611005303547258196512.E-49 34070691656272561422194582630026111767164206858021.E-49 34070691656272561422194582628184366702426127844481.E-49 34070691656272561422194582628275666348866780110705.E-49 34070691656272561422194582628271661692858309901367.E-49 34070691656272561422194582628271810600936280034971.E-49 34070691656272561422194582628271806504336081216950.E-49 34070691656272561422194582628271806529151697629234.E-49 34070691656272561422194582628271806536170465629760.E-49 34070691656272561422194582628271806535499300745912.E-49 34070691656272561422194582628271806535542548688840.E-49 34070691656272561422194582628271806535540241528738.E-49 34070691656272561422194582628271806535540348557090.E-49 34070691656272561422194582628271806535540344239572.E-49 34070691656272561422194582628271806535540344383289.E-49 34070691656272561422194582628271806535540344380187.E-49 34070691656272561422194582628271806535540344380140.E-49 34070691656272561422194582628271806535540344380151.E-49 34070691656272561422194582628271806535540344380150.E-49 34070691656272561422194582628271806535540344380149.E-49 34070691656272561422194582628271806535540344380150.E-49 34070691656272561422194582628271806535540344380150.E-49 34070691656272561422194582628271806535540344380150.E-49 34070691656272561422194582628271806535540344380149.E-49 34070691656272561422194582628271806535540344380150.E-49 34070691656272561422194582628271806535540344380150.E-49 34070691656272561422194582628271806535540344380151.E-49 34070691656272561422194582628271806535540344380150.E-49 34070691656272561422194582628271806535540344380151.E-49 It can be observed that: -- LongFloat numbers are not very user-friendly. -- There is noise in the last digit, so really 49-digit accuracy in this case. -- Rate of converge increases, only about 56 iterations required to confirm 49 digits. RE: [VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5-tier" - Albert Chan - 04-08-2019 03:36 PM 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://lua-users.org/lists/lua-l/2019-04/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 RE: [VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5-tier" - Gerson W. Barbosa - 04-09-2019 06:53 PM (04-07-2019 04:58 PM)John Keith Wrote: It can be observed that: They needn't be so. 34070691656272561422194582628271806535540344380151.E-49 \<< 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. RE: [VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5-tier" - Valentin Albillo - 04-09-2019 11:37 PM . Hi again, all (04-05-2019 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 HP-12C, that is). (04-07-2019 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 high-precision results, much appreciated. (04-08-2019 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 high-precision results fot the tin(x) function I mentioned in the challenge. Again, really appreciated. (04-09-2019 06:53 PM)Gerson W. Barbosa Wrote:(04-07-2019 04:58 PM)John Keith Wrote: It can be observed that: [...] LongFloat numbers are not very user-friendly. 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. . RE: [VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5-tier" - John Keith - 04-10-2019 04:11 PM (04-09-2019 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. RE: [VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5-tier" - Bernd Grubert - 04-13-2019 08:01 PM 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 RE: [VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5-tier" - Albert Chan - 04-13-2019 10:35 PM 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 RE: [VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5-tier" - Valentin Albillo - 04-13-2019 11:27 PM Hi, Bernd Grubert and Albert Chan: (04-13-2019 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. (04-13-2019 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. RE: [VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5-tier" - Bernd Grubert - 04-14-2019 04:39 PM Hello Valentin, Thanks for the explanation. Now everything is clear. Best regards Bernd RE: [VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5-tier" - John Keith - 04-14-2019 07:57 PM 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 built-in 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 prize-winning, and it is sort of cheating as it uses so many pre-existing 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. RE: [VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5-tier" - Bernd Grubert - 04-20-2019 10:47 AM 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 HP-15C. I have done the test runs on the HP-15C emulator on a PC, since the processing time on my DM-15L is far too long... Since the largest integer number the HP-15C can exactly represent is 9,999,999,999. , this implementation of the Miller-Rabin algorithm can check only number up to 99,999. Due to memory limitations, on the real HP-15C 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 HTML-documentation and a txt-file, that can be read into the emulator after changing the extension back to ".15c": [attachment=7172] and [attachment=7171]. Best regards Bernd RE: [VA] Short & Sweet Math Challenge #24: "2019 Spring Special 5-tier" - Gilles - 04-21-2019 07:06 AM 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 » |