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A003001
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Smallest number of multiplicative persistence n.
(Formerly M4687)
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72
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0, 10, 25, 39, 77, 679, 6788, 68889, 2677889, 26888999, 3778888999, 277777788888899
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OFFSET
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0,2
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COMMENTS
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Probably finite.
The persistence of a number (A031346) is the number of times you need to multiply the digits together before reaching a single digit.
For n > 1, the digit 0 doesn't occur. Therefore the digit 1 doesn't occur and all terms have digits in nondecreasing order.
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.
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]
No more up to 10^200. (End)
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:
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.
The largest p(n) with P(p(n))=9 is 2^33 * 3^3 (12 digits).
The largest p(n) with P(p(n))=8 is 2^9 * 3^5 * 7^8 (12 digits).
The largest p(n) with P(p(n))=7 is 2^24 * 3^18 (16 digits).
The largest p(n) with P(p(n))=6 is 2^24 * 3^6 * 7^6 (16 digits).
The largest p(n) with P(p(n))=5 is 2^35 * 3^2 * 7^6 (17 digits).
The largest p(n) with P(p(n))=4 is 2^59 * 3^5 * 7^2 (22 digits).
The largest p(n) with P(p(n))=3 is 2^4 * 3^17 * 7^38 (42 digits).
The largest p(n) with P(p(n))=2 is 2^25 * 3^227 * 7^28 (140 digits).
All p(n) between 10^140 and 10^20000 have a persistence of 1, meaning they contain a 0 digit. (End)
Benjamin Chaffin's comments imply that there are no more terms up to 10^20585. For every number N between 10^200 with 10^20585 with persistence greater than 1, the product of the digits of N is between 10^140 and 10^20000, and each of these products has a persistence of 1. - David Radcliffe, Mar 22 2019
Let p_10(n) be the product of the digits of n in base 10. We can define an equivalence relation DP_10 on n by n DP_10 m if and only if p_10(n) = p_10(m); the name DP_b for the equivalence relation stands for "digits product for representation in base b". A number n is called the class representative number of class n/DP_10 if and only if p_10(n) = p_10(m), m >= n; i.e., if it is the smallest number of that class; it is also called the reduced number.
For any multiplicative persistence, except multiplicative persistence 2, the set of class representative numbers with that multiplicative persistence is conjectured to be finite.
Each class representative number represents an infinite set of numbers with the same multiplicative persistence.
For multiplicative persistence 2, only the set of class representative numbers which end in the digit zero is infinite. The table of numbers of class representative numbers of different multiplicative persistence (mp) is given by:
final digit
mp total 0 1 2 3 4 5 6 7 8 9
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0 10 1 1 1 1 1 1 1 1 1 1
1 10 1 1 1 1 1 1 1 1 1 1
2 inf inf 0 4 0 1 1 5 0 7 0
3 12199 12161 0 8 0 3 3 8 0 16 0
4 408 342 0 14 0 5 4 19 0 24 0
5 151 88 0 9 0 1 3 37 0 13 0
6 41 24 0 1 0 0 0 14 0 2 0
7 13 9 0 0 0 0 0 4 0 0 0
8 8 7 0 0 0 0 0 1 0 0 0
9 5 5 0 0 0 0 0 0 0 0 0
10 2 2 0 0 0 0 0 0 0 0 0
11 2 2 0 0 0 0 0 0 0 0 0
It is observed from this that for the reduced numbers with multiplicative persistence 1, the primes 11, 13, 17 and 19, will not occur in any trajectory of another (larger) number; i.e., all numbers represented by the reduced numbers 11, 13, 17 and 19 have a prime factor of at least 11 (conjectured from the observations).
Example for numbers represented by the reduced number 19: 91 = 7*13, 133 = 7*19, 313 is prime, 331 is prime, 119 = 7*17, 191 is prime, 911 is prime, 1133 = 11*103, 1313 = 13*101, 1331 = 11^3, 3113 = 11*283, 3131 = 31*101 and 3311 = 7*11*43.
In fact all trajectories can be projected to a trajectory in one of the ten trees with reduced numbers with roots 0..9, and the numbers represented by the reduced number of each leaf have a prime factor of at least 11 (as conjectured from the observations).
Example of the trajectory of 277777788888899 (see A121111) in the tree of reduced numbers (the unreduced numbers are given between brackets): 277777788888899 -> 3778888999 (4996238671872) -> 26888999 (438939648) -> 2677889 (4478976) -> 68889 (338688) -> 6788 (27648) -> 2688 (2688) -> 678 (768) -> 69 (336) -> 45 (54) -> 10 (20) -> 0. (End)
New lower bound: if a(12) exists, it must be > 2.67*10^30000. It continues to be the case that the digit products for all candidates with at least 20000 digits (roughly where the last long run reported here stopped) contain a zero digit, so the candidates all have persistence 2. More, the digit products all contain at least one zero in their last 306 digits. An extreme is the digit product 2^13802 * 3^16807 * 7^1757. That has 13659 decimal digits, 1335 of which are zeros. It ends with a zero followed by 305 nonzero digits. So to confirm that the large candidates with no more than 30000 digits have persistence 2, it would suffice to compute digit products modulo 10^306.
Note: by "candidate" I mean a digit string matching one of these eight (pairwise disjoint) simple regular expressions. Each such string gives the smallest integer with its digit product (and viewing the empty string as having digit product 1), and their union covers all digit products that don't end with a zero.
7* 8* 9*
2 7* 8* 9*
3 7* 8* 9*
4 7* 8* 9*
5 5* 7* 9*
6 7* 8* 9*
26 7* 8* 9*
35 5* 7* 9*
There are (8*N^2 + 13*N + 6)*(N + 1)/6 such strings with no more than N digits. A long computer run checked N=30000, a bit over 36*10^12 candidates. The smallest candidate with more than 30000 digits is > 2.67*10^30000, which is the smallest remaining possibility for a(12). (End)
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REFERENCES
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Alex Bellos, Here's Looking at Euclid: A Surprising Excursion Through the Astonishing World of Math, Free Press, 2010, page 176.
M. Gardner, Fractal Music, Hypercards and More, Freeman, NY, 1991, pp. 170, 186.
C. A. Pickover, Wonders of Numbers, "Persistence", Chapter 28, Oxford University Press NY 2001.
Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 66.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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EXAMPLE
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77 -> 49 -> 36 -> 18 -> 8 has persistence 4.
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MATHEMATICA
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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 *)
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PROG
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(PARI) vecprod(w)=prod(i=1, #w, w[i]);
persistence(x)={my(y=digits(x), c=0); while(#y>1, y=digits(vecprod(y)); c++); return(c)}
firstTermsA003001(U)={my(ans=vector(U), k=(U>1), z); while(k+1<=U, if(persistence(z)==k, ans[k++]=z); z++); return(ans)}
\\ Finds the first U terms (is slow); R. J. Cano, Sep 11 2016
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CROSSREFS
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KEYWORD
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nonn,nice,base,more,hard
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AUTHOR
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STATUS
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approved
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