A133500 -id:A133500 - 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

A133501 Number of steps for "powertrain" operation to converge when started at n.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 5, 2, 3, 3, 1, 1, 1, 3, 2, 5, 5, 5, 4, 9, 1, 1, 2, 5, 3, 3, 4, 6, 3, 5, 1, 1, 3, 2, 3, 5, 3, 3, 2, 4, 1, 1, 6, 3, 4, 4, 3, 3, 8, 2, 1, 1, 6, 6, 2, 2, 3, 5, 3, 2, 1, 1, 5, 3, 4, 4, 5, 4, 3, 7, 1, 1, 2, 5, 4, 2, 3, 3, 2, 4, 1, 1, 1, 1, 1

Offset: 0

J. H. Conway and N. J. A. Sloane, Dec 03 2007

See A133500 for definition.

It is conjectured that every number converges to a fixed-point.

			39 -> 19683 -> 1594323 -> 38443359375 -> 59440669655040 -> 0, so a(39) = 5.

N. J. A. Sloane, Table of n, a(n) for n = 0..10000
N. J. A. Sloane, Full trajectories of numbers from 1 to 10000

For the powertrain map itself, see A133500.

See A133508, A133503 for records. See A135381 for high-water marks.

powertrain:=proc(n) local a,i,n1,n2,t1,t2; n1:=abs(n); n2:=sign(n); t1:=convert(n1, base, 10); t2:=nops(t1); a:=1; for i from 0 to floor(t2/2)-1 do a := a*t1[t2-2*i]^t1[t2-2*i-1]; od: if t2 mod 2 = 1 then a:=a*t1[1]; fi; RETURN(n2*a); end;
# Compute trajectory of n under repeated application of the powertrain map of A133500. This will return -1 if the trajectory does not converge to a single number in 100 steps (so it could fail if the trajectory enters a nontrivial loop or takes longer than 100 steps to converge).
PTtrajectory := proc(n) local p,M,t1,t2,i; M:=100; p:=[n]; t1:=n; for i from 1 to M do t2:=powertrain(t1); if t2 = t1 then RETURN(n,i-1,p); fi; t1:=t2; p:=[op(p),t2]; od; RETURN(n,-1,p); end;

A175398 Sequence of resulting numbers after iterations of {((((D_1^D_2)^D_3)^D_4)^... )^D_k, where D_k is the k-th digit D of the number r and k is the digit number of the number r in the decimal expansion of r (A055642)} needed to reach a one-digit number starting at r = n.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 4, 8, 1, 9, 1, 1, 1, 9, 1, 3, 9, 1, 8, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 0

Jaroslav Krizek, May 01 2010

a(n) = 1 - 9 for infinitely many n.
E.g., a(n) = b (b = 1, 2, ..., 9) for numbers n = b*10^k + A002275(k), where k >= 1.
a(n) = 1 for numbers n such that A055642(A133500(n)) = 1 for n >= 1, e.g., if the number n starts with a digit 1 or contains a digit 0 or for n >= 1.
Sequences after k steps of defined iteration (k >= 0):
0th step: A001477: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, ...
1st step: A133500: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 1, 4, 16, 64, 256, 1024, 4096, 16384, 65536, 262144, 1, ...
2nd step: A175399: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 4, 8, 1, 9, 1296, 1, 1073741824, 25, 1, 3, 9, 128, 8, 4096, 1628413597910449, 72057594037927936, 221073919720733357899776, 1, 1, 4, 1, 1296, 1073741824, 1, 1, 1, ...
3rd step: A175400: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 4, 8, 1, 9, 1, 1, 1, 32, 1, 3, 9, 1, 8, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
4th step: A175401: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 4, 8, 1, 9, 1, 1, 1, 9, 1, 3, 9, 1, 8, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
See A175402 and A175403.

			For n = 29: a(29) = 9 because for the number 29 there are 4 steps of defined iteration: {2^9 = 512}, {(5^1)^2 = 25}, {2^5 = 32}, {3^2 = 9}. Resulting number is 9.

