A006050 -id:A006050 - 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

A031286 Additive persistence: number of summations of digits needed to obtain a single digit (the additive digital root).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2

Offset: 0

Eric W. Weisstein

Chai Wah Wu, Table of n, a(n) for n = 0..10000
Antonios Meimaris, On the additive persistence of a number in base p, Preprint, 2015.
N. J. A. Sloane, The persistence of a number, J. Recreational Math., 6 (1973), 97-98.
Eric Weisstein's World of Mathematics, Additive Persistence

Cf. A010888 (additive digital root of n).
Cf. A031347 (multiplicative digital root of n).
Cf. A031346 (multiplicative persistence of n).
Cf. also A006050, A045646.
Cf. Numbers with additive persistence k: A304366 (k=1), A304367 (k=2), A304368 (k=3), A304373 (k=4). - Jaroslav Krizek, May 28 2018

read("transforms") ;
A031286 := proc(n)
    local a,nper;
    nper := n ;
    a := 0 ;
    while nper > 9 do
        nper := digsum(nper) ;
        a := a+1 ;
    end do:
    a ;
end proc:
seq(A031286(n),n=0..80) ; # R. J. Mathar, Jan 02 2018

lst = {}; Do[s = 0; While[n > 9, s++; n = Plus @@ IntegerDigits[n]]; AppendTo[lst, s], {n, 0, 98}]; lst (* Arkadiusz Wesolowski, Oct 17 2012 *)

dsum(n)=my(s);while(n,s+=n%10;n\=10);s
a(n)=my(s);while(n>9,s++;n=dsum(n));s \\ Charles R Greathouse IV, Sep 13 2012

def A031286(n):
    ap = 0
    while n > 9:
        n = sum(int(d) for d in str(n))
        ap += 1
    return ap
# Chai Wah Wu, Aug 23 2014

Corrected by Reinhard Zumkeller, Feb 05 2009

A007013 Catalan-Mersenne numbers: a(0) = 2; for n >= 0, a(n+1) = 2^a(n) - 1.

Original entry on oeis.org

2, 3, 7, 127, 170141183460469231731687303715884105727

Offset: 0

N. J. A. Sloane, Nik Lygeros (webmaster(AT)lygeros.org)

nonn

The next term is too large to include.
Orbit of 2 under iteration of the "Mersenne operator" M: n -> 2^n-1 (0 and 1 are fixed points of M). - M. F. Hasler, Nov 15 2006
Also called the Catalan sequence. - Artur Jasinski, Nov 25 2007
a(n) divides a(n+1)-1 for every n. - Thomas Ordowski, Apr 03 2016
Proof: if 2^a == 2 (mod a), then 2^a = 2 + ka for some k, and 2^(2^a-1) = 2^(1 + ka) = 2*(2^a)^k == 2 (mod 2^a-1). Given that a(1) = 3 satisfies 2^a == 2 (mod a), that gives you all 2^a(n) == 2 (mod a(n)), and since a(n+1) - 1 = 2^a(n) - 2 that says a(n) | a(n+1) - 1. - Robert Israel, Apr 05 2016
All terms shown are primes, the status of the next term is currently unknown. - Joerg Arndt, Apr 03 2016
The next term is a prime or a Fermat pseudoprime to base 2 (i.e., a member of A001567). If it is a pseudoprime, then all succeeding terms are pseudoprimes. - Thomas Ordowski, Apr 04 2016
a(n) is the least positive integer that requires n+1 steps to reach 1 under iteration of the binary weight function A000120. - David Radcliffe, Jun 25 2018
If the next term were prime, it would be a counterexample to the New Mersenne conjecture. It is known that (2^a(4) + 1) / 3 is composite, with factor 886407410000361345663448535540258622490179142922169401 = 5209834514912200*a(4)+1. - William Hu, Jul 30 2024
a(n) is the smallest number of additive persistence n+1 in base 2.  (Similar to A006050 but for binary instead of decimal.) - J. Beach, Nov 17 2024

P. Ribenboim, The Book of Prime Number Records. Springer-Verlag, NY, 2nd ed., 1989, p. 81.
W. Sierpiński, A Selection of Problems in the Theory of Numbers. Macmillan, NY, 1964, p. 91.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Chris K. Caldwell, Mersenne Primes.
Alex Kritov, Explicit Values for Gravitational and Hubble Constants from Cosmological Entropy Bound and Alpha-Quantization of Particle Masses, 2021, see p. 8.
Double Mersennes Prime Search Status of M(M(p)) where M(p) is a Mersenne prime [outdated link of Will Edgington replaced by _Georg Fischer_, Jan 18 2019].
Carlos Rivera, Conjecture 15. The New Mersenne Conjecture, The Prime Puzzles & Problems Connection.
W. Sierpiński, A Selection of Problems in the Theory of Numbers, Macmillan, NY, 1964, p. 91-92. (Annotated scanned copy)
Eric Weisstein's World of Mathematics, Catalan-Mersenne Number
Eric Weisstein's World of Mathematics, Double Mersenne Number.

