A046511 -id:A046511 - cp's OEIS frontend

A003001 Smallest number of multiplicative persistence n.

0, 10, 25, 39, 77, 679, 6788, 68889, 2677889, 26888999, 3778888999, 277777788888899

Offset: 0

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.
From David A. Corneth, Sep 23 2016: (Start)
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)
From Benjamin Chaffin, Sep 29 2016: (Start)
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
From A.H.M. Smeets, Nov 16 2018: (Start)
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
====================================================
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)
From Tim Peters, Sep 19 2023: (Start)
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)

			77 -> 49 -> 36 -> 18 -> 8 has persistence 4.

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.
R. K. Guy, Unsolved Problems in Number Theory, Springer, 1st edition, 1981. See section F25.
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).
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 35.
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 78.

Edson de Faria and Charles Tresser, On Sloane's persistence problem, arXiv preprint arXiv:1307.1188 [math.DS], 2013.
Edson de Faria and Charles Tresser, On Sloane's persistence problem, Experimental Math., 23 (No. 4, 2014), 363-382.
Mark R. Diamond, Multiplicative persistence base 10: some new null results, 2011.
Shyam Sunder Gupta, Digital Root Wonders, Exploring the Beauty of Fascinating Numbers, Springer (2025) Ch. 1, 1-28.
Brady Haran and Matt Parker, What's special about 277777788888899?, Numberphile video, 2019.
T. Lamont-Smith, Multiplicative Persistence and Absolute Multiplicative Persistence, J. Int. Seq., Vol. 24 (2021), Article 21.6.7.
Kevin McElwee, An algorithm for multiplicative persistence research, Jul 13 2019.
S. Perez and R. Styer, Persistence: A Digit Problem.
W. Schneider, The Persistence of a Number.
N. J. A. Sloane, The persistence of a number, J. Recreational Math., 6 (1973), 97-98.
Eric Weisstein's World of Mathematics, Multiplicative Persistence.
Wikipedia, Persistence of a number.
Susan Worst, Multiplicative persistence of base four numbers. [Scanned copy of manuscript and correspondence, May 1980]

Cf. A031346 (persistence), A133500 (powertrain), A133048 (powerback), A006050, A007954, A031286, A031347, A033908, A046511, A121105-A121111.

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 *)

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

A046510 Numbers with multiplicative persistence value 1.

Original entry on oeis.org

10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 30, 31, 32, 33, 40, 41, 42, 50, 51, 60, 61, 70, 71, 80, 81, 90, 91, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 130, 131, 132, 133

Offset: 1

Patrick De Geest, Sep 15 1998

Numbers 0 to 9 have a multiplication persistence of 0, not 1. - Daniel Mondot, Mar 12 2022

			24 -> 2 * 4 = [ 8 ] -> one digit in one step.

Daniel Mondot, Table of n, a(n) for n = 1..10000
N. J. A. Sloane, The persistence of a number, J. Recreational Math., 6 (1973), 97-98.
Eric Weisstein's World of Mathematics, Multiplicative Persistence

Cf. A003001, A014120, A046501.

Numbers with multiplicative persistence m: this sequence (m=1), A046511 (m=2), A046512 (m=3), A046513 (m=4), A046514 (m=5), A046515 (m=6), A046516 (m=7), A046517 (m=8), A046518 (m=9), A352531 (m=10), A352532 (m=11).

Select[Range[10, 121], IntegerLength[Times @@ IntegerDigits[#]] <= 1 &] (* Jayanta Basu, Jun 26 2013 *)

isok(n) = my(d=digits(n)); (#d > 1) && (#digits(prod(k=1, #d, d[k])) <= 1); \\ Michel Marcus, Apr 12 2018 and Mar 13 2022

from math import prod
def ok(n): return n > 9 and prod(map(int, str(n))) < 10
print([k for k in range(134) if ok(k)]) # Michael S. Branicky, Mar 13 2022

Incorrect terms 0 to 9 removed by Daniel Mondot, Mar 12 2022

A350181 Numbers of multiplicative persistence 2 which are themselves the product of digits of a number.

Original entry on oeis.org

25, 27, 28, 35, 36, 45, 48, 54, 56, 63, 64, 72, 84, 125, 126, 128, 135, 144, 162, 192, 216, 224, 225, 243, 245, 252, 256, 315, 324, 375, 432, 441, 512, 525, 567, 576, 588, 625, 675, 735, 756, 875, 945, 1125, 1134, 1152, 1176, 1215, 1225, 1296, 1323, 1372

Offset: 1

Daniel Mondot, Dec 18 2021

The multiplicative persistence of a number mp(n) is the number of times the product of digits function p(n) must be applied to reach a single digit, i.e., A031346(n).
The product of digits function partitions all numbers into equivalence classes. There is a one-to-one correspondence between values in this sequence and equivalence classes of numbers with multiplicative persistence 3.
There are infinitely many numbers with mp of 1 to 11, but the classes of numbers (p(n)) are postulated to be finite for this and subsequent sequences A350182....
Equivalently:
This sequence consists of the numbers A007954(k) such that A031346(k) = 3,
These are the numbers k in A002473 such that A031346(k) = 2,
Or:
- they factor into powers of 2, 3, 5 and 7 exclusively.
- p(n) goes to a single digit in 2 steps.
Postulated to be finite and complete.
The largest known number is 2^25 * 3^227 * 7^28 (140 digits).
No more numbers have been found between 10^140 and probably 10^20000 (according to comment in A003001), and independently verified up to 10^10000.

			25 is in this sequence because:
- 25 goes to a single digit in 2 steps: p(25) = 10, p(10) = 0.
- 25 has ancestors 55, 155, etc. p(55) = 25.
27 is in this sequence because:
- 27 goes to a single digit in 2 steps: p(27) = 14, p(14) = 4.
- 27 has ancestors 39, 93, 333, 139, etc. p(39) = 27.

Daniel Mondot, Table of n, a(n) for n = 1..11994

Cf. A002473, A003001 (smallest number with multiplicative persistence n), A031346 (multiplicative persistence), A031347 (multiplicative digital root), A046511 (all numbers with mp of 2).

Cf. A350180, A350182, A350183, A350184, A350185, A350186, A350187 (numbers with mp 1, and 3 to 10 that are themselves 7-smooth numbers).

Select[Range@1400,AllTrue[First/@FactorInteger@#,#<10&]&&Length@Most@NestWhileList[Times@@IntegerDigits@#&,#,#>9&]==2&] (* Giorgos Kalogeropoulos, Jan 16 2022 *)

from math import prod
from sympy import factorint
def pd(n): return prod(map(int, str(n)))
def ok(n):
    if n <= 9 or max(factorint(n)) > 9: return False
    return (p := pd(n)) > 9 and pd(p) < 10
print([k for k in range(1400) if ok(k)]) # Michael S. Branicky, Jan 16 2022

cp's OEIS Frontend

A003001 Smallest number of multiplicative persistence n.

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A046510 Numbers with multiplicative persistence value 1.

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A350181 Numbers of multiplicative persistence 2 which are themselves the product of digits of a number.

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A046502 Primes with multiplicative persistence value 2.

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A199992 Composite numbers whose multiplicative persistence is 2.

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