A221221 Where powerbacks and powertrains coincide.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 22, 24, 33, 42, 44, 55, 66, 77, 88, 99, 101, 111, 112, 113, 114, 115, 116, 117, 118, 119, 121, 131, 141, 151, 161, 171, 181, 191, 202, 211, 212, 213, 214, 215, 216, 217, 218, 219, 222, 232, 242, 252, 262, 272, 282, 292

Offset: 1

Reinhard Zumkeller, May 27 2013

Numbers m such that A133048(m) = A133500(m);
A133500(a(n)) = A133048(a(n)) = A222493(n);
if m is a term then also its reversal in decimal representation, palindromes are a subsequence, cf. A004086, A002113.

			Some non-palindromic terms:
a(11) = 10: A133500(10) = 1^0 = 1 = A133048(10) = A133048(1) = 1;
a(14) = 24: A133500(24) = 2^4 = 16 = A133048(24) = 4^2;
a(16) = 42: A133500(42) = 4^2 = 16 = A133048(42) = 2^4;
a(25) = 112: A133500(112) = 1^1 * 2 = 2 = A133048(112) = 2^1 * 1;
a(26) = 113: A133500(113) = 1^1 * 3 = 3 = A133048(113) = 3^1 * 1;
a(44) = 213: A133500(213) = 2^1 * 3 = 6 = A133048(213) = 3^1 * 2.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

a221221 n = a221221_list !! (n-1)
a221221_list = filter (\x -> a133500 x == a133048 x) [0..]

A256229 Powering the decimal digits of n (right-associative) with 0^0 = 1 by convention.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 1, 4, 16, 64, 256, 1024, 4096, 16384, 65536, 262144, 1, 5, 25, 125, 625, 3125, 15625, 78125, 390625, 1953125, 1, 6, 36, 216, 1296

Offset: 1

M. F. Hasler, Mar 19 2015

See A075877 for the left-associative version (which grows much more slowly). Usually the "^" operator is considered right-associative (so this is the "natural" version), i.e., a^b^c = a^(b^c) since (a^b)^c could be written a^(b*c) instead, while there is no such simplification for a^(b^c).

If n's first digit is succeeded by an odd number of consecutive 0's, a(n) is 1. If it is by an even number, a(n) is the first digit of n (A000030). - Alex Costea, Mar 27 2019

			a(253) = 2^5^3 = 2^(5^3) = 2^125 = 42535295865117307932921825928971026432.

Alois P. Heinz, Table of n, a(n) for n = 1..237
Wikipedia, Zero to the power of zero

Cf. A007953, A007954, A075877, A133500, A185969, A248907, A256179.

a:= proc(n) local m, r; m, r:= n, 1;
      while m>0 do r:= irem(m, 10, 'm')^r od; r
    end:
seq(a(n), n=1..100);  # Alois P. Heinz, Mar 19 2015

Power @@ IntegerDigits@ # & /@ Range@ 64 /. Indeterminate -> 1 (* Michael De Vlieger, Mar 21 2015 *)

A256229(n,p=1)={until(!n\=10,p=(n%10)^p);p}

def A256229(n):
    y = 1
    for d in reversed(str(n)):
        y = int(d)**y
    return y # Chai Wah Wu, Mar 21 2015

a(n) = A075877(n) for n < 212.
a(n) = A133500(n) for n < 100.
a(10n+1) = a(n).

Incorrect comments deleted by Alex Costea, Mar 24 2019

cp's OEIS Frontend

A287877 Start with n and repeatedly apply the powertrain map x -> A133500(x); a(n) is the fixed point that is eventually reached, or -1 if the trajectory never reaches a fixed point.

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A003001 Smallest number of multiplicative persistence n.

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Mathematica

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A133501 Number of steps for "powertrain" operation to converge when started at n.

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Maple

A175398 Sequence of resulting numbers after iterations of {((((D_1^D_2)^D_3)^D_4)^... )^D_k, where D_k is the k-th digit D of the number r and k is the digit number of the number r in the decimal expansion of r (A055642)} needed to reach a one-digit number starting at r = n.

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A221221 Where powerbacks and powertrains coincide.

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Haskell

A256229 Powering the decimal digits of n (right-associative) with 0^0 = 1 by convention.

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A075877 Powering the decimal digits of n (left-associative).

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A133508 Record numbers of steps associated with the terms of A133503.

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