Cf. A000668, A001567, A014221, A006050, A180094.

M:=n->2^n-1; '(M@@i)(2)'$i=0..4; # M. F. Hasler, Nov 15 2006

NestList[2^#-1&,2,4] (* Harvey P. Dale, Jul 18 2011 *)

a(n)=if(n,2^a(n-1)-1,2) \\ Charles R Greathouse IV, Sep 07 2016

a(n) = M(a(n-1)) = M^n(2) with M: n-> 2^n-1. - M. F. Hasler, Nov 15 2006

A180094(a(n)) = n + 1.

Edited by Henry Bottomley, Nov 07 2002

Amended title name by Marc Morgenegg, Apr 14 2016

A180083 Smallest k such that digit sum of k > previous term.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 19, 299, 3999999999999999999999999999999999

Offset: 1

Dominick Cancilla, Aug 09 2010

a(14) has 444444444444444444444444444444445 digits and is too large to include. - Arkadiusz Wesolowski, Feb 17 2011

			8 is followed by 9 because 9 is the lowest number with a digit sum > 8.
9 is followed by 19 because 19 is the lowest number with a digit sum (1 + 9 = 10) > 9.

Cf. A006050. - Robert G. Wilson v, Aug 15 2010

f[s_List] := Block[{k = prev = s[[ -1]]}, While[ prev >= Plus @@ IntegerDigits@ k, k++ ]; Append[s, k]]; Nest[ f, {0}, 11] (* Robert G. Wilson v, Aug 15 2010 *)

a(13) from Robert G. Wilson v, Aug 15 2010

A131809 Multiplicative persistence of Catalan numbers.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 1, 3, 1, 5, 4, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 0

Paolo P. Lava and Giorgio Balzarotti, Jul 18 2007

After the 57th terms all the numbers have some digits equal to zero thus the persistence is equal to 1.

			Catalan number 429 -> 4*2*9=72 -> 7*2=14 -> 1*4=4 thus persistence is 3

Cf. A003001, A006050, A000108.

P:=proc(n) local i,k,w,ok,cont; for i from 0 by 1 to n do k:=(2*i)!/(i!*(i+1)!); w:=1; ok:=1; if k<10 then print(0); else cont:=1; while ok=1 do while k>0 do w:=w*(k-(trunc(k/10)*10)); k:=trunc(k/10); od; if w<10 then ok:=0; print(cont); else cont:=cont+1; k:=w; w:=1; fi; od; fi; od; end: P(100);

A131810 Additive persistence of Catalan numbers.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 1, 2, 1, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 2, 3, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 2, 2, 2, 3, 2, 3, 2, 2, 2, 3, 3, 2, 2, 2, 3, 2, 4, 3, 2, 3

Offset: 0

Paolo P. Lava and Giorgio Balzarotti, Jul 18 2007

			Catalan number 429 -> 4+2+9=15 -> 1+5=6 thus persistence is 2

Cf. A003001, A006050, A000108, A131809.

with(numtheory): with(combinat): P:=proc(n) local a,t; t:=0; a:=(2*n)!/(n!*(n+1)!); while a>9 do t:=t+1; a:=convert(convert(a,base,10),`+`); od; t;
end: seq(P(i),i=0..10^2);

Corrected entries and Maple code by Paolo P. Lava, Dec 19 2017

A293929 Smallest number in base 10 that cannot be collapsed to a single digit using fewer than n plus signs.

