cp's OEIS Frontend

A002275 is a subsequence, thus this sequence is infinite.

No other terms below 10^300. - Max Alekseyev, May 31 2018

Do not confuse with the digital root of n, A010888 (first term that differs is a(19)).

Also the fixed point of the morphism 0 -> {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, 1 -> {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, 2 -> {2, 3, 4, 5, 6, 7, 8, 9, 10, 11}, etc. - Robert G. Wilson v, Jul 27 2006

For n < 100 equal to (floor(n/10) + n mod 10) = A076314(n). - Hieronymus Fischer, Jun 17 2007

It appears that a(n) is the position of 10*n in the ordered set of numbers obtained by inserting/placing one digit anywhere in the digits of n (except a zero before 1st digit). For instance, for n=2, the resulting set is (12, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 32, 42, 52, 62, 72, 82, 92) where 20 is at position 2, so a(2) = 2. - Michel Marcus, Aug 01 2022

Also the total number of beads required to represent n on a Russian abacus (schoty). - P. Christopher Staecker, Mar 31 2023

a(n) / a(2n) <= 5 with equality iff n is in A169964, while a(n) / a(3n) is unbounded, since if n = (10^k + 2)/3, then a(n) = 3*k+1, a(3n) = 3, so a(n) / a(3n) = k + 1/3 -> oo when k->oo (see Diophante link). - Bernard Schott, Apr 29 2023

Also the number of symbols needed to write number n in Egyptian numerals for n < 10^7. - Wojciech Graj, Jul 10 2025

This is sometimes also called the additive digital root of n.

n mod 9 (A010878) is a very similar sequence.

Partial sums are given by A130487(n-1) + n (for n > 0). - Hieronymus Fischer, Jun 08 2007

Decimal expansion of 13717421/111111111 is 0.123456789123456789123456789... with period 9. - Eric Desbiaux, May 19 2008

Decimal expansion of 13717421 / 1111111110 = 0.0[123456789] (periodic) - Daniel Forgues, Feb 27 2017

a(A005117(n)) < 9. - Reinhard Zumkeller, Mar 30 2010

My friend Jahangeer Kholdi has found that 19 is the smallest prime p such that for each number n, a(p*n) = a(n). In fact we have: a(m*n) = a(a(m)*a(n)) so all numbers with digital root 1 (numbers of the form 9k + 1) have this property. See comment lines of A017173. Also we have a(m+n) = a(a(m) + a(n)). - Farideh Firoozbakht, Jul 23 2010

Complement of A052382.

A168046(a(n)) = 0; A054054(a(n)) = 0; A055640(a(n)) = 0 for n = 1 and A055640(a(n)) > 0 for n > 1; A055641(a(n)) > 0; subsequence of A188643. - Reinhard Zumkeller, Apr 25 2012, Apr 07 2011; corrected by Hieronymus Fischer, Jan 13 2013

A067898(a(n)) > 0. - Reinhard Zumkeller, May 04 2012; corrected by Hieronymus Fischer, Jan 13 2013

From Hieronymus Fischer, Jan 13 2013, May 28 2014; edited by M. F. Hasler; edited by Hieronymus Fischer, Dec 27 2018: (Start)

Zerofree floor: The greatest zerofree number < a(n) is A052382(a(n) + 1 - n).

The greatest zero-containing number (i.e., non-zerofree number, or term of this sequence) less than a given zerofree number A052382(n) is a(A052382(n) + 1 - n).

The ratio n/(a(n) + 1) indicates the relative proportion of zero-containing numbers less than or equal to a(n) compared to all numbers less than or equal to a(n). Since Lim_{n -> infinity} a(n)/n = 1, this can be expressed as "Almost all numbers contain a 0" (in a slightly informal manner).

As an example, for n = 10^100, n/(a(n) + 1) = 0.9999701184..., i.e., 99.997...% of all numbers between 0 and 10^100 contain a zero digit. Only the tiny proportion of 0.0000298816... (less than 0.003%) contain no zero digit. This is in contrast to the behavior for small indices, where the relative portion of numbers that contain no zero digit is significant: for n = 10^3 and even n = 10^7, the proportion of numbers less than or equal to n that contain no zero digit exceeds 81% and 53%, respectively.

Inversion: Given a number z that contains a zero digit, the index n for which a(n) = z is n = (z+1)*probability that a randomly chosen number k from the range 0..z contains a zero digit.

Example 1: z = 10; the probability that a randomly chosen number less than or equal to 10 contains no zero digit is 9/11. The probability that it contains a zero digit is p = 2/11. Thus, n = (z+1)*p = 2 and a(2) = 10.

Example 2: z = 10^6; the probability that a randomly chosen number with m > 1 digits contains no zero digit is (9/10)^(m-1). For m = 1 the probability is 9/10. The probability that a randomly chosen number with 1..m digits contains no zero digit is q = (9/10)*10/(10^m+1) + Sum_{i = 2..m} (9/10)^(i-1)*(10^i - 10^(i-1))/(10^m+1) = (72 + 81*(9^(m-1) - 1))/(8*(10^m+1)). Hence, the probability that the chosen number with 1..m digits contains a zero digit is p = 1 - q = (8*10^m - 9*9^m + 17)/(8*(10^m + 1)). Thus, p = 402131/1000001 (for z = 10^6) and so n = (z+1)*p = 402131, which implies a(402131) = 10^6.

