Smoothly Undulating Palindromic Primes (or SUPP's for short) are numbers that are primes, palindromic in base 10, and the digits alternate, but why smooth one might ask ! The smoothness was added to make a difference with the normal undulating numbers. The description for normal undulating numbers is that the next digits alternately go up and down (or down and up) but the absolute difference values between two adjacent digits may differ. (e.g. 906343609) In a smoothly undulating number the absolute difference values between two adjacent digits are always equal, therefore only two distinct digits can appear in the number. (e.g. 74747474747474747)
[ October 21, 2004 ] Some nontrivial combinations can never produce primes... By Julien Peter Benney (email)
Thus in both last cases only for w of the form 3n+2 is there any chance of a prime !
[ February 9, 2001 ] Jeff Heleen wrote :
" As far as I could see you didn't have a section on your site for these numbers. While I'm sure someone somewhere must have done this before, I have done it also. Within the limitations of the program I believe these are ALL the smoothly undulating palindromic prime numbers with two distinct digits each, smaller than 843 digits long. I used a modified APRT-CLE program in UBASIC to automate and perform the search on a Pentium_II 300 MHz laptop."
That is indeed a very nice and interesting compilation, thanks Jeff. Great job! At the same time it is a topic that might attract other dedicated number crunchers. Perhaps you know a source where larger SUPP's are displayed. Those are welcome as well! Send them in and I'll add them to the table.
[ February 12, 2001 ] Jeff Heleen wrote :
" I have found the following website: http://www.utm.edu/research/primes/lists/top_ten/topten.pdf where, if you will look on page 43 (of 93) you will see the top ten SUPP's as of February 24, 2001. The smallest two on this list are the same as my highest two. It doesn't say whether these are ALL there are up to the highest one shown. However, I suspect not, as they all start and end with the digit 1. So perhaps there are more to discover in this range."
[ February 14, 2001 ] Message from Carlos Rivera
There are several extra terms : (37)k3, is prime for k=424 & 946 (75)k7, is prime for k=539 & 707 (79)k7, is prime for k=838 (92)k9, is prime for k=428 (95)k9, is prime for k=647 (please verify them) In the meanwhile I used PRIMEFORM to get the next pseudoprime following my record from 1997: (12)k1, is pseudoprime for k=3904 (7809 digits) far beyond the current possibilities of rigorous primality testing of the speediest code (TITANIX)
[ April 2001 ] Start of above date I noticed a new entry in G. L. Honaker, Jr.'s Prime Curios! website of Landon Curt Noll. A beautiful SUPP, proved prime with special hardware a few years ago (?), was introduced there, which immediately shattered Carlos Rivera's previous record ! This SUPP has a length of 2883 digits 3(73)1441 See Prime Curios! 37373...37373 You can contact L. C. Noll through his home page at http://www.isthe.com/chongo/
[ May 6, 2001 ] Enters Hans Rosenthal with new and more impressive data !
Here is a probable prime of length 10419 for your SUPP page: 3(13)5209 = 310*(1010418–1)/99+3 I don't know whether this one has been discovered by someone else before (if you know of this, please send me a note).
[ June 26, 2001 ] Carlos Rivera writes the following interesting observations.
1) Any smooth undulating palindrome number composed of two distinct digits can be expressed in any one of the two forms: a(ba)n = (ab)na 2) (ab)na = (ab)nx10+a 3) (ab)n = (ab)xR(2n)/R(2) 4) R(k) = (10k –1)/(10–1) 5) Consequently a(ba)n = (ab)na = (ab)x((102n–1)/99)x10+a 6) But: (ab)x((102n–1)/99)x10+a = [(ab)x102n+1 – (ab)x10 + 99a]/99 = [(ab)x102n+1 – (ab)x10 + 100a – a]/99 = [(ab)x102n+1 – (ba)]/99 7) a(ba)n = (ab)na = (ab)x((102n–1)/99)x10+a = [(ab)x102n+1 – (ba)]/99 8) The form a(ba)n = [(ab)x102n+1 – (ba)]/99 is the one used by you in yourpage and formally is correct. But the second form (ab)na = (ab)x((102n–1)/99)x10+a is a kind of more suitableone form for primality test purposes, especially if: ° a = 1 & ° [(102n –1)/99] can be factorized until certain extent in order to use classical theorems like the Pocklington one.
2) (ab)na = (ab)nx10+a 3) (ab)n = (ab)xR(2n)/R(2) 4) R(k) = (10k –1)/(10–1)
5) Consequently a(ba)n = (ab)na = (ab)x((102n–1)/99)x10+a
6) But:
7) a(ba)n = (ab)na = (ab)x((102n–1)/99)x10+a = [(ab)x102n+1 – (ba)]/99
8) The form a(ba)n = [(ab)x102n+1 – (ba)]/99 is the one used by you in yourpage and formally is correct.
But the second form (ab)na = (ab)x((102n–1)/99)x10+a is a kind of more suitableone form for primality test purposes, especially if: ° a = 1 & ° [(102n –1)/99] can be factorized until certain extent in order to use classical theorems like the Pocklington one.
