%I M4687
%S 0,10,25,39,77,679,6788,68889,2677889,26888999,3778888999,
%T 277777788888899
%N Smallest number of multiplicative persistence n.
%C Probably finite.
%C The persistence of a number (A031346) is the number of times you need to multiply the digits together before reaching a single digit.
%C From _David A. Corneth_, Sep 23 2016: (Start)
%C For n > 1, the digit 0 doesn't occur. Therefore the digit 1 doesn't occur and all terms have digits in nondecreasing order.
%C a(n) consists of at most one three and at most one two but not both. If they contain both, they could be replaced with a single digit 6 giving a lesser number. Two threes can be replaced with a 9. Similarily, there's at most one four and one six but not both. Two sixes can be replaced with 49. A four and a six can be replaced with a three and an eight. For n > 2, an even number and a five don't occur together.
%C Summarizing, a term a(n) for n > 2 consists of 7's, 8's and 9's with a prefix of one of the following sets of digits: {{}, {2}, {3}, {4}, {6}, {2,6}, {3,5}, {5, 5,...}} [Amended by _Kohei Sakai_, May 27 2017]
%C No more up to 10^200. (End)
%C From _Benjamin Chaffin_, Sep 29 2016: (Start)
%C Let p(n) be the product of the digits of n, and P(n) be the multiplicative persistence of n. Any p(n) > 1 must have only prime factors from one of the two sets {2,3,7} or {3,5,7}. The following are true of all p(n) < 10^20000:
%C The largest p(n) with P(p(n))=10 is 2^4 * 3^20 * 7^5. The only other such p(n) known is p(a(11))=2^19 * 3^4 * 7^6.
%C The largest p(n) with P(p(n))=9 is 2^33 * 3^3 (12 digits).
%C The largest p(n) with P(p(n))=8 is 2^9 * 3^5 * 7^8 (12 digits).
%C The largest p(n) with P(p(n))=7 is 2^24 * 3^18 (16 digits).
%C The largest p(n) with P(p(n))=6 is 2^24 * 3^6 * 7^6 (16 digits).
%C The largest p(n) with P(p(n))=5 is 2^35 * 3^2 * 7^6 (17 digits).
%C The largest p(n) with P(p(n))=4 is 2^59 * 3^5 * 7^2 (22 digits).
%C The largest p(n) with P(p(n))=3 is 2^4 * 3^17 * 7^38 (42 digits).
%C The largest p(n) with P(p(n))=2 is 2^25 * 3^227 * 7^28 (140 digits).
%C All p(n) between 10^140 and 10^20000 have a persistence of 1, meaning they contain a 0 digit. (End)
%D Alex Bellos, Here's Looking at Euclid: A Surprising Excursion Through the Astonishing World of Math, Free Press, 2010, page 176.
%D M. Gardner, Fractal Music, Hypercards and More, Freeman, NY, 1991, pp. 170, 186.
%D C. A. Pickover, Wonders of Numbers, "Persistence", Chapter 28, Oxford University Press NY 2001.
%D Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 66.
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H de Faria, Edson, and Charles Tresser, <a href="http://arxiv.org/abs/1307.1188">On Sloane's persistence problem</a>, arXiv preprint arXiv:1307.1188 [math.DS], 2013.
%H de Faria, Edson, and Charles Tresser, <a href="http://dx.doi.org/10.1080/10586458.2014.910849">On Sloane's persistence problem</a>, Experimental Math., 23 (No. 4, 2014), 363382.
%H M. R. Diamond, <a href="http://web.archive.org/web/20160313225019/http://markdiamond.com.au/download/joous311.pdf">Multiplicative persistence base 10: some new null results</a>, 2011.
%H S. Perez, R. Styer, <a href="http://www41.homepage.villanova.edu/robert.styer/MultiplicativePersistence/PersistenceStephPerezJournalArtAug2013.pdf">Persistence: A Digit Problem</a>
%H W. Schneider, <a href="http://web.archive.org/web/2004/www.wschnei.de/digitrelatednumbers/persistence.html">The Persistence of a Number</a>
%H N. J. A. Sloane, <a href="http://neilsloane.com/doc/persistence.html">The persistence of a number</a>, J. Recreational Math., 6 (1973), 9798.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/MultiplicativePersistence.html">Multiplicative Persistence</a>
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Persistence_of_a_number">Persistence of a number</a>
%H Susan Worst, <a href="/A003001/a003001.pdf">Multiplicative persistence of base four numbers</a> [Scanned copy of manuscript and correspondence, May 1980]
%e 77 > 49 > 36 > 18 > 8 has persistence 4.
%t lst = {}; n = 0; Do[While[True, k = n; c = 0; While[k > 9, k = Times @@ IntegerDigits[k]; c++]; If[c == l, Break[]]; n++]; AppendTo[lst, n], {l, 0, 7}]; lst (* _Arkadiusz Wesolowski_, May 01 2012 *)
%o (PARI) vecprod(w)=prod(i=1,#w,w[i]);
%o persistence(x)={my(y=digits(x),c=0);while(#y>1,y=digits(vecprod(y));c++);return(c)}
%o firstTermsA003001(U)={my(ans=vector(U),k=(U>1),z);while(k+1<=U,if(persistence(z)==k,ans[k++]=z);z++);return(ans)}
%o \\ Finds the first U terms (is slow); _R. J. Cano_, Sep 11 2016
%Y Cf. A031346 (persistence), A133500 (powertrain), A133048 (powerback), A006050, A007954, A031286, A031347, A033908, A046511, A121105A121111.
%K nonn,nice,base,hard
%O 0,2
%A _N. J. A. Sloane_