The number of terms z such that k*10^m <= z < (k+1)*10^m is 10^m - 9^m, where 1 <= k < 10 and m >= 0.

The number of terms z such that 10^m <= z < 10^(m+1) is 9*(10^m - 9^m), where m >= 0.

The number of terms z <= 10^m is (8*10^m - 9*9^m + 17)/8 where m>=1 (cf. A217094).

Infinitely many terms are primes, and most primes are zero-containing numbers. Sketch of a proof: The number of zero-containing numbers less than or equal to a(n) is n. Hence there are a(n) + 1 - n zerofree numbers less than or equal to a(n). From the asymptotic behavior of a(n) (see formula section) it follows a(n) + 1 - n < (5/4)*n^log_10(9) for sufficiently large n. By the prime number theorem we have for each fixed d > 0 the relation pi(n) [number of primes less than or equal to n] > (1 - d/4)*(n/log(n)) for sufficiently large n. Thus, for the number of primes less than or equal to a(n) which contain a zero digit [hereafter denoted as P_0(a(n))] we have P_0(a(n)) > pi(a(n)) - (a(n) + 1 - n) > (1 - d/4)*a(n)/log(a(n)) - (5/4)*n^log_10(9) > (1-d/4)*n/log(n) - (5/4)*n^log_10(9) = (1-d/4)*n/log(n) * (1 - (5/4)*(1/(1-d/4))*(1/n) * n^(log_10(9))*log(n)) > (1-d/2)*n/log(n) for sufficiently large n. Because of a(n) = n + o(n) this also implies P_0(a(n)) > (1 - d)*a(n)/log(a(n)) for sufficiently large n. Thus, the proportion of primes less than or equal to a(n) which contain a zero digit compared to the total number of primes less than or equal to a(n) is arbitrarily near to 1 for sufficiently large n.

Sequence inversion:

Given a term m > 0, the index n such that a(n) = m can be calculated with the following procedure: Define k := floor(log_10(m)) and i := digit position of the leftmost '0' in m counted from the right (starting with 0), then:

A011540_inverse(m) = 2 + m mod 10^i + Sum_{j = 1..k} floor((m - 1 - m mod 10^i)/10^j)*9^(j-1) [see PROG section for an implementation in Smalltalk].

Example: m = 905, k = 2, i = 1, A011540_inverse(905) = 2 + 905 mod 10 + floor((905 - 1 - 905 mod 10)/10)*1 + floor((905 - 1 - 905 mod 10)/100)*9 = 2 + 5 + floor(899/10)*1 + floor(899/100)*9 = 2 + 5 + 89*1 + 8*9 = 168.

(End)

For the number of k-digit numbers containing the digit '0', see A229127. - Jon E. Schoenfield, Sep 14 2013

The above "sketch of proof" only compares the relative densities, and since the density of this sequence is 1, the result is "obvious". But the nontrivial part is that there is no correlation between the absence of a digit '0' and primality of the number (cf. A038618). Indeed, consider the set S defined to be the set of primes with all digits '0' replaced by the smallest possible nonzero digit while avoiding duplicates. Having exactly the same density as the set of primes, the argument of the proof applies in the same way and leads to the same conclusion for the number of zero-containing terms; however, there is none in the set S. - M. F. Hasler, Oct 11 2015, example added Feb 11 2019

Numbers written in the dyadic system [Smullyan, Stillwell]. - N. J. A. Sloane, Feb 13 2019

Logic-binary sequence: prefix it by the empty word to have all binary words on the alphabet {1,2}.

The least binary word of length k is a(2^k - 1).

See Mathematica program for logic-binary sequence using (0,1) in place of (1,2); the sequence starts with 0,1,00,01,10. - Clark Kimberling, Feb 09 2012

A007953(a(n)) = A014701(n+1); A007954(a(n)) = A048896(n). - Reinhard Zumkeller, Oct 26 2012

a(n) is n written in base 2 where zeros are not allowed but twos are. The two distinct digits used are 1, 2 instead of 0, 1. To obtain this sequence from the "canonical" base 2 sequence with zeros allowed, just replace any 0 with a 2 and then subtract one from the group of digits situated on the left: (10-->2; 100-->12; 110-->22; 1000-->112; 1010-->122). - Robin Garcia, Jan 31 2014

For numbers made of only two different digits, see also A007088 (digits 0 & 1), A032810 (digits 2 & 3), A032834 (digits 3 & 4), A256290 (digits 4 & 5), A256291 (digits 5 & 6), A256292 (digits 6 & 7), A256340(digits 7 & 8), A256341 (digits 8 & 9), and A032804-A032816 (in other bases). Numbers with exactly two distinct (but unspecified) digits in base 10 are listed in A031955, for other bases in A031948-A031954. - M. F. Hasler, Apr 04 2015

The variant with digits {0, 1} instead of {1, 2} is obtained by deleting all initial digits in sequence A007088 (numbers written in base 2). - M. F. Hasler, Nov 03 2020

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)

a(n) = 0 for almost all n. - Charles R Greathouse IV, Oct 02 2013

More precisely, a(n) = 0 asymptotically almost surely, namely, among others, for all numbers n which have a digit '0', and as n has more and more digits, it becomes increasingly less probable that no digit is equal to zero. (The set A011540 has density 1.) Thus the density of numbers for which a(n) > 0 is zero, although this happens for infinitely many numbers, for example all repunits n = (10^k - 1)/9 = A002275(k). - M. F. Hasler, Oct 11 2015