Thanks Carlos for the interesting observations on the formula formats for the SUPP's. Before Hans Rosenthal entered the stage I used the format you promote in entry 7(highlighted in yellow). But Hans convinced me to use to other one for the following reasons. First the format [(ab)*102n+1–(ba)]/99 displays the exact digitlength of the SUPP namely via the exponent (2n+1). Secondly the (ab) and (ba) coefficients indicate straight away how the SUPP starts and ends !
[ September 4, 2001 ] Hans Rosenthal broke Landon Curt Noll's old record by prime proving the following SUPP of 3015 digits !
3(23)1507 = (32*103015–23)/99
[ October 19, 2001 ] Hans Rosenthal sent in a list of five new records. The largest one he prime proved is the following SUPP of 4859 digits !
" All Primo certificates have been validated with Cert_Val. The proof of the above largest known SUPP (second largest known ECPP prime) took exactly 11 weeks on an Athlon 1.4 GHz, the full validation of this certificate took 20 and a half hours on the same PC. I believe that from now on it's a real challenge (also for myself) to complete/enlarge the SUPP table."
[ October 27, 2002 ] Hans Rosenthal sent a new SUPP record of 4885 digits ! ( Announced at Walter Schneider's site at Undulants )
The proof was done using Marcel Martin's Primo and took 2008 hours and 57 minutes on a AMD Athlon 1.33 GHz. The Primo certificate was then validated with Cert_Val which took on the same PC an additional 25 hours and 11 minutes. See also the Top 20 ECPP records at http://www.ellipsa.eu/public/primo/top20.html
[ December 10, 2002 ] Hans Rosenthal informs :
[ December 22, 2002 ] David Broadhurst announced via a message (http://groups.yahoo.com/group/primeform/message/2937) in the User group for the PrimeForm program that the following SUPP is prime !
[ December 23, 2002 ] Reaction from Hans Rosenthal ¬
" Yes, David informed me, nice result, such a proof won't happen every day. Jim Fougeron double-checked the primality of 1(71)_2442 = (17*10^4885–71)/99 by use of BLS (he only took about 24 hours for that). Both, David and Jim were pretty lucky with finding enough factors in N–1 for their proofs. However, this can only work for the SUPP's that start/end in 1 -- it will never work for the others. I am really glad that I am no longer the only one to contribute new results to the SUPP page. You should update it and also announce the new record on your main page."
[ July 13, 2003 ] Hans Rosenthal announced via a message in Number Theory List (NMBRTHRY@LISTSERV.NODAK.EDU) that the following SUPP is proven prime !
3(23)3479 = (32*106959–23)/99
" I would like to inform you that I have certified the primality of (32*10^6959–23)/99, a smoothly undulating palindromic prime (SUPP) [1] having 6959 decimal digits, with the program Primo [2], Marcel Martin's implementation of the elliptic curve primality proving (ECPP) algorithm. The Primo certificate of primality is available at http://www.ellipsa.eu/public/primo/files/ecpp6959.zip (4457 KB) The certification of this ordinary prime was started on 21 January 2002 with Primo 1.1.0 (tests 1 to 47) and completed on 7 July 2003 with Primo 2.0.0 (tests 48 to 953) on an AMD Athlon 1.4 GHz. There was one relevant interruption of the certification process from 29 March 2003, 6:47, until 3 April 2003, 22:45. So the total running time amounts to approximately 527 days. I thank Marcel Martin for his help and advice, and most of all, for making the ECPP algorithm available to the world of PC users in the most comfortable form I can imagine: his marvellous Primo. Hans Rosenthal " [1] http://www.worldofnumbers.com/undulat.htm# [2] http://www.ellipsa.eu/public/primo/top20.html
The Primo certificate of primality is available at http://www.ellipsa.eu/public/primo/files/ecpp6959.zip (4457 KB)
The certification of this ordinary prime was started on 21 January 2002 with Primo 1.1.0 (tests 1 to 47) and completed on 7 July 2003 with Primo 2.0.0 (tests 48 to 953) on an AMD Athlon 1.4 GHz. There was one relevant interruption of the certification process from 29 March 2003, 6:47, until 3 April 2003, 22:45. So the total running time amounts to approximately 527 days.
I thank Marcel Martin for his help and advice, and most of all, for making the ECPP algorithm available to the world of PC users in the most comfortable form I can imagine: his marvellous Primo.
Hans Rosenthal "
[1] http://www.worldofnumbers.com/undulat.htm# [2] http://www.ellipsa.eu/public/primo/top20.html
CR = Carlos RiveraDB = David BroadhurstHR = Hans RosenthalJH = Jeffrey HeleenLN = Landon Curt NollPDG = Patrick De GeestRC = Ray Chandler
PDG = Patrick De Geest
All of Hans Rosenthal's probable primes above 10000 digits are also submitted to the PRP TOP records table maintained by Henri & Renaud Lifchitz. See : http://www.primenumbers.net/prptop/prptop.php
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