Original entry on oeis.org

0, 10, 19, 118, 3187, 3014173, 3003344034004

Offset: 0

Simon Demers, Oct 19 2017

A253057 considers the number of applications needed to collapse numbers. An alternative is to look at the number of times a plus sign needs to be inserted, disregarding the number of applications.
The sequence is believed to be infinite. Only five terms, a(1)-a(5), were provided by Butler et al. (2016).
Butler et al. (2016) conjectured that collapsing each term x by simply inserting a plus sign in the "middle" of the decimal expansion and adding would require in the order of log (log x) plus signs.
a(5) reveals that all five-digit and six-digit numbers can be collapsed by inserting no more than four plus signs.
a(6) should contain at least 13 digits. After inserting a plus sign in the middle of numbers with 7, 8, 9 and 10 digits and performing the addition, the resulting sums must have at most 5, 5, 6 and 6 digits, respectively. Furthermore, after inserting a plus sign in the middle of any number with 11 or 12 digits and performing the addition, the result must be smaller or equal to, respectively, 1099998 and 1999998 < a(5). This means at most 4 plus signs would be required to collapse the result after the first application. It follows that all 7, 8, 9, 10, 11 and 12-digit numbers can be collapsed using no more than 5 plus signs. - Simon Demers, Oct 30 2017 [Updated Nov 29 2017]
Between 10 and 10^7-1=9999999 inclusively, 270 numbers require only one plus sign, 175803 numbers require two plus signs, 5952451 numbers require three plus signs, 3866392 numbers require four plus signs, and 5074 numbers require five plus signs. - Simon Demers, Oct 29 2017
Conjecture: Digital root for terms a(n) > 0 is 1. - Simon Demers and J. Stauduhar, Nov 16 2017
Using brute-force, no new term less than 10^9 was found. - J. Stauduhar, Nov 20 2017
Proof of claim that all a(n), n > 0, have digital root 1: Assume terms a(1) to a(n) all have digital root 1, but a(n+1) = x does not. Increment a(n) by one until we reach x. Insert one plus sign into x in the optimal way that guarantees that the result of the addition, y, requires exactly n more insertions of a plus sign to arrive at a single digit. Because y requires n insertions it cannot be less than a(n), otherwise we would have found y before a(n). Because x has digital root greater than 1, y cannot equal a(n). So y must be in the range a(n) < y < x, but we already checked these before arriving at x, so no such y can exist, therefore no such x can exist. Clearly, a(n+1) cannot have digital root 0. Since no a(n+1) = x with digital root 0 or 2 through 9 can exist, a(n+1) must have digital root 1. Q.E.D. - J. Stauduhar, Dec 08 2017

			For n=3, the a(3)=118 solution reflects the fact that 1+18 = 19, 1+9 = 10 and 1+0 = 1. Alternatively, 1+1+8 = 10 and 1+0 = 1. Three plus signs are required in both cases. For a(4)=3187, one plus sign is required to obtain 31+87 = 118 = a(3).

S. Butler, R. Graham and R. Stong, Inserting Plus Signs and Adding, The American Mathematical Monthly, 123(3), March 2016, 274-279.
Simon Demers, Minimum number of plus signs needed to collapse every integer 1..10^7-1=9999999. Calculated using brute-force approach. Starting with i=1 and sequentially for each subsequent integer with d digits, all 2^(d-1)-1 possibilities to insert plus signs are considered in the first application. Then, lookup the minimum number of plus signs required to collapse each resulting integer after the first addition is performed. This dataset confirms a(1)=10, a(2)=19, a(3)=118, a(4)=3187, a(5)=3014173.
Simon Demers, The smallest number that cannot be collapsed using fewer than 6 plus signs is 3003344034004, Amer. Math. Monthly, 126 (April 2019) 351.

Cf. A006050, A045646, A253057.

a(n) <= ((a(n-1)-1)^2)/3 + a(n-2) for n > 1 (conjectured). This would provide a relatively tight upper bound on a(n). If the Demers-Stauduhar conjecture in the Comments turns out to be true, this upper bound will always be an integer. - Simon Demers, Nov 29 2017

a(6) (found by Simon Demers) added by Stan Wagon, May 02 2019

cp's OEIS Frontend

A045646 Alternative version of A006050.

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A003001 Smallest number of multiplicative persistence n.

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A031286 Additive persistence: number of summations of digits needed to obtain a single digit (the additive digital root).

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A007013 Catalan-Mersenne numbers: a(0) = 2; for n >= 0, a(n+1) = 2^a(n) - 1.

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A180083 Smallest k such that digit sum of k > previous term.

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A131809 Multiplicative persistence of Catalan numbers.

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A131810 Additive persistence of Catalan numbers.

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A293929 Smallest number in base 10 that cannot be collapsed to a single digit using fewer than n plus signs.